EXERCISE 2A

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question 1.
x² + 7x + 12
Solution:

Question 2.
x² – 2x – 8
Solution:
x² – 2x – 8
Let f(x) = x² – 2x – 8

Question 3.
x² + 3x – 10
Solution:

Question 4.
4x² – 4x – 3 [CBSE 2008C]
Solution:
4x² – 4x – 3

Question 5.
5x² – 4 – 8x [CBSE 2008]
Solution:

Question 6.
2√3 x² – 5x + √3 [CBSE2011]
Solution:

Question 7.
2x² – 11x + 15
Solution:

Question 8.
4x² – 4x + 1
Solution:

Question 9.
x² – 5
Solution:

Question 10.
8x² – 4
Solution:

Question 11.
5y² + 10y
Solution:

Question 12.
3x² – x – 4
Solution:

Question 13.
Find the quadratic polynomial whose zeros are 2 and -6. Verify the relation between the coefficients and the zeros of the polynomial.
Solution:
Zeros of a quadratic polynomial are 2, -6

Question 14.
Find the quadratic polynomial whose zeros are and . Verify the relation between the coefficients and the zeros of the polynomial.
Solution:

Question 15.
Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial. [CBSE2008]
Solution:
Sum of zeros = 8
Product of zeros = 12
Quadratic equation will be x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 8x + 12 = 0
=> x² – 6x – 2x + 12 = 0
=> x (x – 6) – 2 (x – 6) = 0
=> (x – 6) (x – 2) = 0
Either x – 6 = 0, then x = 6
or x – 2 = 0, then x = 2
Zeros are 6, 2
and quadratic polynomial is x² – 8x + 12

Question 16.
Find the quadratic polynomial, the sum of whose zeros is 0 and their product is -1. Hence, find the zeros of the polynomial.
Solution:
Sum of zeros = 0
and product of zeros = -1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 0x – 1 = 0
=> x² – 1= 0
(x + 1)(x – 1) = 0
Either x + 1 = 0, then x = -1 or x – 1 =0, then x = 1
Zeros are 1, -1
and quadratic polynomial is x² – 1

Question 17.
Find the quadratic polynomial, the sum of whose zeros is and their product is 1. Hence, find the zeros of the polynomial.
Solution:
Sum of zeros =
Product of zeros = 1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0

and quadratic polynomial is 2x² – 5x + 2

Question 18.
Find the quadratic polynomial, the sum of whose roots is √2 and their product is
Solution:

Question 19.
If x = and x = -3 are the roots of the quadratic equation ax² + 7x + b = 0 then find the values of a and b. [CBSE2011]
Solution:

Question 20.
If (x + a) is a factor of the polynomial 2x² + 2ax + 5x + 10, find the value of a. [CBSE 2009]
Solution:

Question 21.
One zero of the polynomial 3x³ + 16x² + 15x – 18 is . Find the other zeros of the polynomial.
Solution:
One zero of the given polynomial is

=> (x + 3) (x + 3) = 0
x = -3, -3
Hence, other zeros are -3, -3

EXERCISE 2B

Question 1.
Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = x³ – 2x² – 5x + 6 and verify the relation between its zeros and coefficients.
Solution:
Zeros are 3, -2, 1 of
p(x) = x³ – 2x² – 5x + 6
Here, a = 1, b = -2, c = -5, d = 6
We know that if α, β and γ are the roots of
f(x) = ax³ + bx² + cx + d, then

Question 2.
Verify that 5, -2 and are the zeros of the cubic polynomial p(x) = 3x³ – 10x² – 27x + 10 and verify the relation between its zeros and coefficients.
Solution:
Zeros are 5, -2 and of

Question 3.
Find a cubic polynomial whose zeros are 2, -3 and 4.
Solution:

Question 4.
Find a cubic polynomial whose zeros are , 1 and -3.
Solution:

Question 5.
Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as 5, -2 and -24 respectively.
Solution:

Find the quotient and the remainder when:
Question 6.
f(x) = x³ – 3x² + 5x – 3 is divided by g(x) = x² – 2
Solution:

Question 7.
f(x) = x⁴ – 3x² + 4x + 5 is divided by g(x) = x² + 1 – x.
Solution:

Question 8.
f(x) = x⁴ – 5x + 6 is divided by g(x) = 2 – x².
Solution:

