Mathematics Part i Solutions Solutions for Class 9 Maths Chapter 3 - Polynomials

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Page No 39:

Question 1:

State whether the given algebraic expressions are polynomials? Justify.

(i) $y + \frac{1}{y}$ (ii) $2 - 5 \sqrt{x}$ (iii) $x^{2} + 7 x + 9$ (iv) $2 m^{- 2} + 7 m - 5$ (v) 10

Answer:

In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

(i)
$y + \frac{1}{y} = y + y^{- 1}$

Here, one of the powers of y is −1, which is not a whole number. So, $y + \frac{1}{y}$ is not a polynomial.

(ii)
$2 - 5 \sqrt{x} = 2 - 5 x^{\frac{1}{2}}$

Here, the power of x is $\frac{1}{2}$ , which is not a whole number. So, $2 - 5 \sqrt{x}$ is not a polynomial.

(iii)
$x^{2} + 7 x + 9$

Here, the powers of the variable x are 2, 1 and 0, which are whole numbers. So, $x^{2} + 7 x + 9$ is a polynomial.

(iv)
$2 m^{- 2} + 7 m - 5$

Here, one of the powers of m is −2, which is not a whole number. So, $2 m^{- 2} + 7 m - 5$ is not a polynomial.

(v)
10 = 10 × 1 = 10x⁰

Here, the power of x is 0, which is a whole numbers. So, 10 is a polynomial (or constant polynomial).

Page No 39:

Question 2:

Write the coefficient of m³ in each of the given polynomial.

(i) m³ (ii) $\frac{- 3}{2} + m - \sqrt{3} m^{3}$ (iii) $\frac{- 2}{3} m^{3} - 5 m^{2} + 7 m - 1$

Answer:

(i)
Coefficient of m³ = 1

(ii)
$\frac{- 3}{2} + m - \sqrt{3} m^{3}$

Coefficient of m³ = $- \sqrt{3}$

(iii)
$\frac{- 2}{3} m^{3} - 5 m^{2} + 7 m - 1$

Coefficient of m³ = $- \frac{2}{3}$

Page No 39:

Question 3:

Write the polynomial in x using the given information.
(i) Monomial with degree 7

(ii) Binomial with degree 35

(iii) Trinomial with degree 8

Answer:

(i) A polynomial having only one term is called a monomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

3x⁷is a monomial in x with degree 7.

(ii) A polynomial having only two terms is called a binomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

2x³⁵+1is a binomial in x with degree 35.

(iii) A polynomial having only three terms is called a trinomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

5x⁸+6x⁴ + 7x is a trinomial in x with degree 8.

Page No 40:

Question 4:

Write the degree of the given polynomials.

(i) $\sqrt{5}$ (ii) $x^{°}$ (iii) $x^{2}$ (iv) $\sqrt{2} m^{10} - 7$ (v) $2 p - \sqrt{7}$ (vi) $7 y - y^{3} + y^{5}$ (vii) $x y z + x y - z$ (viii) $m^{3} n^{7} - 3 m^{5} n + m n$

Answer:

The highest power of the variable in a polynomial of one variable is called the degreee of the polynomial. Also, the highest sum of the powers of the variables in each term of the polynomial in more than one variable is the degree of the polynomial.

(i)
$\sqrt{5} = \sqrt{5} \times 1 = \sqrt{5} x^{0}$

The degree of the polynomial $\sqrt{5}$ is 0.

(ii)
The degree of the polynomial x⁰ is 0.

(iii)
The degree of the polynomial x² is 2.

(iv)
The degree of the polynomial $\sqrt{2} m^{10} - 7$ is 10.

(v)
The degree of the polynomial $2 p - \sqrt{7}$ is 1.

(vi)
The degree of the polynomial $7 y - y^{3} + y^{5}$ is 5.

(vii)
The sum of the powers of the variables in the polynomial $x y z + x y - z$ are 1 + 1 + 1 = 3 and 1 + 1 = 2.

The degree of the polynomial $x y z + x y - z$ is 3.

(viii)
The sum of the powers of the variables in the polynomial $m^{3} n^{7} - 3 m^{5} n + m n$ are 3 + 7 = 10, 5 + 1 = 6 and 1 + 1 = 2.

The degree of the polynomial $m^{3} n^{7} - 3 m^{5} n + m n$ is 10.

Page No 40:

Question 5:

Classify the following polynomials as linear, quadratic and cubic polynomial.

(i) 2x² + 3 x + 1 (ii)

5 p

(iii)

\sqrt{2} y - \frac{1}{2}

(iv)

m^{3} + 7 m^{2} + \frac{5}{2} m - \sqrt{7}

(v)

a^{2}

(vi)

3 r^{3}

Answer:

(i)
The degree of the polynomial 2x² + 3x + 1 is 2.

So, the polynomial 2x² + 3x + 1 is a quadratic polynomial.

(ii)
The degree of the polynomial 5p is 1.

So, the polynomial 5p is a linear polynomial.

(iii)
The degree of the polynomial $\sqrt{2} y - \frac{1}{2}$ is 1.

So, the polynomial $\sqrt{2} y - \frac{1}{2}$ is a linear polynomial.

(iv)
The degree of the polynomial $m^{3} + 7 m^{2} + \frac{5}{2} m - \sqrt{7}$ is 3.

So, the polynomial $m^{3} + 7 m^{2} + \frac{5}{2} m - \sqrt{7}$ is a cubic polynomial.

(v)
The degree of the polynomial a² is 2.

So, the polynomial a² is a quadratic polynomial.

(vi)
The degree of the polynomial 3r³ is 3.

So, the polynomial 3r³ is a cubic polynomial.

Page No 40:

Question 6:

Write the following polynomials in standard form.

(i) $m^{3} + 3 + 5 m$ (ii) $- 7 y + y^{5} + 3 y^{3} - \frac{1}{2} + 2 y^{4} - y^{2}$

Answer:

A polynomial written in either descending or ascending powers of its variable is called the standard form of the polynomial.

(i)
The given polynomial is m³ + 3 + 5m.

The standard form of the polynomial is 3 + 5m + m³ or m³ + 5m + 3.

(ii)
The given polynomial is $- 7 y + y^{5} + 3 y^{3} - \frac{1}{2} + 2 y^{4} - y^{2}$ .

The standard form of the polynomial is $y^{5} + 2 y^{4} + 3 y^{3} - y^{2} - 7 y - \frac{1}{2}$ or $- \frac{1}{2} - 7 y - y^{2} + 3 y^{3} + 2 y^{4} + y^{5}$ .

Page No 40:

Question 7:

Write the following polynomials in coefficient form.

(i) $x^{3} - 2$ (ii) $5 y$ (iii) $2 m^{4} - 3 m^{2} + 7$ (iv) $- \frac{2}{3}$

Answer:

(i)
$x^{3} - 2 = x^{3} + 0 x^{2} + 0 x - 2$

The coefficient form of the polynomial is (1, 0, 0, −2).

(ii)
5y = 5y + 0

The coefficient form of the polynomial is (5, 0).

(iii)
$2 m^{4} - 3 m^{2} + 7 = 2 m^{4} + 0 m^{3} - 3 m^{2} + 0 m + 7$

The coefficient form of the polynomial is (2, 0, −3, 0, 7).

