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Page No 28:

Question 1:

The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

(i)

(ii)

(iii)

(iv)

(v)

(v)

Answer:

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.



Page No 33:

Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer:

The value of is zero when x − 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = −2

Therefore, the zeroes of are 4 and −2.

Sum of zeroes =

Product of zeroes

The value of 4s2 − 4s + 1 is zero when 2s − 1 = 0, i.e.,

Therefore, the zeroes of 4s2 − 4s + 1 areand.

Sum of zeroes =

Product of zeroes

The value of 6x2 − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0, i.e., or

Therefore, the zeroes of 6x2 − 3 − 7x are.

Sum of zeroes =

Product of zeroes =

The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = −2

Therefore, the zeroes of 4u2 + 8u are 0 and −2.

Sum of zeroes =

Product of zeroes =

The value of t2 − 15 is zero when or , i.e., when

Therefore, the zeroes of t2 − 15 are and.

Sum of zeroes =

Product of zeroes =

The value of 3x2x − 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e., when or x = −1

Therefore, the zeroes of 3x2x − 4 are and −1.

Sum of zeroes =

Product of zeroes



Page No 33:

Question 2:

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Answer:

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 4x2x − 4.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 3x2x + 1.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be .

Therefore, the quadratic polynomial is.



Page No 36:

Question 1:

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

Answer:

(i)

Therefore, , 1, and −2 are the zeroes of the given polynomial.

Comparing the given polynomial with , we obtain a = 2, b = 1, c = −5, d = 2

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii)

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

Comparing the given polynomial with , we obtain a = 1, b = −4, c = 5, d = −2.

Verification of the relationship between zeroes and coefficient of the given polynomial

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5

Multiplication of zeroes = 2 × 1 × 1 = 2

Hence, the relationship between the zeroes and the coefficients is verified.



Page No 36:

Question 2:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.

Answer:

Let the polynomial be and the zeroes be .

It is given that

If a = 1, then b = −2, c = −7, d = 14

Hence, the polynomial is .



Page No 36:

Question 3:

Obtain all other zeroes of , if two of its zeroes are .

Answer:

Since the two zeroes are ,

is a factor of .

Therefore, we divide the given polynomial by .

We factorize

Therefore, its zero is given by x + 1 = 0

x = −1

As it has the term , therefore, there will be 2 zeroes at x = −1.

Hence, the zeroes of the given polynomial are, −1 and −1.



Page No 36:

Question 4:

On dividing by a polynomial g(x), the quotient and remainder were x − 2 and − 2x + 4, respectively. Find g(x).

Answer:

g(x) = ? (Divisor)

Quotient = (x − 2)

Remainder = (− 2x + 4)

Dividend = Divisor × Quotient + Remainder

g(x) is the quotient when we divide by




Page No 36:

Question 5:

Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Answer:

According to the division algorithm, if p(x) and g(x) are two polynomials with

g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).

Let us assume the division of by 2.

Here, p(x) =

g(x) = 2

q(x) = and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

= 2()

=

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of x3 + x by x2,

Here, p(x) = x3 + x

g(x) = x2

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x3 + x = (x2 ) × x + x

x3 + x = x3 + x

Thus, the division algorithm is satisfied.

(iii)deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x3 + 1by x2.

Here, p(x) = x3 + 1

g(x) = x2

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x3 + 1 = (x2 ) × x + 1

x3 + 1 = x3 + 1

Thus, the division algorithm is satisfied.



Page No 37:

Question 3:

If the zeroes of polynomial are, find a and b.

Answer:

Zeroes are ab, a + a + b

Comparing the given polynomial with , we obtain

p = 1, q = −3, r = 1, t = 1

The zeroes are .

Hence, a = 1 and b = or .



Page No 37:

Question 4:

]It two zeroes of the polynomial are, find other zeroes.

Answer:

Given that 2 + and 2­­ are zeroes of the given polynomial.

Therefore, = x2 + 4 ­­− 4x − 3

= x2 ­− 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing by x2 ­− 4x + 1.

Clearly, =

It can be observed that is also a factor of the given polynomial.

And =

Therefore, the value of the polynomial is also zero when or

Or x = 7 or −5

Hence, 7 and −5 are also zeroes of this polynomial.



Page No 37:

Question 5:

If the polynomial is divided by another polynomial, the remainder comes out to be x + a, find k and a.

Answer:

By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

will be perfectly divisible by .

Let us divide by

It can be observed that will be 0.

Therefore, = 0 and = 0

For = 0,

2 k =10

And thus, k = 5

For = 0

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5



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