Rs Aggarwal 2019 2020 Solutions for Class 8 MATH Chapter 6

Question 1:

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$8 a b - 5 a b 3 a b - a b$
________
$5 a b$
â€‹

Question 2:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$8 a b - 5 a b 3 a b - a b$
________
$5 a b$
â€‹

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$7 x - 3 x 5 x - x - 2 x$
_____
$6 x$

Question 3:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$7 x - 3 x 5 x - x - 2 x$
_____
$6 x$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$3 a - 4 b + 4 c 2 a + 3 b - 8 c a - 6 b + c$
___________
$6 a - 7 b - 3 c$

Question 4:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$3 a - 4 b + 4 c 2 a + 3 b - 8 c a - 6 b + c$
___________
$6 a - 7 b - 3 c$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$5 x - 8 y + 2 z - 2 x - 4 y + 3 z - x + 6 y - z 3 x - 3 y - 2 z 5 x - 9 y + 2 z$

Question 5:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$5 x - 8 y + 2 z - 2 x - 4 y + 3 z - x + 6 y - z 3 x - 3 y - 2 z 5 x - 9 y + 2 z$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$6 a x - 2 b y + 3 c z - 11 a x + 6 b y - c z - 2 a x - 3 b y + 10 c z - 7 a x + b y + 12 c z$

Question 6:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

$6 a x - 2 b y + 3 c z - 11 a x + 6 b y - c z - 2 a x - 3 b y + 10 c z - 7 a x + b y + 12 c z$

Answer:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

$2 x^{3} - 9 x^{2} + 0 x + 8 0 x^{3} + 3 x^{2} - 6 x - 5 7 x^{3} + 0 x^{2} - 10 x + 1 - 4 x^{3} - 5 x^{2} + 2 x + 3 5 x^{3} - 11 x^{2} - 14 x + 7$

Question 7:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

$2 x^{3} - 9 x^{2} + 0 x + 8 0 x^{3} + 3 x^{2} - 6 x - 5 7 x^{3} + 0 x^{2} - 10 x + 1 - 4 x^{3} - 5 x^{2} + 2 x + 3 5 x^{3} - 11 x^{2} - 14 x + 7$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

$6 p + 4 q - r + 3 - 5 p + 0 q + 2 r - 6 - 7 p + 11 q + 2 r - 1 0 p + 2 q - 3 r + 4 - 6 p + 17 q + 0 r + 0 = - 6 p + 17 q$

Question 8:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

$6 p + 4 q - r + 3 - 5 p + 0 q + 2 r - 6 - 7 p + 11 q + 2 r - 1 0 p + 2 q - 3 r + 4 - 6 p + 17 q + 0 r + 0 = - 6 p + 17 q$

Answer:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

$4 x^{2} + 4 y^{2} - 7 x y - 3 x^{2} + 6 y^{2} - 8 x y + 0 2 x^{2} - 5 y^{2} - 2 x y + 6 7 x^{2} + 5 y^{2} - 17 x y + 3$

Question 9:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

$4 x^{2} + 4 y^{2} - 7 x y - 3 x^{2} + 6 y^{2} - 8 x y + 0 2 x^{2} - 5 y^{2} - 2 x y + 6 7 x^{2} + 5 y^{2} - 17 x y + 3$

Answer:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting:

$- 5 a^{2} b 3 a^{2} b - - 8 a^{2} b$

Question 10:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting:

$- 5 a^{2} b 3 a^{2} b - - 8 a^{2} b$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$6 p q - 8 p q + 14 p q$

Question 11:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$6 p q - 8 p q + 14 p q$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$- 8 a b c - 2 a b c + - 6 a b c$

Question 12:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$- 8 a b c - 2 a b c + - 6 a b c$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$- 11 p - 16 p + 5 p$

Question 13:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$- 11 p - 16 p + 5 p$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$3 a - 4 b - c + 6 2 a - 5 b + 2 c - 9 - + - + a + b - 3 c + 15$

Question 14:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$3 a - 4 b - c + 6 2 a - 5 b + 2 c - 9 - + - + a + b - 3 c + 15$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$p - 2 q - 5 r - 8 - 6 p + q + 3 r + 8 + - - - 7 p - 3 q - 8 r - 16$

Question 15:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$p - 2 q - 5 r - 8 - 6 p + q + 3 r + 8 + - - - 7 p - 3 q - 8 r - 16$

Answer:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

$3 x^{3} - x^{2} + 2 x - 4 x^{3} + 3 x^{2} - 5 x + 4 - - + - 2 x^{3} - 4 x^{2} + 7 x - 8$

Question 16:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

$3 x^{3} - x^{2} + 2 x - 4 x^{3} + 3 x^{2} - 5 x + 4 - - + - 2 x^{3} - 4 x^{2} + 7 x - 8$

Answer:

Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

$4 y^{4} - 2 y^{3} - 6 y^{2} - y + 5 5 y^{4} - 3 y^{3} + 2 y^{2} + y - 1 - + - - + - y^{4} + y^{3} - 8 y^{2} - 2 y + 6$

Question 17:

Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

$4 y^{4} - 2 y^{3} - 6 y^{2} - y + 5 5 y^{4} - 3 y^{3} + 2 y^{2} + y - 1 - + - - + - y^{4} + y^{3} - 8 y^{2} - 2 y + 6$

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$3 p^{2} - 4 q^{2} - 5 r^{2} - 6 4 p^{2} + 5 q^{2} - 6 r^{2} + 7 - - + - - p^{2} - 9 q^{2} + r^{2} - 13$

Question 18:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

$3 p^{2} - 4 q^{2} - 5 r^{2} - 6 4 p^{2} + 5 q^{2} - 6 r^{2} + 7 - - + - - p^{2} - 9 q^{2} + r^{2} - 13$

Answer:

Let the required number be $x$ .
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - x = 4 a^{2} - 7 a b - 4 b^{2} + 1$
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - (4 a^{2} - 7 a b - 4 b^{2} + 1) = x$

$3 a^{2} - 6 a b - 3 b^{2} - 1 4 a^{2} - 7 a b - 4 b^{2} + 1 - + + - - a^{2} + a b + b^{2} - 2$

∴ Required number = $- a^{2} + a b + b^{2} - 2$

Question 19:

Let the required number be $x$ .
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - x = 4 a^{2} - 7 a b - 4 b^{2} + 1$
$(3 a^{2} - 6 a b - 3 b^{2} - 1) - (4 a^{2} - 7 a b - 4 b^{2} + 1) = x$

$3 a^{2} - 6 a b - 3 b^{2} - 1 4 a^{2} - 7 a b - 4 b^{2} + 1 - + + - - a^{2} + a b + b^{2} - 2$

∴ Required number = $- a^{2} + a b + b^{2} - 2$

Answer:

Sides of the rectangle are $l$ and $b$ .
$l = 5 x^{2} - 3 y^{2} b = x^{2} + 2 x y$
Perimeter of the rectangle is $(2 l + 2 b)$ .

$Perimeter = 2 (5 x^{2} - 3 y^{2}) + 2 (x^{2} + 2 x y) = 10 x^{2} - 6 y^{2} + 2 x^{2} + 4 x y 10 x^{2} - 6 y^{2} 2 x^{2} + 4 x y 12 x^{2} - 6 y^{2} + 4 x y Hence, the perimeter of the rectangle is 12 x^{2} - 6 y^{2} + 4 x y .$

Question 20:

Sides of the rectangle are $l$ and $b$ .
$l = 5 x^{2} - 3 y^{2} b = x^{2} + 2 x y$
Perimeter of the rectangle is $(2 l + 2 b)$ .

$Perimeter = 2 (5 x^{2} - 3 y^{2}) + 2 (x^{2} + 2 x y) = 10 x^{2} - 6 y^{2} + 2 x^{2} + 4 x y 10 x^{2} - 6 y^{2} 2 x^{2} + 4 x y 12 x^{2} - 6 y^{2} + 4 x y Hence, the perimeter of the rectangle is 12 x^{2} - 6 y^{2} + 4 x y .$

Answer:

Let $a, b a n d c$ be the three sides of the triangle.

∴ Perimeter of the triangle = $(a + b + c)$

Given perimeter of the triangle = $6 p^{2} - 4 p + 9$
One side ( $a$ ) = $p^{2} - 2 p + 1$
Other side ( $b$ ) = $3 p^{2} - 5 p + 3$
Perimeter = $(a + b + c)$
$(6 p^{2} - 4 p + 9) = (p^{2} - 2 p + 1) + (3 p^{2} - 5 p + 3) + c 6 p^{2} - 4 p + 9 - p^{2} + 2 p - 1 - 3 p^{2} + 5 p - 3 = c (6 p^{2} - p^{2} - 3 p^{2}) + (- 4 p + 2 p + 5 p) + (9 - 1 - 3) = c 2 p^{2} + 3 p + 5 = c$

Thus, the third side is $2 p^{2} + 3 p + 5$ .

Question 1:

Let $a, b a n d c$ be the three sides of the triangle.

∴ Perimeter of the triangle = $(a + b + c)$

Given perimeter of the triangle = $6 p^{2} - 4 p + 9$
One side ( $a$ ) = $p^{2} - 2 p + 1$
Other side ( $b$ ) = $3 p^{2} - 5 p + 3$
Perimeter = $(a + b + c)$
$(6 p^{2} - 4 p + 9) = (p^{2} - 2 p + 1) + (3 p^{2} - 5 p + 3) + c 6 p^{2} - 4 p + 9 - p^{2} + 2 p - 1 - 3 p^{2} + 5 p - 3 = c (6 p^{2} - p^{2} - 3 p^{2}) + (- 4 p + 2 p + 5 p) + (9 - 1 - 3) = c 2 p^{2} + 3 p + 5 = c$

Thus, the third side is $2 p^{2} + 3 p + 5$ .