(iv)
The coefficient form of the polynomial $- \frac{2}{3}$ is $(- \frac{2}{3})$ .

Page No 40:

Question 8:

Write the polynomials in index form.

(i) (1, 2, 3) (ii) (5, 0, 0, 0,

-

1) (iii) (

-

2, 2,

-

â€‹2, 2)

Answer:

(i)
The coefficient form of the polynomial is (1, 2, 3).

Therefore, the index form of the polynomial is x²+ 2x + 3.

(ii)
The coefficient form of the polynomial is (5, 0, 0, 0, −1).

Therefore, the index form of the polynomial is 5x⁴+ 0x³ + 0x²+ 0x− 1 or 5x⁴− 1.

(iii)
The coefficient form of the polynomial is (−2, 2, −2, 2).

Therefore, the index form of the polynomial is −2x³ + 2x²−2x+ 2.

Page No 40:

Question 9:

Write the appropriate polynomials in the boxes.

Answer:

Page No 43:

Question 1:

Use the given letters to write the answer.

(i) There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?

(ii) For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?

(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.

Answer:

(i)
Initial number of trees in the village = a

Increase in the number of trees every year = b

∴ Number of trees in the village after x years

= Initial number of trees in the village + Increase in the number of trees every year × x

= a + bx

Thus, the number of trees after x years is a + bx.

(ii)
Number of students in each row = y

Number of rows = x

∴ Total number of students in the parade = Number of students in each row × Number of rows = y × x = yx = xy

Thus, there are in all xy students for the parade.

(iii)
Digit at the tens place = m

Digit at the units place = n

∴ Two digit number = Digit at the tens place × 10 + Digit at the units place = m × 10 + n = 10m + n

Thus, the polynomial which represents the two digit number is 10m + n.

Page No 43:

Question 2:

Add the given polynomials.

(i) $x^{3} - 2 x^{2} - 9; 5 x^{3} + 2 x + 9$ (ii) $- 7 m^{4} + 5 m^{3} + \sqrt{2}; 5 m^{4} - 3 m^{3} + 2 m^{2} + 3 m - 6$ (iii) $2 y^{2} + 7 y + 5; 3 y + 9; 3 y^{2} - 4 y - 3$

Answer:

(i)
$(x^{3} - 2 x^{2} - 9) + (5 x^{3} + 2 x + 9) = x^{3} + 5 x^{3} - 2 x^{2} + 2 x - 9 + 9 = 6 x^{3} - 2 x^{2} + 2 x$
(ii)
$(- 7 m^{4} + 5 m^{3} + \sqrt{2}) + (5 m^{4} - 3 m^{3} + 2 m^{2} + 3 m - 6) = - 7 m^{4} + 5 m^{4} + 5 m^{3} - 3 m^{3} + 2 m^{2} + 3 m + \sqrt{2} - 6 = - 2 m^{4} + 2 m^{3} + 2 m^{2} + 3 m - 6 + \sqrt{2}$
(iii)
$(2 y^{2} + 7 y + 5) + (3 y + 9) + (3 y^{2} - 4 y - 3) = 2 y^{2} + 3 y^{2} + 7 y + 3 y - 4 y + 5 + 9 - 3 = 5 y^{2} + 6 y + 11$

Page No 43:

Question 3:

Subtract the second polynomial from the first.

(i) $x^{2} - 9 x + \sqrt{3}; - 19 x + \sqrt{3} + 7 x^{2}$ (ii) $2 a b^{2} + 3 a^{2} b - 4 a b; 3 a b - 8 a b^{2} + 2 a^{2} b$

Answer:

(i)
$(x^{2} - 9 x + \sqrt{3}) - (- 19 x + \sqrt{3} + 7 x^{2}) = x^{2} - 9 x + \sqrt{3} + 19 x - \sqrt{3} - 7 x^{2} = x^{2} - 7 x^{2} - 9 x + 19 x + \sqrt{3} - \sqrt{3} = - 6 x^{2} + 10 x$
(ii)
$(2 a b^{2} + 3 a^{2} b - 4 a b) - (3 a b - 8 a b^{2} + 2 a^{2} b) = 2 a b^{2} + 3 a^{2} b - 4 a b - 3 a b + 8 a b^{2} - 2 a^{2} b = 2 a b^{2} + 8 a b^{2} + 3 a^{2} b - 2 a^{2} b - 4 a b - 3 a b = 10 a b^{2} + a^{2} b - 7 a b$

Page No 43:

Question 4:

Multiply the given polynomials.

(i) $2 x; x^{2} - 2 x - 1$ (ii) $x^{5} - 1; x^{3} + 2 x^{2} + 2$ (iii) $2 y + 1; y^{2} - 2 y^{3} + 3 y$

Answer:

(i)
$2 x (x^{2} - 2 x - 1) = 2 x \times x^{2} + 2 x \times (- 2 x) + 2 x \times (- 1) = 2 x^{3} - 4 x^{2} - 2 x$
(ii)
$(x^{5} - 1) \times (x^{3} + 2 x^{2} + 2) = x^{5} (x^{3} + 2 x^{2} + 2) - 1 (x^{3} + 2 x^{2} + 2) = x^{8} + 2 x^{7} + 2 x^{5} - x^{3} - 2 x^{2} - 2$
(iii)
$(2 y + 1) \times (y^{2} - 2 y^{3} + 3 y) = 2 y (y^{2} - 2 y^{3} + 3 y) + 1 (y^{2} - 2 y^{3} + 3 y) = 2 y^{3} - 4 y^{4} + 6 y^{2} + y^{2} - 2 y^{3} + 3 y = - 4 y^{4} + 2 y^{3} - 2 y^{3} + 6 y^{2} + y^{2} + 3 y = - 4 y^{4} + 7 y^{2} + 3 y$

Page No 43:

Question 5:

Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor

\times

Quotient + Remainder’.

(i)

x^{3} - 64; x - 4

(ii)

5 x^{5} + 4 x^{4} - 3 x^{3} + 2 x^{2} + 2; x^{2} - x

Answer:

(i)
$x^{3} - 64 = x^{3} + 0 x^{2} + 0 x - 64$

Using long division method,

Dividend = Divisor ×â€‹ Quotient + Remainder

$∴ x^{3} - 64 = (x - 4) \times (x^{2} + 4 x + 16) + 0$

(ii)
$5 x^{5} + 4 x^{4} - 3 x^{3} + 2 x^{2} + 2 = 5 x^{5} + 4 x^{4} - 3 x^{3} + 2 x^{2} + 0 x + 2$

Using long division method,

Dividend = Divisor ×â€‹ Quotient + Remainder

$∴ 5 x^{5} + 4 x^{4} - 3 x^{3} + 2 x^{2} + 2 = (x^{2} - x) \times (5 x^{3} + 9 x^{2} + 6 x + 8) + (8 x + 2)$

Page No 43:

Question 6:

Write down the information in the form of algebraic expression and simplify.

There is a rectangular farm with length

(2 a^{2} + 3 b^{2})

metre and breadth (

a^{2} + b^{2}

) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (

a^{2} - b^{2}

) metre. What is the area of the remaining part of the farm ?