Question 9.
By actual division, show that x² – 3 is a factor of 2x⁴ + 3x³ – 2x² – 9x – 12.
Solution:
f(x) = 2x⁴ + 3x³ – 2x² – 9x – 12
g(x) = x² – 3
Quotient [q(x)] = 2x² + 3x + 4
Remainder [r(x)] = 0
Remainder is zero.
x² – 3 is a factor of f(x)

Question 10.
On dividing 3x³ + x² + 2x + 5 by a polynomial g(x), the quotient and remainder are (3x – 5) and (9x +10) respectively. Find g(x).
Solution:

Question 11.
Verify division algorithm for the polynomials f(x) = 8 + 20x + x² – 6x³ and g(x) = 2 + 5x – 3x².
Solution:

Question 12.
It is given that -1 is one of the zeros of the polynomial x³ + 2x² – 11x – 12. Find all the zeros of the given polynomial.
Solution:

= (x + 1) (x + 4) (x – 3)
If x + 4 = 0, then x = -4
If x – 3 = 0, then x = 3
Zeros are -1, -4, 3

Question 13.
If 1 and -2 are two zeros of the polynomial (x³ – 4x² – 7x + 10), find its third zero.
Solution:

Question 14.
If 3 and -3 are two zeros of the polynomial (x⁴ + x³ – 11x² – 9x + 18), find all the zeros of the given polynomial.
Solution:

=> (x – 3) (x + 3) [x (x + 2) – 1 (x + 2)]
=> (x – 3) (x + 3) (x + 2) (x – 1)
Other zeros will be
If x + 2 = 0 then x = -2
and if x – 1 =0, then x = 1
Zeros are 3, -3, -2, 1

Question 15.
If 2 and -2 are two zeros of the polynomial (x⁴ + x³ – 34x² – 4x + 120), find all the zeros of the given polynomial. [CBSE 2008]
Solution:
2 and -2 are the two zeros of the polynomial
f(x) = x⁴ + x³ – 34x² – 4x + 120,
Then (x – 2) (x + 2) or x² – 4 will its the factor of f(x)
Now dividing f(x) by x² – 4, we get

f(x) = (x – 2) (x + 2) (x² + x – 30)
= (x – 2)(x + 2)[x² + 6x – 5x – 30]
= (x – 2) (x + 2)[x(x + 6) – 5(x + 6)]
= (x – 2) (x + 2) (x + 6) (x – 5)
Other two zeros are
If x + 6 = 0, then x = -6 and
if x – 5 = 0, then x = 5
Roots of f(x) are 2, -2, -6, 5

Question 16.
Find all the zeros of (x⁴ + x³ – 23x² – 3x + 60), if it is given that two of its zeros are √3 and -√3. [CBSE2009C]
Solution:

Other two zeros are :
if x + 5 = 0, then x = -5
and if x – 4 = 0, then x = 4
Hence, all the zeros of f(x) are : √3, – √3, 4, -5

Question 17.
Find all the zeros of (2x⁴ – 3x³ – 5x² + 9x – 3), it being given that two of its zeros are √3 and – √3
Solution:
√3 and – √3 are the zeros of the polynomial
f(x) = 2x⁴ – 3x³ – 5x² + 9x – 3
=> (x – √3) (x + √3) or (x² – 3) is a factor of f(x)
Now, dividing f(x) by x² – 3, we get

Question 18.
Obtain all other zeros of (x⁴ + 4x³ – 2x² – 20x – 15) if two of its zeros are √5 and – √5 [CBSE2009C]
Solution:

Question 19.
Find all the zeros of the polynomial (2x⁴ – 11x³ + 7x² + 13x – 7), it being given that two of its zeros are (3 + √2)and (3 – √2 )
Solution:

EXERCISE 2C

Very-Short-Answer Questions
Question 1.
If one zero of the polynomial x² – 4x + 1 is (2 + √3), write the other zero. [CBSE2010]
Solution:
Let other zero of x² – 4x + 1 be a, then
Sum of zeros = = = 4
But one zero is 2 + √3
Second zero = 4 – (2 + √3) =4 – 2 – √3 = 2 – √3

Question 2.
Find the zeros of the polynomial x² + x – p(p + 1). [CBSE2011]
Solution:
Let f(x) = x² + x – p(p + 1)
= x² + (p + 1) x – px – p(p + 1)
= x(x + p + 1) – p(x + p + 1)
= (x + p + 1) (x – p)
Either x + p + 1 = 0, then x = -(p + 1)
or x – p = 0, then x = p
Hence, zeros are p and -(p + 1)