Answer:

By horizontal method:
$(5 x + 7) \times (3 x + 4) = 5 x (3 x + 4) + 7 (3 x + 4) = 15 x^{2} + 20 x + 21 x + 28 = 15 x^{2} + 41 x + 28$

Question 2:

By horizontal method:
$(5 x + 7) \times (3 x + 4) = 5 x (3 x + 4) + 7 (3 x + 4) = 15 x^{2} + 20 x + 21 x + 28 = 15 x^{2} + 41 x + 28$

Answer:

By horizontal method:

$(4 x + 9) \times (x - 6) = 4 x (x - 6) + 9 (x - 6) = 4 x^{2} - 24 x + 9 x - 54 = 4 x^{2} - 15 x - 54$

Question 3:

By horizontal method:

$(4 x + 9) \times (x - 6) = 4 x (x - 6) + 9 (x - 6) = 4 x^{2} - 24 x + 9 x - 54 = 4 x^{2} - 15 x - 54$

Answer:

By horizontal method:

$(2 x + 5) \times (4 x - 3) = 2 x (4 x - 3) + 5 (4 x - 3) = 8 x^{2} - 6 x + 20 x - 15 = 8 x^{2} + 14 x - 15$

Question 4:

By horizontal method:

$(2 x + 5) \times (4 x - 3) = 2 x (4 x - 3) + 5 (4 x - 3) = 8 x^{2} - 6 x + 20 x - 15 = 8 x^{2} + 14 x - 15$

Answer:

By horizontal method:

$(3 y - 8) \times (5 y - 1) = 3 y (5 y - 1) - 8 (5 y - 1) = 15 y^{2} - 3 y - 40 y + 8 = 15 y^{2} - 43 y + 8$

Question 5:

By horizontal method:

$(3 y - 8) \times (5 y - 1) = 3 y (5 y - 1) - 8 (5 y - 1) = 15 y^{2} - 3 y - 40 y + 8 = 15 y^{2} - 43 y + 8$

Answer:

By horizontal method:

$(7 x + 2 y) \times (x + 4 y) = 7 x (x + 4 y) + 2 y (x + 4 y) = 7 x^{2} + 28 x y + 2 x y + 8 y^{2} = 7 x^{2} + 30 x y + 8 y^{2}$

Question 6:

By horizontal method:

$(7 x + 2 y) \times (x + 4 y) = 7 x (x + 4 y) + 2 y (x + 4 y) = 7 x^{2} + 28 x y + 2 x y + 8 y^{2} = 7 x^{2} + 30 x y + 8 y^{2}$

Answer:

By horizontal method:

$(9 x + 5 y) \times (4 x + 3 y) 9 x (4 x + 3 y) + 5 y (4 x + 3 y) = 36 x^{2} + 27 x y + 20 x y + 15 y^{2} = 36 x^{2} + 47 x y + 15 y^{2}$

Question 7:

By horizontal method:

$(9 x + 5 y) \times (4 x + 3 y) 9 x (4 x + 3 y) + 5 y (4 x + 3 y) = 36 x^{2} + 27 x y + 20 x y + 15 y^{2} = 36 x^{2} + 47 x y + 15 y^{2}$

Answer:

By horizontal method:

$(3 m - 4 n) \times (2 m - 3 n) = 3 m (2 m - 3 n) - 4 n (2 m - 3 n) = 6 m^{2} - 9 m n - 8 m n + 12 n^{2} = 6 m^{2} - 17 m n + 12 n^{2}$

Question 8:

By horizontal method:

$(3 m - 4 n) \times (2 m - 3 n) = 3 m (2 m - 3 n) - 4 n (2 m - 3 n) = 6 m^{2} - 9 m n - 8 m n + 12 n^{2} = 6 m^{2} - 17 m n + 12 n^{2}$

Answer:

By horizontal method:

$(x^{2} - a^{2}) \times (x - a) = x^{2} (x - a) - a^{2} (x - a) = x^{3} - a x^{2} - a^{2} x + a^{3}$
i.e $(x^{3} + a^{3}) - a x (x - a)$

Question 9:

By horizontal method:

$(x^{2} - a^{2}) \times (x - a) = x^{2} (x - a) - a^{2} (x - a) = x^{3} - a x^{2} - a^{2} x + a^{3}$
i.e $(x^{3} + a^{3}) - a x (x - a)$

Answer:

By horizontal method:

$(x^{2} - y^{2}) \times (x + 2 y) = x^{2} (x + 2 y) - y^{2} (x + 2 y) = x^{3} + 2 x^{2} y - x y^{2} - 2 y^{3} i . e (x^{3} - 2 y^{3}) + x y (2 x - y)$

Question 10:

By horizontal method:

$(x^{2} - y^{2}) \times (x + 2 y) = x^{2} (x + 2 y) - y^{2} (x + 2 y) = x^{3} + 2 x^{2} y - x y^{2} - 2 y^{3} i . e (x^{3} - 2 y^{3}) + x y (2 x - y)$

Answer:

By horizontal method:

$(3 p^{2} + q^{2}) \times (2 p^{2} - 3 q^{2}) = 3 p^{2} (2 p^{2} - 3 q^{2}) + q^{2} (2 p^{2} - 3 q^{2}) = 6 p^{4} - 9 p^{2} q^{2} + 2 p^{2} q^{2} - 3 q^{4} i . e 6 p^{4} - 7 p^{2} q^{2} - 3 q^{4}$

Question 11:

By horizontal method:

$(3 p^{2} + q^{2}) \times (2 p^{2} - 3 q^{2}) = 3 p^{2} (2 p^{2} - 3 q^{2}) + q^{2} (2 p^{2} - 3 q^{2}) = 6 p^{4} - 9 p^{2} q^{2} + 2 p^{2} q^{2} - 3 q^{4} i . e 6 p^{4} - 7 p^{2} q^{2} - 3 q^{4}$

Answer:

By horizontal method:

$(2 x^{2} - 5 y^{2}) \times (x^{2} + 3 y^{2}) = 2 x^{2} (x^{2} + 3 y^{2}) - 5 y^{2} (x^{2} + 3 y^{2}) = 2 x^{4} + 6 x^{2} y^{2} - 5 x^{2} y^{2} - 15 y^{4} = 2 x^{4} + x^{2} y^{2} - 15 y^{4}$

Question 12:

By horizontal method:

$(2 x^{2} - 5 y^{2}) \times (x^{2} + 3 y^{2}) = 2 x^{2} (x^{2} + 3 y^{2}) - 5 y^{2} (x^{2} + 3 y^{2}) = 2 x^{4} + 6 x^{2} y^{2} - 5 x^{2} y^{2} - 15 y^{4} = 2 x^{4} + x^{2} y^{2} - 15 y^{4}$

Answer:

By horizontal method:

$(x^{3} - y^{3}) \times (x^{2} + y^{2}) = x^{3} (x^{2} + y^{2}) - y^{3} (x^{2} + y^{2}) = x^{5} + x^{3} y^{2} - x^{2} y^{3} - y^{5} = (x^{5} - y^{5}) + x^{2} y^{2} (x - y)$

Question 13:

By horizontal method:

$(x^{3} - y^{3}) \times (x^{2} + y^{2}) = x^{3} (x^{2} + y^{2}) - y^{3} (x^{2} + y^{2}) = x^{5} + x^{3} y^{2} - x^{2} y^{3} - y^{5} = (x^{5} - y^{5}) + x^{2} y^{2} (x - y)$

Answer:

By horizontal method:
$(x^{4} + y^{4}) \times (x^{2} - y^{2}) = x^{4} (x^{2} - y^{2}) + y^{4} (x^{2} - y^{2}) = x^{6} - x^{4} y^{2} + y^{4} x^{2} - y^{6} = (x^{6} - y^{6}) - x^{2} y^{2} (x^{2} - y^{2})$

Question 14:

By horizontal method:
$(x^{4} + y^{4}) \times (x^{2} - y^{2}) = x^{4} (x^{2} - y^{2}) + y^{4} (x^{2} - y^{2}) = x^{6} - x^{4} y^{2} + y^{4} x^{2} - y^{6} = (x^{6} - y^{6}) - x^{2} y^{2} (x^{2} - y^{2})$

Answer:

By horizontal method:

$(x^{4} + \frac{1}{x^{4}}) \times (x + \frac{1}{x}) = x^{4} (x + \frac{1}{x}) + \frac{1}{x^{4}} (x + \frac{1}{x}) = x^{5} + x^{3} + \frac{1}{x^{3}} + \frac{1}{x^{5}} i . e x^{3} (x^{2} + 1) + \frac{1}{x^{3}} (1 + \frac{1}{x^{2}})$

Question 15:

By horizontal method:

$(x^{4} + \frac{1}{x^{4}}) \times (x + \frac{1}{x}) = x^{4} (x + \frac{1}{x}) + \frac{1}{x^{4}} (x + \frac{1}{x}) = x^{5} + x^{3} + \frac{1}{x^{3}} + \frac{1}{x^{5}} i . e x^{3} (x^{2} + 1) + \frac{1}{x^{3}} (1 + \frac{1}{x^{2}})$

Answer:

By horizontal method:

$(x^{2} - 3 x + 7) \times (2 x + 3) = 2 x (x^{2} - 3 x + 7) + 3 (x^{2} - 3 x + 7) = 2 x^{3} - 6 x^{2} + 14 x + 3 x^{2} - 9 x + 21 = 2 x^{3} - 3 x^{2} + 5 x + 21$

Question 16:

By horizontal method:

$(x^{2} - 3 x + 7) \times (2 x + 3) = 2 x (x^{2} - 3 x + 7) + 3 (x^{2} - 3 x + 7) = 2 x^{3} - 6 x^{2} + 14 x + 3 x^{2} - 9 x + 21 = 2 x^{3} - 3 x^{2} + 5 x + 21$

Answer:

By horizontal method:
$(3 x^{2} + 5 x - 9) \times (3 x - 5) = 3 x (3 x^{2} + 5 x - 9) - 5 (3 x^{2} + 5 x - 9) = 9 x^{3} + 15 x^{2} - 27 x - 15 x^{2} - 25 x + 45 = 9 x^{3} - 52 x + 45$