Answer:

Lenght of the rectangular farm = (2a² + 3b²) m

Breadth of the rectangular farm = (a² + b²) m

Side of the square plot = (a² − b²) m

∴ Area of the remaining part of the farm

= Total area of the farm − Area of the square plot

= Lenght of the rectangular farm × Breadth of the rectangular farm − (Side of the square plot)²

= (2a² + 3b²) × (a² + b²) − (a² − b²)²

= 2a²(a² + b²) + 3b²(a² + b²) − (a⁴ + b⁴− 2a²b²)

= 2a⁴+ 2a²b²+3a²b² + 3b⁴ − a⁴ − b⁴+ 2a²b²

= 2a⁴ − a⁴+ 2a²b²+3a²b²+ 2a²b²+ 3b⁴− b⁴

= (a⁴ + 7a²b²+ 2b⁴) m²

Thus, the area of the remaining part of the farm is (a⁴ + 7a²b²+ 2b⁴) m².

Page No 46:

Question 1:

Divide each of the following polynomials by synthetic division method and also by

linear division method. Write the quotient and the remainder.

(i)

(2 m^{2} - 3 m + 10) \div (m - 5)

(ii)

(x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5) \div (x + 2)

(iii)

(y^{3} - 216) \div (y - 6)

(iv)

(2 x^{4} + 3 x^{3} + 4 x - 2 x^{2}) \div (x + 3)

(v)

(x^{4} - 3 x^{2} - 8) \div (x + 4)

(vi)

(y^{3} - 3 y^{2} + 5 y - 1) \div (y - 1)

Answer:

(i)
Synthetic Division:

Dividend = $2 m^{2} - 3 m + 10$

Divisor = $m - 5$

Opposite of −5 = 5

The coefficient form of the quotient is (2, 7).

∴ Quotient = 2m + 7 and Remainder = 45

Linear Method:

$2 m^{2} - 3 m + 10 = 2 m (m - 5) + 10 m - 3 m + 10 = 2 m (m - 5) + 7 (m - 5) + 35 + 10 = (m - 5) \times (2 m + 7) + 45$
(ii)
Synthetic Division:

Dividend = $x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5$

Divisor = $x + 2$

Opposite of 2 = −2

The coefficient form of the quotient is (1, 0, 3, −2).

∴ Quotient = x³+ 3x − 2 and Remainder = 9

Linear Method:

$x^{4} + 2 x^{3} + 3 x^{2} + 4 x + 5 = x^{3} (x + 2) + 3 x (x + 2) - 6 x + 4 x + 5 = x^{3} (x + 2) + 3 x (x + 2) - 2 x + 5 = x^{3} (x + 2) + 3 x (x + 2) - 2 (x + 2) + 4 + 5 = (x + 2) \times (x^{3} + 3 x - 2) + 9$
(iii)
Synthetic Division:

Dividend = $y^{3} - 216 = y^{3} + 0 y^{2} + 0 y - 216$

Divisor = $y - 6$

Opposite of −6 = 6

The coefficient form of the quotient is (1, 6, 36).

∴ Quotient = y²+ 6y + 36 and Remainder = 0

Linear Method:

$y^{3} - 216 = y^{2} (y - 6) + 6 y^{2} - 216 = y^{2} (y - 6) + 6 y (y - 6) + 36 y - 216 = y^{2} (y - 6) + 6 y (y - 6) + 36 (y - 6) + 216 - 216 = y^{2} (y - 6) + 6 y (y - 6) + 36 (y - 6) = (y - 6) \times (y^{2} + 6 y + 36)$
(iv)
Synthetic Division:

Dividend = $2 x^{4} + 3 x^{3} + 4 x - 2 x^{2} = 2 x^{4} + 3 x^{3} - 2 x^{2} + 4 x + 0$

Divisor = $x + 3$

Opposite of 3 = −3

The coefficient form of the quotient is (2, −3, 7, −17).

∴ Quotient = 2x³− 3x² + 7x − 17 and Remainder = 51

Linear Method:

$2 x^{4} + 3 x^{3} - 2 x^{2} + 4 x = 2 x^{3} (x + 3) - 6 x^{3} + 3 x^{3} - 2 x^{2} + 4 x = 2 x^{3} (x + 3) - 3 x^{2} (x + 3) + 9 x^{2} - 2 x^{2} + 4 x = 2 x^{3} (x + 3) - 3 x^{2} (x + 3) + 7 x (x + 3) - 21 x + 4 x = 2 x^{3} (x + 3) - 3 x^{2} (x + 3) + 7 x (x + 3) - 17 (x + 3) + 51 = (x + 3) \times (2 x^{3} - 3 x^{2} + 7 x - 17) + 51$
(v)
Synthetic Division:

Dividend = $x^{4} - 3 x^{2} - 8 = x^{4} + 0 x^{3} - 3 x^{2} + 0 x - 8$

Divisor = $x + 4$

Opposite of 4 = −4

The coefficient form of the quotient is (1, −4, 13, −52).

∴ Quotient = x³− 4x² + 13x − 52 and Remainder = 200

Linear Method:

$x^{4} - 3 x^{2} - 8 = x^{3} (x + 4) - 4 x^{3} - 3 x^{2} - 8 = x^{3} (x + 4) - 4 x^{2} (x + 4) + 16 x^{2} - 3 x^{2} - 8 = x^{3} (x + 4) - 4 x^{2} (x + 4) + 13 x (x + 4) - 52 x - 8 = x^{3} (x + 4) - 4 x^{2} (x + 4) + 13 x (x + 4) - 52 (x + 4) + 208 - 8 = (x + 4) \times (x^{3} - 4 x^{2} + 13 x - 52) + 200$
(vi)
Synthetic Division:

Dividend = $y^{3} - 3 y^{2} + 5 y - 1$

Divisor = $y - 1$

Opposite of −1 = 1

The coefficient form of the quotient is (1, −2, 3).

∴ Quotient = y²− 2y + 3 and Remainder = 2

Linear Method:

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

Page No 48:

Question 1:

For $x$ = 0 find the value of the polynomial $x^{2} - 5 x + 5$ .

Answer:

Let $p (x) = x^{2} - 5 x + 5$ .

$∴ p (0) = {(0)}^{2} - 5 \times 0 + 5 = 0 - 0 + 5 = 5$

Hence, for x = 0 the value of the polynomial is 5.

Page No 48:

Question 2:

If $p (y) = y^{2} - 3 \sqrt{2} y + 1$ then find $p (3 \sqrt{2})$ .

Answer:

$p (y) = y^{2} - 3 \sqrt{2} y + 1$

$∴ p (3 \sqrt{2}) = {(3 \sqrt{2})}^{2} - 3 \sqrt{2} \times 3 \sqrt{2} + 1 = 18 - 18 + 1 = 1$

Page No 48:

Question 3:

If $p (m) = m^{3} + 2 m^{2} - m + 10$ then $p (a) + p (- a) = ?$

Answer:

$p (m) = m^{3} + 2 m^{2} - m + 10$

$∴ p (a) = a^{3} + 2 a^{2} - a + 10$ .....(1)

Also,

$p (- a) = {(- a)}^{3} + 2 {(- a)}^{2} - (- a) + 10$

$\Rightarrow p (- a) = - a^{3} + 2 a^{2} + a + 10$ .....(2)

Adding (1) and (2), we get

$p (a) + p (- a) = (a^{3} + 2 a^{2} - a + 10) + (- a^{3} + 2 a^{2} + a + 10) = a^{3} - a^{3} + 2 a^{2} + 2 a^{2} - a + a + 10 + 10 = 4 a^{2} + 20$
$∴ p (a) + p (- a) = 4 a^{2} + 20$

Page No 48:

Question 4:

If $p (y) = 2 y^{3} - 6 y^{2} - 5 y + 7$ then find $p (2) .$

Answer:

$p (y) = 2 y^{3} - 6 y^{2} - 5 y + 7$

$∴ p (2) = 2 \times {(2)}^{3} - 6 \times {(2)}^{2} - 5 \times 2 + 7 = 16 - 24 - 10 + 7 = - 11$

Page No 53:

Question 1:

Find the value of the polynomial $2 x - 2 x^{3} + 7$ using given values for $x$ .