Question 3.
Find the zeros of the polynomial x² – 3x – m(m + 3). [CBSE2011]
Solution:
p(x) = x² – 3x – m(m + 3)
= x² – (m + 3)x + mx – m(m + 3)
= x(x – m – 3) + m(x – m – 3)
= (x – m – 3)(x + m)
Either x – m – 3 = 0, then x = m + 3
or x + m = 0, then x = -m
Zeros are (m + 3), -m

Question 4.
If α, β are the zeros of a polynomial such that α + β = 6 and αβ = 4 then write the polynomial. [CBSE2010]
Solution:
a and p are the zeros of a polynomial
and α + β = 6, αβ = 4
Polynomial = x² – (α + β)x + αβ = x² – (6)x + 4 = x² – 6x + 4

Question 5.
If one zero of the quadratic polynomial kx² + 3x + k is 2 then find the value of k.
Solution:
One zero of kx² + 3x + k is 2
x = 2 will satisfy it
⇒ k(2)² + 3 x 2 + k = 0
⇒ 4k + 6 + k= 0
⇒5k + 6 = 0
⇒ 5k = -6
⇒ k =
Hence, k =

Question 6.
If 3 is a zero of the polynomial 2x² + x + k, find the value of k. [CBSE2010]
Solution:
3 is a zero of the polynomial 2x² + x + k
Then 3 will satisfy it
2x² + x + k = 0
⇒ 2(3)² + 3 + k = 0
⇒ 18 + 3+ k = 0
⇒ 21 + k = 0
⇒ k = -21
Hence, k = -21

Question 7.
If -4 is a zero of the polynomial x² – x – (2k + 2) then find the value of k. [CBSE 2009]
Solution:
-4 is a zero of polynomial x² – x – (2k + 2)
Then it will satisfy the equation
x² – x – (2k + 2) = 0
⇒ (-4)2 – (-4) – 2k – 2 = 0
⇒ 16 + 4 – 2k – 2 = 0
⇒ -2k + 18 = 0
⇒ 2k = 18
k = 9

Question 8.
If 1 is a zero of the polynomial ax² – 3(a – 1) x – 1 then find the value of a.
Solution:
1 is a zero of the polynomial ax² – 3(a – 1)x – 1
Then 1 will satisfy the equation ax² – 3(a – 1) x – 1 = 0
a(1)² – 3(a – 1) x 1 – 1 = 0
⇒ a x 1 – 3a + 3 – 1 = 0
⇒ a – 3a + 2 = 0
⇒ -2a + 2 = 0
⇒ 2a = 2
⇒ a = 1

Question 9.
If -2 is a zero of the polynomial 3x² + 4x + 2k then find the value of k. [CBSE 2010]
Solution:
-2 is a zero of 3x² + 4x + 2k
It will satisfy the equation 3x² + 4x + 2k = 5
3(-2)² + 4(-2) + 2k = 0
⇒ 3 x 4 + 4(-2) + 2k = 0
⇒ 12 – 8 + 2k = 0
⇒ 4 + 2k=0
⇒ 2k = -4
⇒ k = -2
k = -2

Question 10.
Write the zeros of the polynomial x² – x – 6. [CBSE 2008]
Solution:
Let f(x) = x² – x – 6
= x² – 3x + 2x – 6
= x(x – 3) + 2(x – 3)
= (x – 3)(x + 2)
(x – 3)(x + 2) = 0
Either x – 3 = 0, then x = 3
or x + 2 = 0, then x = -2
Zeros are 3, -2

Question 11.
If the sum of the zeros of the quadratic polynomial kx² – 3x + 5 is 1, write the value of k.
Solution:
Sum of zeros = 1
and polynomial is kx² – 3x + 5
Sum of zeros = = =
= 1
⇒ k = 3
Hence, k = 3

Question 12.
If the product of the zeros of the quadratic polynomial x² – 4x + k is 3 then write the value of k.
Solution:
Product of zeros of polynomial x² – 4x + k is 3
Product of zeros =
⇒ = 3
⇒ k = 3

Question 13.
If (x + a) is a factor of (2x² + 2ax + 5x + 10), find the value of a. [CBSE 2010]
Solution:
x + a is a factor of
f(x) = 2x² + (2a + 5) x + 10
Let x + a = 0, then
Zero of f(x) = -a
Now f(-a) = 2 (-a)² + (2a + 5)(-a) + 10 = 0
2a² – 2a² – 5a + 10 = 0
⇒ 5a = 10
⇒ a = 2