Question 17:

By horizontal method:
$(3 x^{2} + 5 x - 9) \times (3 x - 5) = 3 x (3 x^{2} + 5 x - 9) - 5 (3 x^{2} + 5 x - 9) = 9 x^{3} + 15 x^{2} - 27 x - 15 x^{2} - 25 x + 45 = 9 x^{3} - 52 x + 45$

Answer:

By horizontal method:
$(x^{2} - x y + y^{2}) \times (x + y) = x (x^{2} - x y + y^{2}) + y (x^{2} - x y + y^{2}) = x^{3} - x^{2} y + y^{2} x + x^{2} y - x y^{2} + y^{3} = x^{3} + y^{3}$

Question 18:

By horizontal method:
$(x^{2} - x y + y^{2}) \times (x + y) = x (x^{2} - x y + y^{2}) + y (x^{2} - x y + y^{2}) = x^{3} - x^{2} y + y^{2} x + x^{2} y - x y^{2} + y^{3} = x^{3} + y^{3}$

Answer:

By horizontal method:

$(x^{2} + x y + y^{2}) \times (x - y) x (x^{2} + x y + y^{2}) - y (x^{2} + x y + y^{2}) = x^{3} + x^{2} y + x y^{2} - x^{2} y - x y^{2} - y^{3} = x^{3} - y^{3}$

Question 19:

By horizontal method:

$(x^{2} + x y + y^{2}) \times (x - y) x (x^{2} + x y + y^{2}) - y (x^{2} + x y + y^{2}) = x^{3} + x^{2} y + x y^{2} - x^{2} y - x y^{2} - y^{3} = x^{3} - y^{3}$

Answer:

By horizontal method:

$(x^{3} - 2 x^{2} + 5) \times (4 x - 1) = 4 x (x^{3} - 2 x^{2} + 5) - 1 (x^{3} - 2 x^{2} + 5) = 4 x^{4} - 8 x^{3} + 20 x - x^{3} + 2 x^{2} - 5 = 4 x^{4} - 9 x^{3} + 2 x^{2} + 20 x - 5$

Question 20:

By horizontal method:

$(x^{3} - 2 x^{2} + 5) \times (4 x - 1) = 4 x (x^{3} - 2 x^{2} + 5) - 1 (x^{3} - 2 x^{2} + 5) = 4 x^{4} - 8 x^{3} + 20 x - x^{3} + 2 x^{2} - 5 = 4 x^{4} - 9 x^{3} + 2 x^{2} + 20 x - 5$

Answer:

By horizontal method:

$(9 x^{2} - x + 15) \times (x^{2} - 3) = x^{2} (9 x^{2} - x + 15) - 3 (9 x^{2} - x + 15) = 9 x^{4} - x^{3} + 15 x^{2} - 27 x^{2} + 3 x - 45 = 9 x^{4} - x^{3} - 12 x^{2} + 3 x - 45$

Question 21:

By horizontal method:

$(9 x^{2} - x + 15) \times (x^{2} - 3) = x^{2} (9 x^{2} - x + 15) - 3 (9 x^{2} - x + 15) = 9 x^{4} - x^{3} + 15 x^{2} - 27 x^{2} + 3 x - 45 = 9 x^{4} - x^{3} - 12 x^{2} + 3 x - 45$

Answer:

By horizontal method:

$(x^{2} - 5 x + 8) \times (x^{2} + 2) = x^{2} (x^{2} - 5 x + 8) + 2 (x^{2} - 5 x + 8) = x^{4} - 5 x^{3} + 8 x^{2} + 2 x^{2} - 10 x + 16 = x^{4} - 5 x^{3} + 10 x^{2} - 10 x + 16$

Question 22:

By horizontal method:

$(x^{2} - 5 x + 8) \times (x^{2} + 2) = x^{2} (x^{2} - 5 x + 8) + 2 (x^{2} - 5 x + 8) = x^{4} - 5 x^{3} + 8 x^{2} + 2 x^{2} - 10 x + 16 = x^{4} - 5 x^{3} + 10 x^{2} - 10 x + 16$

Answer:

By horizontal method:

$(x^{3} - 5 x^{2} + 3 x + 1) \times (x^{2} - 3) = x^{2} (x^{3} - 5 x^{2} + 3 x + 1) - 3 (x^{3} - 5 x^{2} + 3 x + 1) = x^{5} - 5 x^{4} + 3 x^{3} + x^{2} - 3 x^{3} + 15 x^{2} - 9 x - 3 = x^{5} - 5 x^{4} + 16 x^{2} - 9 x - 3$

Question 23:

By horizontal method:

$(x^{3} - 5 x^{2} + 3 x + 1) \times (x^{2} - 3) = x^{2} (x^{3} - 5 x^{2} + 3 x + 1) - 3 (x^{3} - 5 x^{2} + 3 x + 1) = x^{5} - 5 x^{4} + 3 x^{3} + x^{2} - 3 x^{3} + 15 x^{2} - 9 x - 3 = x^{5} - 5 x^{4} + 16 x^{2} - 9 x - 3$

Answer:

By horizontal method:

$(3 x + 2 y - 4) \times (x - y + 2) x (3 x + 2 y - 4) - y (3 x + 2 y - 4) + 2 (3 x + 2 y - 4) = 3 x^{2} + 2 x y - 4 x - 3 x y - 2 y^{2} + 4 y + 6 x + 4 y - 8 = 3 x^{2} - 2 y^{2} - x y + 2 x + 8 y - 8$

Question 24:

By horizontal method:

$(3 x + 2 y - 4) \times (x - y + 2) x (3 x + 2 y - 4) - y (3 x + 2 y - 4) + 2 (3 x + 2 y - 4) = 3 x^{2} + 2 x y - 4 x - 3 x y - 2 y^{2} + 4 y + 6 x + 4 y - 8 = 3 x^{2} - 2 y^{2} - x y + 2 x + 8 y - 8$

Answer:

By horizontal method:

$(x^{2} - 5 x + 8) \times (x^{2} + 2 x - 3) = x^{2} (x^{2} - 5 x + 8) + 2 x (x^{2} - 5 x + 8) - 3 (x^{2} - 5 x + 8) = x^{4} - 5 x^{3} + 8 x^{2} + 2 x^{3} - 10 x^{2} + 16 x - 3 x^{2} + 15 x - 24 = x^{4} - 3 x^{3} - 5 x^{2} + 31 x - 24$

Question 25:

By horizontal method:

$(x^{2} - 5 x + 8) \times (x^{2} + 2 x - 3) = x^{2} (x^{2} - 5 x + 8) + 2 x (x^{2} - 5 x + 8) - 3 (x^{2} - 5 x + 8) = x^{4} - 5 x^{3} + 8 x^{2} + 2 x^{3} - 10 x^{2} + 16 x - 3 x^{2} + 15 x - 24 = x^{4} - 3 x^{3} - 5 x^{2} + 31 x - 24$

Answer:

By horizontal method:

$(2 x^{2} + 3 x - 7) \times (3 x^{2} - 5 x + 4) = 2 x^{2} (3 x^{2} - 5 x + 4) + 3 x (3 x^{2} - 5 x + 4) - 7 (3 x^{2} - 5 x + 4) = 6 x^{4} - 10 x^{3} + 8 x^{2} + 9 x^{3} - 15 x^{2} + 12 x - 21 x^{2} + 35 x - 28 = 6 x^{4} - x^{3} - 28 x^{2} + 47 x - 28$

Question 26:

By horizontal method:

$(2 x^{2} + 3 x - 7) \times (3 x^{2} - 5 x + 4) = 2 x^{2} (3 x^{2} - 5 x + 4) + 3 x (3 x^{2} - 5 x + 4) - 7 (3 x^{2} - 5 x + 4) = 6 x^{4} - 10 x^{3} + 8 x^{2} + 9 x^{3} - 15 x^{2} + 12 x - 21 x^{2} + 35 x - 28 = 6 x^{4} - x^{3} - 28 x^{2} + 47 x - 28$

Answer:

By horizontal method:

$(9 x^{2} - x + 15) \times (x^{2} - x - 1) = x^{2} (9 x^{2} - x + 15) - x (9 x^{2} - x + 15) - 1 (9 x^{2} - x + 15) = 9 x^{4} - x^{3} + 15 x^{2} - 9 x^{3} + x^{2} - 15 x - 9 x^{2} + x - 15 = 9 x^{4} - 10 x^{3} + 7 x^{2} - 14 x - 15$

Question 1:

By horizontal method:

$(9 x^{2} - x + 15) \times (x^{2} - x - 1) = x^{2} (9 x^{2} - x + 15) - x (9 x^{2} - x + 15) - 1 (9 x^{2} - x + 15) = 9 x^{4} - x^{3} + 15 x^{2} - 9 x^{3} + x^{2} - 15 x - 9 x^{2} + x - 15 = 9 x^{4} - 10 x^{3} + 7 x^{2} - 14 x - 15$

Answer:

(i) 24x²y³ by 3xy

$\frac{24 x^{2} y^{3}}{3 x y} \Rightarrow (\frac{24}{3}) (x^{2 - 1}) (y^{3 - 1}) \Rightarrow 8 x y^{2} .$

Therefore, the quotient is 8xy².

(ii) 36xyz² by −9xz

$\frac{36 x y z^{2}}{- 9 x z} \Rightarrow (\frac{36}{- 9}) (x^{1 - 1}) (y^{1 - 0}) (z^{2 - 1}) \Rightarrow - 4 y z$

Therefore, the quotient is −4yz.

(iii)
$- 72 x^{2} y^{2} z b y - 12 x y z \frac{- 72 x^{2} y^{2} z}{- 12 x y z} \Rightarrow (\frac{- 72}{- 12}) (x^{2 - 1}) (y^{2 - 1}) (z^{1 - 1}) \Rightarrow 6 x y$

Therefore, the quotient is 6xy.