(i) $x = 3$ (ii) $x = - 1$ (iii) x = 0

Answer:

Let $p (x) = 2 x - 2 x^{3} + 7$ .

(i)
$p (3) = 2 \times 3 - 2 \times {(3)}^{3} + 7 = 6 - 2 \times 27 + 7 = 6 - 54 + 7 = - 41$
Thus, the value of polynomial for x = 3 is −41.

(ii)
$p (- 1) = 2 \times (- 1) - 2 \times {(- 1)}^{3} + 7 = - 2 - 2 \times (- 1) + 7 = - 2 + 2 + 7 = 7$
Thus, the value of polynomial for x = −1 is 7.

(iii)
$p (0) = 2 \times 0 - 2 \times {(0)}^{3} + 7 = 0 - 0 + 7 = 7$
Thus, the value of polynomial for x = 0 is 7.

Page No 53:

Question 2:

For each of the following polynomial, find $p (1), p (0)$ and $p (- 2)$ .

(i) $p (x) = x^{3}$ (ii) $p (y) = y^{2} - 2 y + 5$ (iii) $p (x) = x^{4} - 2 x^{2} - x$

Answer:

(i)
$p (x) = x^{3} ∴ p (1) = {(1)}^{3} = 1 p (0) = {(0)}^{3} = 0 p (- 2) = {(- 2)}^{3} = - 8$
(ii)
$p (y) = y^{2} - 2 y + 5 ∴ p (1) = {(1)}^{2} - 2 \times 1 + 5 = 1 - 2 + 5 = 4 p (0) = {(0)}^{2} - 2 \times 0 + 5 = 0 - 0 + 5 = 5 p (- 2) = {(- 2)}^{2} - 2 \times (- 2) + 5 = 4 + 4 + 5 = 13$
(iii)
$p (x) = x^{4} - 2 x^{2} - x ∴ p (1) = {(1)}^{4} - 2 \times {(1)}^{2} - 1 = 1 - 2 - 1 = - 2 p (0) = {(0)}^{4} - 2 \times {(0)}^{2} - 0 = 0 - 0 - 0 = 0 p (- 2) = {(- 2)}^{4} - 2 \times {(- 2)}^{2} - (- 2) = 16 - 2 \times 4 + 2 = 16 - 8 + 2 = 10$

Page No 53:

Question 3:

If the value of the polynomial $m^{3} + 2 m + a$ is 12 for $m = 2$ , then find the value of a.

Answer:

Let $p (m) = m^{3} + 2 m + a$

For m = 2, p(2) = 12.

$∴ {(2)}^{3} + 2 \times 2 + a = 12 \Rightarrow 8 + 4 + a = 12 \Rightarrow 12 + a = 12 \Rightarrow a = 12 - 12 = 0$
Thus, the value of a is 0.

Page No 53:

Question 4:

For the polynomial $m x^{2} - 2 x + 3$ if $p (- 1) = 7$ then find m.

Answer:

Let $p (x) = m x^{2} - 2 x + 3$ .

$∴ p (- 1) = 7 \Rightarrow m \times {(- 1)}^{2} - 2 \times (- 1) + 3 = 7 \Rightarrow m + 2 + 3 = 7 \Rightarrow m = 7 - 5 = 2$
Thus, the value of m is 2.

Page No 53:

Question 5:

Divide the first polynomial by the second polynomial and find the remainder using factor theorem .

(i) $(x^{2} - 7 x + 9); (x + 1)$ (ii) $(2 x^{3} - 2 x^{2} + a x - a); (x - a)$ (iii) $(54 m^{3} + 18 m^{2} - 27 m + 5); (m - 3)$

Answer:

(i)
By synthetic division:

Dividend = $x^{2} - 7 x + 9$

Divisor = x + 1

Opposite of 1 = −1

The coefficient form of the quotient is (1, −8).

∴ Quotient = x − 8

Remainder = 17

By remainder theorem:

Let $p (x) = x^{2} - 7 x + 9$ .

Divisor = x + 1

By remainder theorem,

Remainder = p(−1) = ${(- 1)}^{2} - 7 \times (- 1) + 9 = 1 + 7 + 9 = 19$

(ii)
By synthetic division:

Dividend = $2 x^{3} - 2 x^{2} + a x - a$

Divisor = x − a

Opposite of −a = a

The coefficient form of the quotient is (2, 2a − 2, 2a²â€‹ − a).

∴ Quotient = 2x² + (2a − 2)x + 2a²â€‹ − a

Remainder = 2a³â€‹ − a²− a

By remainder theorem:

Let $p (x) = 2 x^{3} - 2 x^{2} + a x - a$ .

Divisor = x − a

By remainder theorem,

Remainder = p(a) = $2 \times a^{3} - 2 \times a^{2} + a \times a - a = 2 a^{3} - 2 a^{2} + a^{2} - a = 2 a^{3} - a^{2} - a$

(iii)
By synthetic division:

Dividend = $54 m^{3} + 18 m^{2} - 27 m + 5$

Divisor = m − 3

Opposite of −3 = 3

The coefficient form of the quotient is (54, 180, 513).

∴ Quotient = 54x² + 180x + 513

Remainder = 1544

By remainder theorem:

Let $p (m) = 54 m^{3} + 18 m^{2} - 27 m + 5$ .

Divisor = m − 3

By remainder theorem,

Remainder = p(3) = $54 \times {(3)}^{3} + 18 \times {(3)}^{2} - 27 \times 3 + 5 = 54 \times 27 + 18 \times 9 - 27 \times 3 + 5 = 1458 + 162 - 81 + 5 = 1544$

Page No 53:

Question 6:

If the polynomial $y^{3} - 5 y^{2} + 7 y + m$ is divided by y + 2 and the remainder is 50 then find the value of m.

Answer:

Let $p (y) = y^{3} - 5 y^{2} + 7 y + m$ .

When the polynomial is divided by (y + 2), the remainder is 50. This means that the value of the polynomial when y = −2 is 50.

By remainder theorem,

Remainder = p(−2) = 50

$∴ {(- 2)}^{3} - 5 \times {(- 2)}^{2} + 7 \times (- 2) + m = 50 \Rightarrow - 8 - 5 \times 4 - 14 + m = 50 \Rightarrow - 8 - 20 - 14 + m = 50 \Rightarrow - 42 + m = 50 \Rightarrow m = 50 + 42 = 92$
Thus, the value of m is 92.

Page No 53:

Question 7:

Use factor theorem to determine whether x + 3 is factor of x² + 2x − 3 or not.

Answer:

Let p(x) = x² + 2x − 3.

Divisor = x + 3

$∴ p (- 3) = {(- 3)}^{2} + 2 \times (- 3) - 3 = 9 - 6 - 3 = 0$

So, by factor theorem, (x + 3) is a factor of x² + 2x − 3.

Page No 53:

Question 8:

If ( x $- 2$ ) is a factor of $x^{3} - m x^{2} + 10 x - 20$ then find the value of m.

Answer:

Let $p (x) = x^{3} - m x^{2} + 10 x - 20$ .

It is given that (x − 2) is a factor of $p (x) = x^{3} - m x^{2} + 10 x - 20$ .

$∴ p (2) = 0 \Rightarrow {(2)}^{3} - m \times {(2)}^{2} + 10 \times (2) - 20 = 0 \Rightarrow 8 - 4 m + 20 - 20 = 0 \Rightarrow 8 - 4 m = 0 \Rightarrow 4 m = 8 \Rightarrow m = 2$
Thus, the value of m is 2.

Page No 53:

Question 9:

By using factor theorem in the following examples, determine whether q ( x ) is a factor p ( x ) or not.

(i) $p (x) = x^{3} - x^{2} - x - 1, q (x) = x - 1$

(ii) $p (x) = 2 x^{3} - x^{2} - 45, q (x) = x - 3$

Answer:

(i)
$p (x) = x^{3} - x^{2} - x - 1$

Divisor = $q (x) = x - 1$

$∴ p (1) = {(1)}^{3} - {(1)}^{2} - 1 - 1 = 1 - 1 - 1 - 1 = - 2$

Since p(1) ≠ 0, so by factor theorem $q (x) = x - 1$ is not a factor of polynomial $p (x) = x^{3} - x^{2} - x - 1$ .

(ii)
$p (x) = 2 x^{3} - x^{2} - 45$

Divisor = $q (x) = x - 3$

$∴ p (3) = 2 \times {(3)}^{3} - {(3)}^{2} - 45 = 2 \times 27 - 9 - 45 = 54 - 54 = 0$

So, by factor theorem $q (x) = x - 3$ is a factor of polynomial $p (x) = 2 x^{3} - x^{2} - 45$ .

Page No 53:

Question 10:

If ( x³¹ + 31) is divided by (x + 1) then find the remainder.

Answer:

Let p(x) = x³¹ + 31.

Divisor = x + 1

By remainder theorem, we have

Remainder = p(−1) = (−1)³¹+ 31 = −1 + 31 = 30

Thus, the remainder when (x³¹ + 31) is divided by (x + 1) is 30.

Page No 53:

Question 11:

Show that m $-$ 1 is a factor of m²¹ $-$ 1 and m²² $-$ 1.

Answer:

Let p(m) = m²¹ − 1 and q(m) = m²² − 1.

Divisor = m − 1

Now,

p(1) = (1)²¹ − 1 = 1 − 1 = 0

Therefore, by factor theorem (m − 1) is a factor of p(m) = m²¹ − 1.

Also,

q(1) = (1)²² − 1 = 1 − 1 = 0

Therefore, by factor theorem (m − 1) is a factor of q(m) = m²² − 1.

Hence, (m − 1) is a factor of m²¹ − 1 and m²² − 1.

Page No 53:

Question 12:

If $x - 2$ and $x - \frac{1}{2}$ both are the factors of the polynomial nx² − 5x + m, then show that $m = n = 2$

Answer:

Let p(x) = nx² − 5x + m.

It given that (x − 2) and $(x - \frac{1}{2})$ are the factors of the polynomial p(x) = nx² − 5x + m.

∴ By factor theorem, p(2) = 0 and $p (\frac{1}{2}) = 0$ .

$p (2) = 0 \Rightarrow n \times {(2)}^{2} - 5 \times 2 + m = 0 \Rightarrow 4 n - 10 + m = 0 \Rightarrow 4 n + m = 10 . . . . . (1)$
Also,

$p (\frac{1}{2}) = 0 \Rightarrow n {(\frac{1}{2})}^{2} - 5 \times \frac{1}{2} + m = 0 \Rightarrow \frac{n}{4} + m = \frac{5}{2} \Rightarrow n + 4 m = 10 . . . . . (2)$
From (1) and (2), we have

4n + m = n + 4m

⇒ 4n − n = 4m − m

⇒ 3n = 3m

⇒ n = m

Putting n = m in (1), we have

4m + m = 10

⇒ 5m = 10

⇒ m = 2

∴ n = m = 2

Page No 53:

Question 13:

(i) If $p (x) = 2 + 5 x$ then $p (2) + p (- 2)$ $- p (1)$ .

(ii) $p (x) = 2 x^{2} - 5 \sqrt{3} x + 5$ then $p (5 \sqrt{3})$ .

Answer:

(i)
$p (x) = 2 + 5 x ∴ p (2) = 2 + 5 \times 2 = 2 + 10 = 12 p (- 2) = 2 + 5 \times (- 2) = 2 - 10 = - 8 p (1) = 2 + 5 \times 1 = 2 + 5 = 7$
$∴ p (2) + p (- 2) - p (1) = 12 + (- 8) - 7 = 12 - 8 - 7 = 12 - 15 = - 3$

(ii)
$p (x) = 2 x^{2} - 5 \sqrt{3} x + 5$
$∴ p (5 \sqrt{3}) = 2 \times {(5 \sqrt{3})}^{2} - 5 \sqrt{3} \times 5 \sqrt{3} + 5 = 2 \times 75 - 25 \times 3 + 5 = 150 - 75 + 5 = 80$

Page No 54:

Question 1:

Find the factors of the polynomials given below.
(i) 2x² + x – 1

(ii) 2m² + 5m – 3

(iii) 12x² + 61x + 77

(iv) 3y² – 2y – 1

(v) $\sqrt{3} x^{2} + 4 x + \sqrt{3}$

(vi) $\frac{1}{2} x^{2} - 3 x + 4$

Answer:

(i)
$2 x^{2} + x - 1 = 2 x^{2} + 2 x - x - 1 = 2 x (x + 1) - 1 (x + 1) = (x + 1) (2 x - 1)$
(ii)
$2 m^{2} + 5 m - 3 = 2 m^{2} + 6 m - m - 3 = 2 m (m + 3) - 1 (m + 3) = (m + 3) (2 m - 1)$
(iii)
$12 x^{2} + 61 x + 77 = 12 x^{2} + 28 x + 33 x + 77 = 4 x (3 x + 7) + 11 (3 x + 7) = (3 x + 7) (4 x + 11)$
(iv)
$3 y^{2} - 2 y - 1 = 3 y^{2} - 3 y + y - 1 = 3 y (y - 1) + 1 (y - 1) = (y - 1) (3 y + 1)$
(v)
$\sqrt{3} x^{2} + 4 x + \sqrt{3} = \sqrt{3} x^{2} + 3 x + x + \sqrt{3} = \sqrt{3} x (x + \sqrt{3}) + 1 (x + \sqrt{3}) = (x + \sqrt{3}) (\sqrt{3} x + 1)$
(vi)
$\frac{1}{2} x^{2} - 3 x + 4 = \frac{1}{2} x^{2} - 2 x - x + 4 = \frac{1}{2} x (x - 4) - 1 (x - 4) = (x - 4) (\frac{1}{2} x - 1)$