(iv) −56mnp² by 7mnp

$\frac{- 56 m n p^{2}}{7 m n p} \Rightarrow (\frac{- 56}{7}) (m^{1 - 1}) (n^{1 - 1}) (p^{2 - 1}) \Rightarrow - 8 p$

Therefore, the quotient is −8p.

Question 2:

(i) 24x²y³ by 3xy

$\frac{24 x^{2} y^{3}}{3 x y} \Rightarrow (\frac{24}{3}) (x^{2 - 1}) (y^{3 - 1}) \Rightarrow 8 x y^{2} .$

Therefore, the quotient is 8xy².

(ii) 36xyz² by −9xz

$\frac{36 x y z^{2}}{- 9 x z} \Rightarrow (\frac{36}{- 9}) (x^{1 - 1}) (y^{1 - 0}) (z^{2 - 1}) \Rightarrow - 4 y z$

Therefore, the quotient is −4yz.

(iii)
$- 72 x^{2} y^{2} z b y - 12 x y z \frac{- 72 x^{2} y^{2} z}{- 12 x y z} \Rightarrow (\frac{- 72}{- 12}) (x^{2 - 1}) (y^{2 - 1}) (z^{1 - 1}) \Rightarrow 6 x y$

Therefore, the quotient is 6xy.

(iv) −56mnp² by 7mnp

$\frac{- 56 m n p^{2}}{7 m n p} \Rightarrow (\frac{- 56}{7}) (m^{1 - 1}) (n^{1 - 1}) (p^{2 - 1}) \Rightarrow - 8 p$

Therefore, the quotient is −8p.

Answer:

(i) 5m³− 30m² + 45m by 5m

$(5 m^{3} - 30 m^{2} + 45 m) \div 5 m \Rightarrow \frac{5 m^{3}}{5 m} - \frac{30 m^{2}}{5 m} + \frac{45 m}{5 m} \Rightarrow m^{2} - 6 m + 9$

Therefore, the quotient is m²− 6m + 9.

(ii) 8x²y² − 6xy² + 10x²y³ by 2xy

$(8 x^{2} y^{2} - 6 x y^{2} + 10 x^{2} y^{3}) \div 2 x y \Rightarrow \frac{8 x^{2} y^{2}}{2 x y} - \frac{6 x y^{2}}{2 x y} + \frac{10 x^{2} y^{3}}{2 x y} \Rightarrow 4 x y - 3 y + 5 x y^{2}$

Therefore, the quotient is 4xy − 3y + 5xy².

(iii) 9x²y − 6xy + 12xy² by − 3xy

$(9 x^{2} y - 6 x y + 12 x y^{2}) \div - 3 x y \Rightarrow \frac{9 x^{2} y}{- 3 x y} - \frac{6 x y}{- 3 x y} + \frac{12 x y^{2}}{- 3 x y} \Rightarrow - 3 x + 2 - 4 y$

Therefore, the quotient is −3x + 2 − 4y.

(iv) 12x⁴ + 8x³ − 6x² by − 2x²

^{$(12 x^{4} + 8 x^{3} - 6 x^{2}) \div - 2 x^{2} \Rightarrow \frac{12 x^{4}}{- 2 x^{2}} + \frac{8 x^{3}}{- 2 x^{2}} - \frac{6 x^{2}}{- 2 x^{2}} \Rightarrow - 6 x$ ²-4x+3²}

Therefore the quotient is −6x² − 4x + 3.

Question 3:

(i) 5m³− 30m² + 45m by 5m

$(5 m^{3} - 30 m^{2} + 45 m) \div 5 m \Rightarrow \frac{5 m^{3}}{5 m} - \frac{30 m^{2}}{5 m} + \frac{45 m}{5 m} \Rightarrow m^{2} - 6 m + 9$

Therefore, the quotient is m²− 6m + 9.

(ii) 8x²y² − 6xy² + 10x²y³ by 2xy

$(8 x^{2} y^{2} - 6 x y^{2} + 10 x^{2} y^{3}) \div 2 x y \Rightarrow \frac{8 x^{2} y^{2}}{2 x y} - \frac{6 x y^{2}}{2 x y} + \frac{10 x^{2} y^{3}}{2 x y} \Rightarrow 4 x y - 3 y + 5 x y^{2}$

Therefore, the quotient is 4xy − 3y + 5xy².

(iii) 9x²y − 6xy + 12xy² by − 3xy

$(9 x^{2} y - 6 x y + 12 x y^{2}) \div - 3 x y \Rightarrow \frac{9 x^{2} y}{- 3 x y} - \frac{6 x y}{- 3 x y} + \frac{12 x y^{2}}{- 3 x y} \Rightarrow - 3 x + 2 - 4 y$

Therefore, the quotient is −3x + 2 − 4y.

(iv) 12x⁴ + 8x³ − 6x² by − 2x²

^{$(12 x^{4} + 8 x^{3} - 6 x^{2}) \div - 2 x^{2} \Rightarrow \frac{12 x^{4}}{- 2 x^{2}} + \frac{8 x^{3}}{- 2 x^{2}} - \frac{6 x^{2}}{- 2 x^{2}} \Rightarrow - 6 x$ ²-4x+3²}

Therefore the quotient is −6x² − 4x + 3.

Answer:

Therefore, the quotient is $(x - 2)$ and the remainder is 0.

Question 4:

Therefore, the quotient is $(x - 2)$ and the remainder is 0.

Answer:

Therefore, the quotient is $x$ −2 and the remainder is 0.

Question 5:

Therefore, the quotient is $x$ −2 and the remainder is 0.

Answer:

(x² + 12x + 35) by (x + 7)

Therefore, the quotient is $(x + 5)$ and the remainder is 0.

Question 6:

(x² + 12x + 35) by (x + 7)

Therefore, the quotient is $(x + 5)$ and the remainder is 0.

Answer:

Therefore, the quotient is $(5 x - 3)$ and the remainder is 0.

Question 7:

Therefore, the quotient is $(5 x - 3)$ and the remainder is 0.

Answer:

Therefore, the quotient is $(2 x - 5)$ and the remainder is 0.

Question 8:

Therefore, the quotient is $(2 x - 5)$ and the remainder is 0.

Answer:

Therefore, the quotient is $(3 x - 8)$ and the remainder is 7.

Question 9:

Therefore, the quotient is $(3 x - 8)$ and the remainder is 7.

Answer:

Therefore, the quotient is $(x^{2} - x - 1)$ and the remainder is 1.

Question 10:

Therefore, the quotient is $(x^{2} - x - 1)$ and the remainder is 1.

Answer:

Therefore, the quotient is $x^{2}$ -x+1 and the remainder is 0.

Question 11:

Therefore, the quotient is $x^{2}$ -x+1 and the remainder is 0.

Answer:

Therefore, the quotient is ( x² - 3x + 4) and remainder is 0.

Question 12:

Therefore, the quotient is ( x² - 3x + 4) and remainder is 0.

Answer:

Therefore, the quotient is (x-1) and the remainder is 0.

Question 13:

Therefore, the quotient is (x-1) and the remainder is 0.

Answer:

Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

Question 14:

Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

Answer:

Therefore, the quotient is (x-1) and the remainder is 0.

Question 15:

Therefore, the quotient is (x-1) and the remainder is 0.

Answer:

Therefore, the quotient is ( 4x²+ 3x -2) and the remainder is ( x-1).

Question 1:

Therefore, the quotient is ( 4x²+ 3x -2) and the remainder is ( x-1).

Answer:

(i) We have:

$(x + 6) (x + 6) = (x + 6)^{2} = x^{2} + 6^{2} + 2 \times x \times 6 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{2} + 36 + 12 x$

(ii) We have:

$(4 x + 5 y) (4 x + 5 y) = (4 x + 5 y)^{2} = {(4 x)}^{2} + {(5 y)}^{2} + 2 \times 4 x \times 5 y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 16 x^{2} + 25 y^{2} + 40 x y$

(iii) We have:
$(7 a + 9 b) (7 a + 9 b) = (7 a + 9 b)^{2} = {(7 a)}^{2} + {(9 b)}^{2} + 2 \times 7 a \times 9 b [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 49 a^{2} + 81 b^{2} + 126 a b$

(iv) We have:
$(\frac{2}{3} x + \frac{4}{5} y) (\frac{2}{3} x + \frac{4}{5} y) = {(\frac{2}{3} x + \frac{4}{5} y)}^{2} = {(\frac{2}{3} x)}^{2} + {(\frac{4}{5} y)}^{2} + 2 \times \frac{2}{3} x \times \frac{4}{5} y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{4}{9} x^{2} + \frac{16}{25} y^{2} + \frac{16}{15} x y$

(v) We have:
$(x^{2} + 7) (x^{2} + 7) = (x^{2} + 7)^{2} = {(x^{2})}^{2} + 7^{2} + 2 \times x^{2} \times 7 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{4} + 49 + 14 x^{2}$

(vi) We have:
$(\frac{5}{6} a^{2} + 2) (\frac{5}{6} a^{2} + 2) = {(\frac{5}{6} a^{2} + 2)}^{2} = {(\frac{5}{6} a^{2})}^{2} + {(2)}^{2} + 2 \times \frac{5}{6} a^{2} \times 2 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{25}{36} a^{4} + 4 + \frac{10}{3} a^{2}$

Question 2:

(i) We have:

$(x + 6) (x + 6) = (x + 6)^{2} = x^{2} + 6^{2} + 2 \times x \times 6 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{2} + 36 + 12 x$

(ii) We have:

$(4 x + 5 y) (4 x + 5 y) = (4 x + 5 y)^{2} = {(4 x)}^{2} + {(5 y)}^{2} + 2 \times 4 x \times 5 y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 16 x^{2} + 25 y^{2} + 40 x y$

(iii) We have:
$(7 a + 9 b) (7 a + 9 b) = (7 a + 9 b)^{2} = {(7 a)}^{2} + {(9 b)}^{2} + 2 \times 7 a \times 9 b [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 49 a^{2} + 81 b^{2} + 126 a b$