Page No 55:

Question 2:

Factorize the following polynomials.
(i) (x² – x)² – 8 (x² – x) + 12
(ii) (x – 5)² – (5x – 25) – 24
(iii) (x² – 6x)² – 8 (x² – 6x + 8) – 64
(iv) (x² – 2x + 3) (x² – 2x + 5) – 35
(v) (y + 2) (y – 3) (y + 8) (y + 3) + 56
(vi) (y² + 5y) (y² + 5y – 2) – 24
(vii) (x – 3) (x – 4)² (x – 5) – 6

Answer:

(i)
(x² – x)² – 8(x² – x) + 12
Let x² – x = z.
$∴ {(x^{2} - x)}^{2} - 8 (x^{2} - x) + 12 = z^{2} - 8 z + 12 = z^{2} - 6 z - 2 z + 12 = z (z - 6) - 2 (z - 6) = (z - 6) (z - 2)$
$= (x^{2} - x - 6) (x^{2} - x - 2) (Replace z = x^{2} - x) = (x^{2} - 3 x + 2 x - 6) (x^{2} - 2 x + x - 2) = [x (x - 3) + 2 (x - 3)] [x (x - 2) + 1 (x - 2)] = (x - 3) (x + 2) (x - 2) (x + 1)$
(ii)
(x – 5)² – (5x – 25) – 24
= (x – 5)² – 5(x – 5) – 24
Let x – 5 = z.
$∴ {(x - 5)}^{2} - 5 (x - 5) - 24 = z^{2} - 5 z - 24 = z^{2} - 8 z + 3 z - 24 = z (z - 8) + 3 (z - 8) = (z - 8) (z + 3)$
$= (x - 5 - 8) (x - 5 + 3) (Replace z = x - 5) = (x - 13) (x - 2)$
(iii)
(x² – 6x)² – 8(x² – 6x + 8) – 64
= (x² – 6x)² – 8(x² – 6x) – 64 – 64
= (x² – 6x)² – 8(x² – 6x) – 128
Let x² – 6x = z.
$∴ {(x^{2} - 6 x)}^{2} - 8 (x^{2} - 6 x) - 128 = z^{2} - 8 z - 128 = z^{2} - 16 z + 8 z - 128 = z (z - 16) + 8 (z - 16) = (z - 16) (z + 8)$
$= (x^{2} - 6 x - 16) (x^{2} - 6 x + 8) (Replace z = x^{2} - 6 x) = (x^{2} - 8 x + 2 x - 16) (x^{2} - 4 x - 2 x + 8) = [x (x - 8) + 2 (x - 8)] [x (x - 4) - 2 (x - 4)] = (x - 8) (x + 2) (x - 4) (x - 2)$

(iv)
(x² – 2x + 3)(x² – 2x + 5) – 35
Let x² – 2x = z.
$∴ (x^{2} - 2 x + 3) (x^{2} - 2 x + 5) - 35 = (z + 3) (z + 5) - 35 = z^{2} + 5 z + 3 z + 15 - 35 = z^{2} + 8 z - 20 = z^{2} + 10 z - 2 z - 20$
$= z (z + 10) - 2 (z + 10) = (z + 10) (z - 2) = (x^{2} - 2 x + 10) (x^{2} - 2 x - 2) (Replace z = x^{2} - 2 x)$
(v)
(y + 2)(y – 3)(y + 8)(y + 3) + 56
= (y + 2)(y + 3)(y + 8)(y – 3) + 56
= (y² + 5y + 6)(y² + 5y – 24) + 56
Let y² + 5y = z.
$∴ (y^{2} + 5 y + 6) (y^{2} + 5 y - 24) + 56 = (z + 6) (z - 24) + 56 = z^{2} - 18 z - 144 + 56 = z^{2} - 18 z - 88$
$= z^{2} - 22 z + 4 z - 88 = z (z - 22) + 4 (z - 22) = (z - 22) (z + 4) = (y^{2} + 5 y - 22) (y^{2} + 5 y + 4) (Replace z = y^{2} + 5 y)$
$= (y^{2} + 5 y - 22) (y^{2} + 4 y + y + 4) = (y^{2} + 5 y - 22) [y (y + 4) + 1 (y + 4)] = (y^{2} + 5 y - 22) (y + 4) (y + 1)$
(vi)
(y² + 5y)(y² + 5y – 2) – 24
Let y² + 5y = z.
$∴ (y^{2} + 5 y) (y^{2} + 5 y - 2) - 24 = z (z - 2) - 24 = z^{2} - 2 z - 24 = z^{2} - 6 z + 4 z - 24 = z (z - 6) + 4 (z - 6) = (z - 6) (z + 4)$
$= (y^{2} + 5 y - 6) (y^{2} + 5 y + 4) (Replace z = y^{2} + 5 y) = (y^{2} + 6 y - y - 6) (y^{2} + 4 y + y + 4) = [y (y + 6) - 1 (y + 6)] [y (y + 4) + 1 (y + 4)] = (y + 6) (y - 1) (y + 4) (y + 1)$
(vii)
(x – 3)(x – 4)²(x – 5) – 6
= (x – 3)(x – 5)(x – 4)² – 6
= (x² – 8x + 15)(x² – 8x + 16) – 6
Let x² – 8x = z.
$∴ (x^{2} - 8 x + 15) (x^{2} - 8 x + 16) - 6 = (z + 15) (z + 16) - 6 = z^{2} + 31 z + 240 - 6 = z^{2} + 31 z + 234$
$= z^{2} + 18 z + 13 z + 234 = z (z + 18) + 13 (z + 18) = (z + 18) (z + 13) = (x^{2} - 8 x + 18) (x^{2} - 8 x + 13) (Replace z = x^{2} - 8 x)$

Page No 55:

Question 1:

Write the correct alternative answer for each of the following questions.

(i) Which of the following is a polynomial ?
(A)

\frac{x}{y}

(B)

x^{\sqrt{2}} - 3 x

(C)

x^{- 2} + 7

(D)

\sqrt{2} x^{2} + \frac{1}{2}

(ii) What is the degree of the polynomial

\sqrt{7}

?
(A)

\frac{1}{2}

(B) 5 (C) 2 (D) 0

(iii) What is the degree of the 0 polynomial ?

(A) 0 (B) 1 (C) undefined (D) any real number

(iv) What is the degree of the polynomial 2x²+ 5x³+ 7?
(A) 3 (B) 2 (C) 5 (D) 7

(v) What is the coefficient form of

x^{3} - 1

?
(A) (1,

-

1) (B) (3,

-

1) (C) (1, 0, 0,

-

1) (D) (1, 3,

-

1)

(vi)

p (x) = x^{2} - 7 \sqrt{7} x + 3

then

p (7 \sqrt{7}) = ?

(A) 3 (B)

7 \sqrt{7}

(C)

42 \sqrt{7} + 3

(D)

49 \sqrt{7}

(vii) When

x = - 1

, what is the value of the polynomial 2x³+ 2x ?
(A) 4 (B) 2 (C)

- 2

(D)

- 4

(viii) If

x - 1

, what is a factor of the polynomial

3 x^{2} + m x

then find the value of m.
(A) 2 (B)

-

2 (C)

- 3

(D) 3

(ix) Multiply ( x²

-

3) (2x

-

7x³+ 4) and write the degree of the product.
(A) 5 (B) 3 (C) 2 (D) 0

(x) Which of the following is a linear polynomial ?