(iv) We have:
$(\frac{2}{3} x + \frac{4}{5} y) (\frac{2}{3} x + \frac{4}{5} y) = {(\frac{2}{3} x + \frac{4}{5} y)}^{2} = {(\frac{2}{3} x)}^{2} + {(\frac{4}{5} y)}^{2} + 2 \times \frac{2}{3} x \times \frac{4}{5} y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{4}{9} x^{2} + \frac{16}{25} y^{2} + \frac{16}{15} x y$

(v) We have:
$(x^{2} + 7) (x^{2} + 7) = (x^{2} + 7)^{2} = {(x^{2})}^{2} + 7^{2} + 2 \times x^{2} \times 7 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{4} + 49 + 14 x^{2}$

(vi) We have:
$(\frac{5}{6} a^{2} + 2) (\frac{5}{6} a^{2} + 2) = {(\frac{5}{6} a^{2} + 2)}^{2} = {(\frac{5}{6} a^{2})}^{2} + {(2)}^{2} + 2 \times \frac{5}{6} a^{2} \times 2 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{25}{36} a^{4} + 4 + \frac{10}{3} a^{2}$

Answer:

(i) We have:
$(x - 4) (x - 4) = (x - 4)^{2} = x^{2} - 2 \times x \times 4 + 4^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 8 x + 16$

(ii) We have:
$(2 x - 3 y) (2 x - 3 y) = (2 x - 3 y)^{2} = {(2 x)}^{2} - 2 \times 2 x \times 3 y + {(3 y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = 4 x^{2} - 12 x y + 9 y^{2}$

(iii) We have:
$(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y) = {(\frac{3}{4} x - \frac{5}{6} y)}^{2} = {(\frac{3}{4} x)}^{2} - 2 \times \frac{3}{4} x \times \frac{5}{6} y + {(\frac{5}{6} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{9}{16} x^{2} - \frac{15}{12} x y + \frac{25}{36} y^{2}$

(iv) We have:
$(x - \frac{3}{x}) (x - \frac{3}{x}) = {(x - \frac{3}{x})}^{2} = {(x)}^{2} - 2 \times x \times \frac{3}{x} + {(\frac{3}{x})}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 6 + \frac{9}{x^{2}}$

(v) We have:
$(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9) = {(\frac{1}{3} x^{2} - 9)}^{2} = {(\frac{1}{3} x^{2})}^{2} - 2 \times \frac{1}{3} x^{2} \times 9 + {(9)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{9} x^{4} - 6 x^{2} + 81$

(vi) We have:
$(\frac{1}{2} y^{2} - \frac{1}{3} y) (\frac{1}{2} y^{2} - \frac{1}{3} y) = {(\frac{1}{2} y^{2} - \frac{1}{3} y)}^{2} = {(\frac{1}{2} y^{2})}^{2} - 2 \times \frac{1}{2} y^{2} \times \frac{1}{3} y + {(\frac{1}{3} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{4} y^{4} - \frac{1}{3} y^{3} + \frac{1}{9} y^{2}$

Question 3:

(i) We have:
$(x - 4) (x - 4) = (x - 4)^{2} = x^{2} - 2 \times x \times 4 + 4^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 8 x + 16$

(ii) We have:
$(2 x - 3 y) (2 x - 3 y) = (2 x - 3 y)^{2} = {(2 x)}^{2} - 2 \times 2 x \times 3 y + {(3 y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = 4 x^{2} - 12 x y + 9 y^{2}$

(iii) We have:
$(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y) = {(\frac{3}{4} x - \frac{5}{6} y)}^{2} = {(\frac{3}{4} x)}^{2} - 2 \times \frac{3}{4} x \times \frac{5}{6} y + {(\frac{5}{6} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{9}{16} x^{2} - \frac{15}{12} x y + \frac{25}{36} y^{2}$

(iv) We have:
$(x - \frac{3}{x}) (x - \frac{3}{x}) = {(x - \frac{3}{x})}^{2} = {(x)}^{2} - 2 \times x \times \frac{3}{x} + {(\frac{3}{x})}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 6 + \frac{9}{x^{2}}$

(v) We have:
$(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9) = {(\frac{1}{3} x^{2} - 9)}^{2} = {(\frac{1}{3} x^{2})}^{2} - 2 \times \frac{1}{3} x^{2} \times 9 + {(9)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{9} x^{4} - 6 x^{2} + 81$

(vi) We have:
$(\frac{1}{2} y^{2} - \frac{1}{3} y) (\frac{1}{2} y^{2} - \frac{1}{3} y) = {(\frac{1}{2} y^{2} - \frac{1}{3} y)}^{2} = {(\frac{1}{2} y^{2})}^{2} - 2 \times \frac{1}{2} y^{2} \times \frac{1}{3} y + {(\frac{1}{3} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{4} y^{4} - \frac{1}{3} y^{3} + \frac{1}{9} y^{2}$

Answer:

We shall use the identities (a+b)² =a² +b² +2ab and (a-b)² =a² +b² -2ab.

(i) We have:
$(8 a + 3 b)^{2} = {(8 a)}^{2} + 2 \times 8 a \times 3 b + {(3 b)}^{2} = 64 a^{2} + 48 a b + 9 b^{2}$

(ii)We have:
$(7 x + 2 y)^{2} = {(7 x)}^{2} + 2 \times 7 x \times 2 y + {(2 y)}^{2} = 49 x^{2} + 28 x y + 4 y^{2}$

(iii) We have :
$(5 x + 11)^{2} = {(5 x)}^{2} + 2 \times 5 x \times 11 + {(11)}^{2} = 25 x^{2} + 110 x + 121$

(iv) We have:
${(\frac{a}{2} + \frac{2}{a})}^{2} = {(\frac{a}{2})}^{2} + 2 \times \frac{a}{2} \times \frac{2}{a} + {(\frac{2}{a})}^{2} = {\frac{a}{4}}^{2} + 2 + \frac{4}{a^{2}}$

(v) We have:
${(\frac{3 x}{4} + \frac{2 y}{9})}^{2} = {(\frac{3 x}{4})}^{2} + 2 \times \frac{3 x}{4} \times \frac{2 y}{9} + {(\frac{2 y}{9})}^{2} = {\frac{9 x}{16}}^{2} + \frac{1}{3} x y + \frac{4 y^{2}}{81}$

(vi) We have:
$(9 x - 10)^{2} {(9 x)}^{2} - 2 \times 9 x \times 10 + {(10)}^{2} = 81 x^{2} - 180 x + 100$

(vii) We have:
$(x^{2} y - y z^{2})^{2} {(x^{2} y)}^{2} - 2 \times x^{2} y \times y z^{2} + {(y z^{2})}^{2} = x^{4} y^{2} - 2 x^{2} y^{2} z^{2} + y^{2} z^{4}$

(viii) We have:
${(\frac{x}{y} - \frac{y}{x})}^{2} = {(\frac{x}{y})}^{2} - 2 \times \frac{x}{y} \times \frac{y}{x} + {(\frac{y}{x})}^{2} = \frac{x^{2}}{y^{2}} - 2 + \frac{y^{2}}{x^{2}}$

(ix) We have:
${(3 m - \frac{4}{5} n)}^{2} = {(3 m)}^{2} - 2 \times 3 m \times \frac{4}{5} n + {(\frac{4}{5} n)}^{2} = 9 m^{2} - \frac{24 m n}{5} + \frac{16}{25} n^{2}$

Question 4:

We shall use the identities (a+b)² =a² +b² +2ab and (a-b)² =a² +b² -2ab.

(i) We have:
$(8 a + 3 b)^{2} = {(8 a)}^{2} + 2 \times 8 a \times 3 b + {(3 b)}^{2} = 64 a^{2} + 48 a b + 9 b^{2}$

(ii)We have:
$(7 x + 2 y)^{2} = {(7 x)}^{2} + 2 \times 7 x \times 2 y + {(2 y)}^{2} = 49 x^{2} + 28 x y + 4 y^{2}$

(iii) We have :
$(5 x + 11)^{2} = {(5 x)}^{2} + 2 \times 5 x \times 11 + {(11)}^{2} = 25 x^{2} + 110 x + 121$

(iv) We have:
${(\frac{a}{2} + \frac{2}{a})}^{2} = {(\frac{a}{2})}^{2} + 2 \times \frac{a}{2} \times \frac{2}{a} + {(\frac{2}{a})}^{2} = {\frac{a}{4}}^{2} + 2 + \frac{4}{a^{2}}$

(v) We have:
${(\frac{3 x}{4} + \frac{2 y}{9})}^{2} = {(\frac{3 x}{4})}^{2} + 2 \times \frac{3 x}{4} \times \frac{2 y}{9} + {(\frac{2 y}{9})}^{2} = {\frac{9 x}{16}}^{2} + \frac{1}{3} x y + \frac{4 y^{2}}{81}$

(vi) We have:
$(9 x - 10)^{2} {(9 x)}^{2} - 2 \times 9 x \times 10 + {(10)}^{2} = 81 x^{2} - 180 x + 100$

(vii) We have:
$(x^{2} y - y z^{2})^{2} {(x^{2} y)}^{2} - 2 \times x^{2} y \times y z^{2} + {(y z^{2})}^{2} = x^{4} y^{2} - 2 x^{2} y^{2} z^{2} + y^{2} z^{4}$

(viii) We have:
${(\frac{x}{y} - \frac{y}{x})}^{2} = {(\frac{x}{y})}^{2} - 2 \times \frac{x}{y} \times \frac{y}{x} + {(\frac{y}{x})}^{2} = \frac{x^{2}}{y^{2}} - 2 + \frac{y^{2}}{x^{2}}$

(ix) We have:
${(3 m - \frac{4}{5} n)}^{2} = {(3 m)}^{2} - 2 \times 3 m \times \frac{4}{5} n + {(\frac{4}{5} n)}^{2} = 9 m^{2} - \frac{24 m n}{5} + \frac{16}{25} n^{2}$

Answer:

(i) We have:

$(x + 3) (x - 3) = x^{2} - 9 [using (a + b) (a - b) = a^{2} - b^{2}]$

(ii) We have:

$(2 x + 5) (2 x - 5) = 4 x^{2} - 25 [using (a + b) (a - b) = a^{2} - b^{2}]$

(iii) We have:

$(8 + x) (8 - x) = 64 - x^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(iv) We have:

$(7 x + 11 y) (7 x - 11 y) = 49 x^{2} - 121 y^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(v) We have:

$(5 x^{2} + \frac{3}{4} y^{2}) (5 x^{2} - \frac{3}{4} y^{2}) = 25 x^{4} - \frac{9}{16} y^{4} [using (a + b) (a - b) = a^{2} - b^{2}]$

(vi) We have:

$(\frac{4 x}{5} - \frac{5 y}{3}) (\frac{4 x}{5} + \frac{5 y}{3}) = \frac{16 x^{2}}{25} - \frac{25 y^{2}}{9} [using (a + b) (a - b) = a^{2} - b^{2})]$

(vii) We have:
$(x + \frac{1}{x}) (x - \frac{1}{x}) = x^{2} - \frac{1}{x^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(viii) We have:
$(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y}) = \frac{1}{x^{2}} - \frac{1}{y^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(ix) We have:
$(2 a + \frac{3}{b}) (2 a - \frac{3}{b}) = 4 a^{2} - \frac{9}{b^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

Question 5:

(i) We have:

$(x + 3) (x - 3) = x^{2} - 9 [using (a + b) (a - b) = a^{2} - b^{2}]$

(ii) We have:

$(2 x + 5) (2 x - 5) = 4 x^{2} - 25 [using (a + b) (a - b) = a^{2} - b^{2}]$

(iii) We have:

$(8 + x) (8 - x) = 64 - x^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(iv) We have:

$(7 x + 11 y) (7 x - 11 y) = 49 x^{2} - 121 y^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(v) We have:

$(5 x^{2} + \frac{3}{4} y^{2}) (5 x^{2} - \frac{3}{4} y^{2}) = 25 x^{4} - \frac{9}{16} y^{4} [using (a + b) (a - b) = a^{2} - b^{2}]$

(vi) We have:

$(\frac{4 x}{5} - \frac{5 y}{3}) (\frac{4 x}{5} + \frac{5 y}{3}) = \frac{16 x^{2}}{25} - \frac{25 y^{2}}{9} [using (a + b) (a - b) = a^{2} - b^{2})]$

(vii) We have:
$(x + \frac{1}{x}) (x - \frac{1}{x}) = x^{2} - \frac{1}{x^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(viii) We have:
$(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y}) = \frac{1}{x^{2}} - \frac{1}{y^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(ix) We have:
$(2 a + \frac{3}{b}) (2 a - \frac{3}{b}) = 4 a^{2} - \frac{9}{b^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

Answer:

We shall use the identity (a+b)² =a² +b² +2ab.

(i)
${(54)}^{2} = (50 + 4)^{2} = {(50)}^{2} + 2 \times 50 \times 4 + {(4)}^{2} = 2500 + 400 + 16 = 2916$

(ii)
${(82)}^{2} = (80 + 2)^{2} = {(80)}^{2} + 2 \times 80 \times 2 + {(2)}^{2} = 6400 + 320 + 4 = 6724$

(iii)
${(103)}^{2} = (100 + 3)^{2} = {(100)}^{2} + 2 \times 100 \times 3 + {(3)}^{2} = 10000 + 600 + 9 = 10609$

(iv)
${(704)}^{2} = (700 + 4)^{2} = {(700)}^{2} + 2 \times 700 \times 4 + {(4)}^{2} = 490000 + 5600 + 16 = 495616$

Question 6:

We shall use the identity (a+b)² =a² +b² +2ab.

(i)
${(54)}^{2} = (50 + 4)^{2} = {(50)}^{2} + 2 \times 50 \times 4 + {(4)}^{2} = 2500 + 400 + 16 = 2916$

(ii)
${(82)}^{2} = (80 + 2)^{2} = {(80)}^{2} + 2 \times 80 \times 2 + {(2)}^{2} = 6400 + 320 + 4 = 6724$

(iii)
${(103)}^{2} = (100 + 3)^{2} = {(100)}^{2} + 2 \times 100 \times 3 + {(3)}^{2} = 10000 + 600 + 9 = 10609$

(iv)
${(704)}^{2} = (700 + 4)^{2} = {(700)}^{2} + 2 \times 700 \times 4 + {(4)}^{2} = 490000 + 5600 + 16 = 495616$

Answer:

We shall use the identity (a-b)² = a² +b² -2ab.

(i)
${(69)}^{2} = (70 - 1)^{2} = {(70)}^{2} - 2 \times 70 \times 1 + 1 = 4900 - 140 + 1 = 4761$

(ii)
${(78)}^{2} = (80 - 2)^{2} = {(80)}^{2} - 2 \times 80 \times 2 + 4 = 6400 - 320 + 4 = 6084$

(iii)
${(197)}^{2} = (200 - 3)^{2} = {(200)}^{2} - 2 \times 200 \times 3 + 9 = 40000 - 1200 + 9 = 38809$

(iv)
${(999)}^{2} = (1000 - 1)^{2} = {(1000)}^{2} - 2 \times 1000 \times 1 + 1 = 1000000 - 2000 + 1 = 998001$

Question 7:

We shall use the identity (a-b)² = a² +b² -2ab.

(i)
${(69)}^{2} = (70 - 1)^{2} = {(70)}^{2} - 2 \times 70 \times 1 + 1 = 4900 - 140 + 1 = 4761$

(ii)
${(78)}^{2} = (80 - 2)^{2} = {(80)}^{2} - 2 \times 80 \times 2 + 4 = 6400 - 320 + 4 = 6084$

(iii)
${(197)}^{2} = (200 - 3)^{2} = {(200)}^{2} - 2 \times 200 \times 3 + 9 = 40000 - 1200 + 9 = 38809$

(iv)
${(999)}^{2} = (1000 - 1)^{2} = {(1000)}^{2} - 2 \times 1000 \times 1 + 1 = 1000000 - 2000 + 1 = 998001$

Answer:

We shall use the identity (a-b) (a+b)=a² - b².

(i)
$(82)^{2} - (18)^{2} = (82 - 18) (82 + 18) = (64) (100) = 6400$

(ii)
$(128)^{2} - (72)^{2} = (128 - 72) (128 + 72) = (56) (200) = 11200$

(iii)
$197 \times 203 = (200 - 3) (200 + 3) = {(200)}^{2} - {(3)}^{2} = 40000 - 9 = 39991$

(iv)
$\frac{198 \times 198 - 102 \times 102}{96} = \frac{{(198)}^{2} - {(102)}^{2}}{96} = \frac{(198 - 102) (198 + 102)}{96} = \frac{(96) (300)}{96} = 300$

(v)
$(14.7 \times 15.3) = (15 - 0.3) \times (15 + 0.3) = (15)^{2} - (0.3)^{2} = 225 - 0.09 = 224.91$

(vi)
$(8.63)^{2} - (1.37)^{2} = (8.63 - 1.37) (8.63 + 1.37) = (7.26) (10) = 72.6$

Question 8:

We shall use the identity (a-b) (a+b)=a² - b².

(i)
$(82)^{2} - (18)^{2} = (82 - 18) (82 + 18) = (64) (100) = 6400$

(ii)
$(128)^{2} - (72)^{2} = (128 - 72) (128 + 72) = (56) (200) = 11200$

(iii)
$197 \times 203 = (200 - 3) (200 + 3) = {(200)}^{2} - {(3)}^{2} = 40000 - 9 = 39991$

(iv)
$\frac{198 \times 198 - 102 \times 102}{96} = \frac{{(198)}^{2} - {(102)}^{2}}{96} = \frac{(198 - 102) (198 + 102)}{96} = \frac{(96) (300)}{96} = 300$

(v)
$(14.7 \times 15.3) = (15 - 0.3) \times (15 + 0.3) = (15)^{2} - (0.3)^{2} = 225 - 0.09 = 224.91$

(vi)
$(8.63)^{2} - (1.37)^{2} = (8.63 - 1.37) (8.63 + 1.37) = (7.26) (10) = 72.6$

Answer:

$(9 x^{2} + 24 x + 16) G i v e n, x = 12 \Rightarrow {(3 x)}^{2} + 2 (3 x) (4) + {(4)}^{2} \Rightarrow {(3 x + 4)}^{2} \Rightarrow {(3 (12) + 4)}^{2} \Rightarrow {(36 + 4)}^{2} \Rightarrow {(40)}^{2} = 1600$

Therefore, the value of the expression (9x² + 24x + 16), when x = 12, is 1600.

Question 9:

$(9 x^{2} + 24 x + 16) G i v e n, x = 12 \Rightarrow {(3 x)}^{2} + 2 (3 x) (4) + {(4)}^{2} \Rightarrow {(3 x + 4)}^{2} \Rightarrow {(3 (12) + 4)}^{2} \Rightarrow {(36 + 4)}^{2} \Rightarrow {(40)}^{2} = 1600$

Therefore, the value of the expression (9x² + 24x + 16), when x = 12, is 1600.

Answer:

$(64 x^{2} + 81 y^{2} + 144 x y) G i v e n : x = 11 y = \frac{4}{3} \Rightarrow {(8 x)}^{2} + {(9 y)}^{2} + 2 (8 x) (9 y) \Rightarrow {(8 x + 9 y)}^{2} \Rightarrow {(8 (11) + 9 (\frac{4}{3}))}^{2} \Rightarrow {(88 + 12)}^{2} \Rightarrow {(100)}^{2} \Rightarrow 10000$

Therefore, the value of the expression (64x² + 81y² + 144xy), when x = 11 and y = $\frac{4}{3}$ , is 10000.y=43

Question 10:

$(64 x^{2} + 81 y^{2} + 144 x y) G i v e n : x = 11 y = \frac{4}{3} \Rightarrow {(8 x)}^{2} + {(9 y)}^{2} + 2 (8 x) (9 y) \Rightarrow {(8 x + 9 y)}^{2} \Rightarrow {(8 (11) + 9 (\frac{4}{3}))}^{2} \Rightarrow {(88 + 12)}^{2} \Rightarrow {(100)}^{2} \Rightarrow 10000$

Therefore, the value of the expression (64x² + 81y² + 144xy), when x = 11 and y = $\frac{4}{3}$ , is 10000.y=43

Answer:

$(36 x^{2} + 25 y^{2} - 60 x y) \Rightarrow x = \frac{2}{3}, y = \frac{1}{5} = {(6 x)}^{2} + {(5 y)}^{2} - 2 (6 x) (5 y) = {(6 x - 5 y)}^{2} = {(6 (\frac{2}{3}) - 5 (\frac{1}{5}))}^{2} = {(4 - 1)}^{2} = {(3)}^{2} \Rightarrow 9$

Question 11:

$(36 x^{2} + 25 y^{2} - 60 x y) \Rightarrow x = \frac{2}{3}, y = \frac{1}{5} = {(6 x)}^{2} + {(5 y)}^{2} - 2 (6 x) (5 y) = {(6 x - 5 y)}^{2} = {(6 (\frac{2}{3}) - 5 (\frac{1}{5}))}^{2} = {(4 - 1)}^{2} = {(3)}^{2} \Rightarrow 9$

Answer:

$(i) (x + \frac{1}{x}) = 4 Squaring both the sides : \Rightarrow {(x + \frac{1}{x})}^{2} = {(4)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} + 2 (x) (\frac{1}{x})) = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) + 2 = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 16 - 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 14$

Therefore, the value of x²+ $\frac{1}{x^{2}}$ is 14.

$(x^{2} + \frac{1}{x^{2}}) = 14 S q u a r i n g b o t h t h e s i d e s : \Rightarrow (x^{4} + \frac{1}{x^{4}} + 2 (x^{2}) (\frac{1}{x^{2}})) = {(14)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) + 2 = 196 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 196 - 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 194$

Therefore, the value of x⁴+ $\frac{1}{x^{4}}$ is 194.

Question 12:

$(i) (x + \frac{1}{x}) = 4 Squaring both the sides : \Rightarrow {(x + \frac{1}{x})}^{2} = {(4)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} + 2 (x) (\frac{1}{x})) = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) + 2 = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 16 - 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 14$

Therefore, the value of x²+ $\frac{1}{x^{2}}$ is 14.

$(x^{2} + \frac{1}{x^{2}}) = 14 S q u a r i n g b o t h t h e s i d e s : \Rightarrow (x^{4} + \frac{1}{x^{4}} + 2 (x^{2}) (\frac{1}{x^{2}})) = {(14)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) + 2 = 196 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 196 - 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 194$

Therefore, the value of x⁴+ $\frac{1}{x^{4}}$ is 194.

Answer:

$(i) (x - \frac{1}{x}) = 5 \Rightarrow Squaring both the sides : \Rightarrow {(x - \frac{1}{x})}^{2} = {(5)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} - 2 (x) (\frac{1}{x})) = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 2 = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 25 + 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 27 Therefore, the value of (x^{2} + \frac{1}{x^{2}}) is 27 .$

$(x^{2} + \frac{1}{x^{2}}) = 27 \Rightarrow Squaring both the sides : \Rightarrow (x^{4} + \frac{1}{x^{4}} - 2 (x^{2}) (\frac{1}{x^{2}})) = {(27)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) - 2 = 729 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 729 + 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 731 Therefore, the value of (x^{4} + \frac{1}{x^{4}}) is 731 .$

Question 13:

$(i) (x - \frac{1}{x}) = 5 \Rightarrow Squaring both the sides : \Rightarrow {(x - \frac{1}{x})}^{2} = {(5)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} - 2 (x) (\frac{1}{x})) = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 2 = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 25 + 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 27 Therefore, the value of (x^{2} + \frac{1}{x^{2}}) is 27 .$

$(x^{2} + \frac{1}{x^{2}}) = 27 \Rightarrow Squaring both the sides : \Rightarrow (x^{4} + \frac{1}{x^{4}} - 2 (x^{2}) (\frac{1}{x^{2}})) = {(27)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) - 2 = 729 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 729 + 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 731 Therefore, the value of (x^{4} + \frac{1}{x^{4}}) is 731 .$

Answer:

$(i) (x + 1) (x - 1) (x^{2} + 1) \Rightarrow (x^{2} - x + x - 1) (x^{2} + 1) \Rightarrow (x^{2} - 1) (x^{2} + 1) \Rightarrow {(x^{2})}^{2} - {(1^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 1 . Therefore, the product of (x + 1) (x - 1) (x^{2} + 1) is x^{4} - 1 .$

$(i i) (x - 3) (x + 3) (x^{2} + 9) \Rightarrow ({(x)}^{2} - {(3)}^{2}) (x^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (x^{2} - 9) (x^{2} + 9) \Rightarrow {(x^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 81 Therefore, the product of (x - 3) (x + 3) (x^{2} + 9) is x^{4} - 81 .$

$(iii) (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) \Rightarrow ({(3 x)}^{2} - {(2 y)}^{2}) (9 x^{2} + 4 y^{2}) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (9 x^{2} - 4 y^{2}) (9 x^{2} + 4 y^{2}) \Rightarrow {(9 x^{2})}^{2} - {(4 y^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 81 x^{4} - 16 y^{4} . Therefore, the product of (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) is 81 x^{4} - 16 y^{4} .$

$(iv) (2 p + 3) (2 p - 3) (4 p^{2} + 9) \Rightarrow ({(2 p)}^{2} - {(3)}^{2}) (4 p^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (4 p^{2} - 9) (4 p^{2} + 9) \Rightarrow {(4 p^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 16 p^{4} - 81 . Therefore, the product of (2 p + 3) (2 p - 3) (4 p^{2} + 9) is 16 p^{4} - 81 .$

Question 14:

$(i) (x + 1) (x - 1) (x^{2} + 1) \Rightarrow (x^{2} - x + x - 1) (x^{2} + 1) \Rightarrow (x^{2} - 1) (x^{2} + 1) \Rightarrow {(x^{2})}^{2} - {(1^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 1 . Therefore, the product of (x + 1) (x - 1) (x^{2} + 1) is x^{4} - 1 .$

$(i i) (x - 3) (x + 3) (x^{2} + 9) \Rightarrow ({(x)}^{2} - {(3)}^{2}) (x^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (x^{2} - 9) (x^{2} + 9) \Rightarrow {(x^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 81 Therefore, the product of (x - 3) (x + 3) (x^{2} + 9) is x^{4} - 81 .$

$(iii) (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) \Rightarrow ({(3 x)}^{2} - {(2 y)}^{2}) (9 x^{2} + 4 y^{2}) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (9 x^{2} - 4 y^{2}) (9 x^{2} + 4 y^{2}) \Rightarrow {(9 x^{2})}^{2} - {(4 y^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 81 x^{4} - 16 y^{4} . Therefore, the product of (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) is 81 x^{4} - 16 y^{4} .$

$(iv) (2 p + 3) (2 p - 3) (4 p^{2} + 9) \Rightarrow ({(2 p)}^{2} - {(3)}^{2}) (4 p^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (4 p^{2} - 9) (4 p^{2} + 9) \Rightarrow {(4 p^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 16 p^{4} - 81 . Therefore, the product of (2 p + 3) (2 p - 3) (4 p^{2} + 9) is 16 p^{4} - 81 .$

Answer:

$x + y = 12 On squaring both the sides : \Rightarrow {(x + y)}^{2} = {(12)}^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 144 \Rightarrow x^{2} + y^{2} = 144 - 2 x y Given : x y = 14 \Rightarrow x^{2} + y^{2} = 144 - 2 (14) \Rightarrow x^{2} + y^{2} = 144 - 28 \Rightarrow x^{2} + y^{2} = 116 Therefore, the value of x^{2} + y^{2} is 116 .$

Question 15:

$x + y = 12 On squaring both the sides : \Rightarrow {(x + y)}^{2} = {(12)}^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 144 \Rightarrow x^{2} + y^{2} = 144 - 2 x y Given : x y = 14 \Rightarrow x^{2} + y^{2} = 144 - 2 (14) \Rightarrow x^{2} + y^{2} = 144 - 28 \Rightarrow x^{2} + y^{2} = 116 Therefore, the value of x^{2} + y^{2} is 116 .$

Answer:

$x - y = 7 \Rightarrow On squaring both the sides : \Rightarrow {(x - y)}^{2} = {(7)}^{2} \Rightarrow x^{2} + y^{2} - 2 x y = 49 \Rightarrow x^{2} + y^{2} = 49 + 2 x y Given : x y = 9 \Rightarrow x^{2} + y^{2} = 49 + 2 (9) \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67 . Therefore, the value of x^{2} + y^{2} i s 67 .$

Question 1:

$x - y = 7 \Rightarrow On squaring both the sides : \Rightarrow {(x - y)}^{2} = {(7)}^{2} \Rightarrow x^{2} + y^{2} - 2 x y = 49 \Rightarrow x^{2} + y^{2} = 49 + 2 x y Given : x y = 9 \Rightarrow x^{2} + y^{2} = 49 + 2 (9) \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67 . Therefore, the value of x^{2} + y^{2} i s 67 .$

Answer:

(c) (−6a + 17b)

$6 a + 4 b - c + 3 + 2 b - 3 c + 4 - 7 a + 11 b + 2 c - 1 - 5 a + 2 c - 6 \bar{- 6 a + 17 b + 0 c + 0}$

Question 2:

(c) (−6a + 17b)

$6 a + 4 b - c + 3 + 2 b - 3 c + 4 - 7 a + 11 b + 2 c - 1 - 5 a + 2 c - 6 \bar{- 6 a + 17 b + 0 c + 0}$

Answer:

(d) (3p² + 5q − 9r³ +7)