(A) x + 5

(B) x²+ 5

(D) x⁴ + 5

Answer:

(i)
In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

In the expression $\sqrt{2} x^{2} + \frac{1}{2}$ , the power of the variable x is 2, which is a whole number. So, the expression $\sqrt{2} x^{2} + \frac{1}{2}$ is a polynomial.

Hence, the correct answer is option (D).

(ii)
$\sqrt{7} = \sqrt{7} \times 1 = \sqrt{7} x^{0}$

The degree of the polynomial $\sqrt{7}$ is 0.

Hence, the correct answer is option (D).

(iii)
The degree of 0 polynomial is not defined.

Hence, the correct answer is option (C).

(iv)
The highest power of the variable in a polynomial is called the degree of the polynomial.

The degree of the polynomial 2x²+ 5x³+ 7 is 3.

Hence, the correct answer is option (A).

(v)
$x^{3} - 1 = x^{3} + 0 x^{2} + 0 x - 1$

The coefficient form of the polynomial $x^{3} - 1$ is (1, 0, 0, −1).

Hence, the correct answer is option (C).

(vi)
$p (x) = x^{2} - 7 \sqrt{7} x + 3$

$∴ p (7 \sqrt{7}) = {(7 \sqrt{7})}^{2} - 7 \sqrt{7} \times 7 \sqrt{7} + 3 = 343 - 343 + 3 = 3$

Hence, the correct answer is option (A).

(vii)
Let p(x) = 2x³+ 2x

∴ p(−1) = 2 × (−1)³ + 2 × (−1) = 2 × (−1) − 2 = −2 − 2 = −4

Thus, the value of the polynomial when x = −1 is −4.

Hence, the correct answer is option (D).

(viii)
Let p(x) = 3x²+ mx.

(x − 1) is a factor of p(x).

∴ p(1) = 0

⇒ 3 × (1)² + m × 1 = 0

⇒ 3 + m = 0

⇒ m = −3

Hence, the correct answer is option (C).

(ix)
$(x^{2} - 3) (2 x - 7 x^{3} + 4) = x^{2} (2 x - 7 x^{3} + 4) - 3 (2 x - 7 x^{3} + 4) = 2 x^{3} - 7 x^{5} + 4 x^{2} - 6 x + 21 x^{3} - 12 = - 7 x^{5} + 2 x^{3} + 21 x^{3} + 4 x^{2} - 6 x - 12 = - 7 x^{5} + 23 x^{3} + 4 x^{2} - 6 x - 12$
Thus, the degree of the resultant polynomial is 5.

Hence, the correct answer is option (A).

(x)
A polynomial with degree one is called a linear polynomial.

Thus, the polynomial x + 5 is a linear polynomial.

Hence, the correct answer is option (A).

Page No 56:

Question 2:

Write the degree of the polynomial for each of the following.

(i) 5 + 3x⁴ (ii) 7 (iii) ax⁷+ bx⁹( a, b are constants.)

Answer:

The highest power of the variable in a polynomial in one variable is called the degree of the polynomial.

(i)
The degree of the polynomial 5 + 3x⁴is 4.

(ii)
7 = 7 × 1 = 7x⁰

The degree of the constant polynomial 7 is 0.

(iii)
The degree of the polynomial ax⁷+ bx⁹ is 9.

Page No 56:

Question 3:

Write the following polynomials in standard form.

(i) 4x² + 7x⁴

-

x³

-

x + 9

(ii) p + 2 p³ + 10 p² + 5 p⁴

-

Answer:

A polynomial written in either descending or ascending power of its variable is called the standard form of the polynomial.

(i)
The given polynomial is 4x² + 7x⁴− x³− x + 9.
The standard form of the polynomial is 7x⁴− x³ + 4x² − x + 9 or 9 − x + 4x² − x³ + 7x⁴.

(ii)
The given polynomial is p + 2p³ + 10p² + 5p⁴ − 8.

The standard form of the polynomial is 5p⁴+ 2p³ + 10p² + p − 8 or −8 + p + 10p² + 2p³+5p⁴.

Page No 56:

Question 4:

Write the following polynomial in coefficient form.

(i) x⁴ + 16 (ii) m⁵+ 2m² + 3m + 15

Answer:

(i)
$x^{4} + 16 = x^{4} + 0 x^{3} + 0 x^{2} + 0 x + 16$

Therefore, the given polynomial in coefficient form is (1, 0, 0, 0, 16).

(ii)
$m^{5} + 2 m^{2} + 3 m + 15 = m^{5} + 0 m^{4} + 0 m^{3} + 2 m^{2} + 3 m + 15$

Therefore, the given polynomial in coefficient form is (1, 0, 0, 2, 3, 15).

Page No 56:

Question 5:

Write the index form of the polynomial using variable x from its coefficient form.

(i) (3,

-

2, 0, 7, 18)

(ii) (6, 1, 0, 7)

(iii) (4, 5,

-

3, 0)

Answer:

(i)
The coefficient form of the polynomial is (3, −2, 0, 7, 18).

Therefore, the index form the polynomial is 3x⁴ − 2x³+ 0x² + 7x + 18 or 3x⁴ − 2x³+ 7x + 18.

(ii)
The coefficient form of the polynomial is (6, 1, 0, 7).

Therefore, the index form the polynomial is 6x³+ x² + 0x + 7 or 6x³+ x² + 7.

(iii)
The coefficient form of the polynomial is (4, 5, −3, 0).

Therefore, the index form the polynomial is 4x³+ 5x² − 3x + 0 or 4x³+ 5x² − 3x.

Page No 56:

Question 6:

Add the following polynomials.

(i) 7x⁴

-

2x³+ x + 10 ; 3x⁴ + 15x³ + 9x²

-

8x + 2

(ii) 3p³q+ 2p²q + 7 ; 2p²q + 4pq

-

2p³q

Answer:

(i)
$(7 x^{4} - 2 x^{3} + x + 10) + (3 x^{4} + 15 x^{3} + 9 x^{2} - 8 x + 2) = 7 x^{4} + 3 x^{4} - 2 x^{3} + 15 x^{3} + 9 x^{2} + x - 8 x + 10 + 2 = 10 x^{4} + 13 x^{3} + 9 x^{2} - 7 x + 12$
(ii)
$(3 p^{3} q + 2 p^{2} q + 7) + (2 p^{2} q + 4 p q - 2 p^{3} q) = 3 p^{3} q - 2 p^{3} q + 2 p^{2} q + 2 p^{2} q + 4 p q + 7 = p^{3} q + 4 p^{2} q + 4 p q + 7$

Page No 56:

Question 7:

Subtract the second polynomial from the first.

(i) 5x²

-

2y + 9 ; 3x² + 5y

-

7 (ii) 2x²+ 3x + 5 ; x ²

-

2x + 3

Answer:

(i)
$(5 x^{2} - 2 y + 9) - (3 x^{2} + 5 y - 7) = 5 x^{2} - 2 y + 9 - 3 x^{2} - 5 y + 7 = 5 x^{2} - 3 x^{2} - 2 y - 5 y + 9 + 7 = 2 x^{2} - 7 y + 16$
(ii)
$(2 x^{2} + 3 x + 5) - (x^{2} - 2 x + 3) = 2 x^{2} + 3 x + 5 - x^{2} + 2 x - 3 = 2 x^{2} - x^{2} + 3 x + 2 x + 5 - 3 = x^{2} + 5 x + 2$

Page No 56:

Question 8:

Multiply the following polynomials.