$7 p^{2} + 3 q - 2 r^{3} + 4 4 p^{2} - 2 q + 7 r^{3} - 3 - + - + \bar{3 p^{2} + 5 q - 9 r^{3} + 7}$

Question 3:

(d) (3p² + 5q − 9r³ +7)

$7 p^{2} + 3 q - 2 r^{3} + 4 4 p^{2} - 2 q + 7 r^{3} - 3 - + - + \bar{3 p^{2} + 5 q - 9 r^{3} + 7}$

Answer:

(d) x² + 2x − 15

$(x + 5) (x - 3) \Rightarrow (x) (x - 3) + (5) (x - 3) \Rightarrow x^{2} - 3 x + 5 x - 15 \Rightarrow x^{2} + 2 x - 15$

Question 4:

(d) x² + 2x − 15

$(x + 5) (x - 3) \Rightarrow (x) (x - 3) + (5) (x - 3) \Rightarrow x^{2} - 3 x + 5 x - 15 \Rightarrow x^{2} + 2 x - 15$

Answer:

(b) (6x² + 7x − 3)

$(2 x + 3) (3 x - 1) \Rightarrow (2 x) (3 x - 1) + (3) (3 x - 1) \Rightarrow 6 x^{2} - 2 x + 9 x - 3 \Rightarrow 6 x^{2} + 7 x - 3$

Question 5:

(b) (6x² + 7x − 3)

$(2 x + 3) (3 x - 1) \Rightarrow (2 x) (3 x - 1) + (3) (3 x - 1) \Rightarrow 6 x^{2} - 2 x + 9 x - 3 \Rightarrow 6 x^{2} + 7 x - 3$

Answer:

(c) (x² + 8x + 16)

$(x + 4) (x + 4) \Rightarrow {(x + 4)}^{2} (according to the formula {(a + b)}^{2} = a^{2} + 2 a b + b^{2}) \Rightarrow (x^{2}) + 2 (x) (4) + {(4)}^{2} \Rightarrow x^{2} + 8 x + 16$

Question 6:

(c) (x² + 8x + 16)

$(x + 4) (x + 4) \Rightarrow {(x + 4)}^{2} (according to the formula {(a + b)}^{2} = a^{2} + 2 a b + b^{2}) \Rightarrow (x^{2}) + 2 (x) (4) + {(4)}^{2} \Rightarrow x^{2} + 8 x + 16$

Answer:

(d) (x² − 12x + 36)

$(x - 6) (x - 6) \Rightarrow {(x - 6)}^{2} (according to the formula {(a - b)}^{2} = a^{2} - 2 a b + b^{2}) \Rightarrow (x^{2}) - 2 (x) (6) + {(6)}^{2} \Rightarrow x^{2} - 12 x + 36$

Question 7:

(d) (x² − 12x + 36)

$(x - 6) (x - 6) \Rightarrow {(x - 6)}^{2} (according to the formula {(a - b)}^{2} = a^{2} - 2 a b + b^{2}) \Rightarrow (x^{2}) - 2 (x) (6) + {(6)}^{2} \Rightarrow x^{2} - 12 x + 36$

Answer:

(b) (4x² − 25)

$(2 x + 5) (2 x - 5) \Rightarrow {(2 x)}^{2} - {(5)}^{2} (according to the formula (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow 4 x^{2} - 25$

Question 8:

(b) (4x² − 25)

$(2 x + 5) (2 x - 5) \Rightarrow {(2 x)}^{2} - {(5)}^{2} (according to the formula (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow 4 x^{2} - 25$

Answer:

(c) −4ab²

$8 a^{2} b^{3} \div (- 2 a b) \Rightarrow (\frac{8}{- 2}) (a^{2 - 1}) (b^{3 - 1}) \Rightarrow - 4 a b^{2}$

Question 9:

(c) −4ab²

$8 a^{2} b^{3} \div (- 2 a b) \Rightarrow (\frac{8}{- 2}) (a^{2 - 1}) (b^{3 - 1}) \Rightarrow - 4 a b^{2}$

Answer:

(b) (2x + 1)

Question 10:

(b) (2x + 1)

Answer:

(a) (x − 2)

Question 11:

(a) (x − 2)

Answer:

(c) (a⁴ − 1)

$(i) (a + 1) (a - 1) (a^{2} + 1) \Rightarrow ({(a)}^{2} - {(1)}^{2}) (a^{2} + 1) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (a^{2} - 1) (a^{2} + 1) \Rightarrow {(a^{2})}^{2} - {(1^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow a^{4} - 1$

Question 12:

(c) (a⁴ − 1)

$(i) (a + 1) (a - 1) (a^{2} + 1) \Rightarrow ({(a)}^{2} - {(1)}^{2}) (a^{2} + 1) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (a^{2} - 1) (a^{2} + 1) \Rightarrow {(a^{2})}^{2} - {(1^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow a^{4} - 1$

Answer:

a) $(\frac{1}{x^{2}} - \frac{1}{y^{2}})$

$(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y}) \Rightarrow A c c o r d i n g t o t h e f o r m u l a (a + b) (a - b) = {(a)}^{2} - {(b)}^{2} : \Rightarrow (\frac{1}{x^{2}} - \frac{1}{y^{2}})$ (1x2−1y2)

Question 13:

a) $(\frac{1}{x^{2}} - \frac{1}{y^{2}})$

$(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y}) \Rightarrow A c c o r d i n g t o t h e f o r m u l a (a + b) (a - b) = {(a)}^{2} - {(b)}^{2} : \Rightarrow (\frac{1}{x^{2}} - \frac{1}{y^{2}})$ (1x2−1y2)

Answer:

(c) 23

$(x + \frac{1}{x}) = 5 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(x + \frac{1}{x})}^{2} = {(5)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} + 2 (x) (\frac{1}{x})) = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) + 2 = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 25 - 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 23$

Question 14:

(c) 23

$(x + \frac{1}{x}) = 5 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(x + \frac{1}{x})}^{2} = {(5)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} + 2 (x) (\frac{1}{x})) = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) + 2 = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 25 - 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 23$

Answer:

(b) 38

$(x - \frac{1}{x}) = 6 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(x - \frac{1}{x})}^{2} = {(6)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} - 2 (x) (\frac{1}{x})) = 36 \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 2 = 36 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 36 + 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 38$

Question 15:

(b) 38

$(x - \frac{1}{x}) = 6 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(x - \frac{1}{x})}^{2} = {(6)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} - 2 (x) (\frac{1}{x})) = 36 \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 2 = 36 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 36 + 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 38$

Answer:

(c) 6400

${(82)}^{2} - {(18)}^{2} = (82 + 18) (82 - 18) = (100) (64) = 6400$ [using the identity (a-b)(a+b)=a² -b²]

Question 16:

(c) 6400

${(82)}^{2} - {(18)}^{2} = (82 + 18) (82 - 18) = (100) (64) = 6400$ [using the identity (a-b)(a+b)=a² -b²]

Answer:

(a) 39991

$(197) \times (203) \Rightarrow (200 - 3) (200 + 3) \Rightarrow {(200)}^{2} - {(3)}^{2} \Rightarrow 40000 - 9 \Rightarrow 39991$ [using the identity (a+b) (a-b) = a²-b²]

Question 17:

(a) 39991

$(197) \times (203) \Rightarrow (200 - 3) (200 + 3) \Rightarrow {(200)}^{2} - {(3)}^{2} \Rightarrow 40000 - 9 \Rightarrow 39991$ [using the identity (a+b) (a-b) = a²-b²]

Answer:

(b) 116

$(a + b) = 12 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(a + b)}^{2} = {(12)}^{2} \Rightarrow (a^{2} + b^{2} + 2 a b) = 144 \Rightarrow (a^{2} + b^{2}) = 144 - 2 a b \Rightarrow (a^{2} + b^{2}) = 144 - 2 (14) \Rightarrow (a^{2} + b^{2}) = 144 - 28 \Rightarrow (a^{2} + b^{2}) = 116$

Question 18:

(b) 116

$(a + b) = 12 \Rightarrow S q u a r i n g b o t h t h e s i d e s : \Rightarrow {(a + b)}^{2} = {(12)}^{2} \Rightarrow (a^{2} + b^{2} + 2 a b) = 144 \Rightarrow (a^{2} + b^{2}) = 144 - 2 a b \Rightarrow (a^{2} + b^{2}) = 144 - 2 (14) \Rightarrow (a^{2} + b^{2}) = 144 - 28 \Rightarrow (a^{2} + b^{2}) = 116$

Answer:

(a) 67

$(a - b) = 7 \Rightarrow Squaring both the sides : \Rightarrow {(a - b)}^{2} = {(7)}^{2} \Rightarrow (a^{2} + b^{2} - 2 a b) = 49 \Rightarrow (a^{2} + b^{2}) = 49 + 2 a b \Rightarrow (a^{2} + b^{2}) = 49 + 2 (9) \Rightarrow (a^{2} + b^{2}) = 49 + 18 \Rightarrow (a^{2} + b^{2}) = 67$

Question 19:

(a) 67

$(a - b) = 7 \Rightarrow Squaring both the sides : \Rightarrow {(a - b)}^{2} = {(7)}^{2} \Rightarrow (a^{2} + b^{2} - 2 a b) = 49 \Rightarrow (a^{2} + b^{2}) = 49 + 2 a b \Rightarrow (a^{2} + b^{2}) = 49 + 2 (9) \Rightarrow (a^{2} + b^{2}) = 49 + 18 \Rightarrow (a^{2} + b^{2}) = 67$

Answer:

(c) 625

$(4 x^{2} + 20 x + 25) \Rightarrow {(2 x)}^{2} + 2 (2 x) (5) + {(5)}^{2} \Rightarrow {(2 x + 5)}^{2} \Rightarrow {(2 (10) + 5)}^{2} \Rightarrow {(20 + 5)}^{2} \Rightarrow {(25)}^{2} \Rightarrow 625$

RS Aggarwal 2019 2020 Solutions for Class 8 Math Chapter 6 - Operations On Algebraic Expressions

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