(i) (m³

-

2m + 3)(m⁴

-

2m²+ 3m + 2)

(ii) (5m³

-

2)(m²

-

m + 3)

Answer:

(i)
$(m^{3} - 2 m + 3) (m^{4} - 2 m^{2} + 3 m + 2) = m^{3} (m^{4} - 2 m^{2} + 3 m + 2) - 2 m (m^{4} - 2 m^{2} + 3 m + 2) + 3 (m^{4} - 2 m^{2} + 3 m + 2) = m^{7} - 2 m^{5} + 3 m^{4} + 2 m^{3} - 2 m^{5} + 4 m^{3} - 6 m^{2} - 4 m + 3 m^{4} - 6 m^{2} + 9 m + 6 = m^{7} - 2 m^{5} - 2 m^{5} + 3 m^{4} + 3 m^{4} + 2 m^{3} + 4 m^{3} - 6 m^{2} - 6 m^{2} - 4 m + 9 m + 6 = m^{7} - 4 m^{5} + 6 m^{4} + 6 m^{3} - 12 m^{2} + 5 m + 6$
(ii)
$(5 m^{3} - 2) (m^{2} - m + 3) = 5 m^{3} (m^{2} - m + 3) - 2 (m^{2} - m + 3) = 5 m^{5} - 5 m^{4} + 15 m^{3} - 2 m^{2} + 2 m - 6$

Page No 56:

Question 9:

Divide polynomial 3x³

-

8x² + x + 7 by x

-

3 using synthetic method and write the quotient and remainder.

Answer:

Dividend = $3 x^{3} - 8 x^{2} + x + 7$

Divisor = $x - 3$

Opposite of −3 = 3

The coefficient form of the quotient is (3, 1, 4).

∴ Quotient = 3x²+ x + 4 and Remainder = 19

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Question 10:

For which the value of m, x + 3 is the factor of the polynomial x³ $-$ 2mx + 21?

Answer:

Let $p (x) = x^{3} - 2 m x + 21$ .

(x + 3) is the factor of the polynomial $p (x) = x^{3} - 2 m x + 21$ .

$∴ p (- 3) = 0 \Rightarrow {(- 3)}^{3} - 2 m \times (- 3) + 21 = 0 \Rightarrow - 27 + 6 m + 21 = 0 \Rightarrow 6 m - 6 = 0 \Rightarrow 6 m = 6 \Rightarrow m = 1$
Thus, the value of m is 1.

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Question 11:

At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x² $-$ 3 y² , 7 y²+ 2 xy and 9 x²+ 4 xy respectively. At the beginning of the year 2017, x²+ xy $-$ y² , 5 xy and 3 x² + xy persons from each of the three villages respectively went to another village for education then what is the remaining total population of these three villages ?

Answer:

Total population of the three villages
$= (5 x^{2} - 3 y^{2}) + (7 y^{2} + 2 x y) + (9 x^{2} + 4 x y) = 5 x^{2} + 9 x^{2} - 3 y^{2} + 7 y^{2} + 2 x y + 4 x y = 14 x^{2} + 4 y^{2} + 6 x y$
Total number of persons who went to another village for education
$= (x^{2} + x y - y^{2}) + 5 x y + (3 x^{2} + x y) = x^{2} + 3 x^{2} - y^{2} + x y + 5 x y + x y = 4 x^{2} - y^{2} + 7 x y$
∴ Remaining total population of the three villages = Total population of the three villages − Total number of persons who went to another village for education
$= (14 x^{2} + 4 y^{2} + 6 x y) - (4 x^{2} - y^{2} + 7 x y) = 14 x^{2} + 4 y^{2} + 6 x y - 4 x^{2} + y^{2} - 7 x y = 14 x^{2} - 4 x^{2} + 4 y^{2} + y^{2} + 6 x y - 7 x y = 10 x^{2} + 5 y^{2} - x y$
Thus, the remaining total population of these three villages is 10x² + 5y² − xy.

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Question 12:

Polynomials bx²+ x + 5 and bx³ $-$ 2x + 5 are divided by polynomial x $-$ 3 and the remainders are m and n respectively. If m $-$ n = 0 then find the value of b.

Answer:

Let $p (x) = b x^{2} + x + 5$ and $q (x) = b x^{3} - 2 x + 5$ .

The remainder when $p (x) = b x^{2} + x + 5$ is divided by (x − 3) is m.

By remainder theorem,

Remainder = $p (3) = m$

$∴ b \times {(3)}^{2} + 3 + 5 = m \Rightarrow m = 9 b + 8 . . . . . (1)$
The remainder when $q (x) = b x^{3} - 2 x + 5$ is divided by (x − 3) is n.

By remainder theorem,

Remainder = $q (3) = n$

$∴ b \times {(3)}^{3} - 2 \times 3 + 5 = n \Rightarrow n = 27 b - 6 + 5 \Rightarrow n = 27 b - 1 . . . . . (2)$
Now,
$m - n = 0 \Rightarrow (9 b + 8) - (27 b - 1) = 0 [Using (1) and (2)] \Rightarrow 9 b - 27 b + 8 + 1 = 0 \Rightarrow - 18 b + 9 = 0 \Rightarrow - 18 b = - 9 \Rightarrow b = \frac{- 9}{- 18} = \frac{1}{2}$
Thus, the value of b is $\frac{1}{2}$ .

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Question 13:

Simplify

(8m² + 3m $-$ 6) $-$ (9m $-$ 7) + (3m² $-$ 2m + 4)

Answer:

$(8 m^{2} + 3 m - 6) - (9 m - 7) + (3 m^{2} - 2 m + 4) = 8 m^{2} + 3 m - 6 - 9 m + 7 + 3 m^{2} - 2 m + 4 = 8 m^{2} + 3 m^{2} + 3 m - 9 m - 2 m - 6 + 7 + 4 = 11 m^{2} - 8 m + 5$

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Question 14:

Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x²+ 5x - 4?

Answer:

Let p(x) be the polynomial which is to be subtracted from x² + 13x + 7 to get the polynomial 3x²+ 5x − 4.

∴ (x² + 13x + 7) − p(x) = 3x²+ 5x − 4

⇒ p(x) = (x² + 13x + 7) − (3x²+ 5x − 4)

⇒ p(x) = x² + 13x + 7 − 3x²− 5x + 4

⇒ p(x) = x² − 3x²+ 13x − 5x + 7 + 4

⇒ p(x) = −2x²+ 8x + 11

Thus, the required polynomial is −2x²+ 8x + 11.

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Question 15:

Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?

Answer:

The required polynomial can be obtained by subtracting the polynomial 4m + 2n + 3 from 6m + 3n + 10.

∴ Required polynomial

= (6m + 3n + 10) − (4m + 2n + 3)

= 6m + 3n + 10 − 4m − 2n − 3

= 6m − 4m + 3n − 2n + 10 − 3

= 2m + n + 7

Thus, the polynomial 2m + n + 7 is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10.

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