Question 14.
If (a – b), a and (a + b) are zeros of the polynomial 2x³ – 6x² + 5x – 7, write the value of a.
Solution:
(a – b), a, (a + b) are the zeros of 2x³ – 6x² + 5x – 7
Sum of zeros =
⇒ a – b + a + a + b =
⇒ 3a =
⇒ 3a = 3
⇒ a = 1

Question 15.
If x³ + x² – ax + b is divisible by (x² – x), write the values of a and b.
Solution:
f(x) = x³ + x² – ax + 6 is divisible by x² – x
Remainder will be zero
Now dividing f(x) by x² – x
Remainder = (2 – a) x + b

(2 – a) x + b = 0
2 – a = 0
⇒ a = 2 and b = 0
Hence, a = 2, b = 0

Question 16.
If α and β are the zeros of the polynomial 2x² + 7x + 5, write the value of α + β + αβ. [CBSE 2010]
Solution:
α and β are the zeros of polynomial f(x) = 2x² + 7x + 5

Question 17.
State division algorithm for polynomials.
Solution:
Division algorithm for polynomials:
If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomial q(x) and r(x).
f(x) = q(x) x g(x) + r(x)
where r (x) = 0
or [degree of r(x) < degree of g(x)]
or Dividend=Quotient x Division + Remainder

Question 18.
The sum of the zeros and the product of zeros of a quadratic polynomial are and -3 respectively. Write the polynomial.
Solution:
Sum of zeros =
Product of zeros = -3
Polynomial: x² – (Sum of zeros) x + product of zeros

Short-Answer Questions
Question 19.
Write the zeros of the quadratic polynomial f(x) = 6x² – 3.
Solution:

Question 20.
Write the zeros of the quadratic polynomial f(x) = 4√3 x² + 5x – 2√3.
Solution:

Question 21.
If α and β are the zeros of the polynomial f(x) = x² – 5x + k such that α – β = 1, find the value of k.
Solution:
α and β are the zeros of polynomial f(x) = x² – 5x + k

(1)² = (5)² – 4 k
1 ⇒ 25 – 4k
⇒ 4k = 25 – 1 = 24
Hence, k = 6

Question 22.
If α and β are the zeros of the polynomial f(x) = 6x² + x – 2, find the value of
Solution:

Question 23.
If α and β are the zeros of the polynomial f(x) = 5x² – 7x + 1, find the value of
Solution:
α and β are the zeros of polynomial
f(x) = 5x² – 7x + 1

Question 24.
If α and β are the zeros of the polynomial f(x) = x² + x – 2, find the value of
Solution:

Question 25.
If the zeros of the polynomial f(x) = x³ – 3x² + x + 1 are (a – b), a and (a + b), find a and b.
Solution:
(a – b), a and (a + b) are the zeros of the polynomial
f(x) = x³ – 3x² + x + 1

MULTIPLE CHOICE QUESTIONS

Choose the correct answer in each of the following questions.
Question 1.
Which of the following is a polynomial?

Solution:
(d) √2 x² – 3√3 x + √6 is polynomial, others are not polynomial.

Question 2.
Which of the following is not a polynomial?
(a) √3 x² – 2√3 x + 5
(b) 9x² – 4x + √2

Solution:
(d) x + is not a polynomial, other are polynomial.

Question 3.
The zeros of the polynomial x² – 2x – 3 are
(a) -3, 1
(b) -3, -1
(c) 3, -1
(d) 3, 1
Solution:
(c) Let f(x) = x² – 2x – 3
= x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3)(x + 1)
If x – 3 = 0, then x – 3
and if x + 1 = 0, then x = -1
Zeros are 3, -1

Question 4.
The zeros of the polynomial x² – √2 x – 12 are
(a) √2, – √2
(b) 3√2, -2√2
(c) -3√2, 2√2
(d) 3√2, 2√2
Solution:
(b)

Question 5.
The zeros of the polynomial 4x² + 5√2x – 3 are

Solution:
(c)

Question 6.
The zeros of the polynomial x² + x – 2 are
(a) -3, 4
(b) ,
(c) ,
(d) none of these
Solution:
(b) Polynomial is x² + x – 2

Question 7.

Solution:
(a)

Question 8.
The sum and the product of the zeros of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is
(a) x² – 3x + 10
(b) x² + 3x – 10
(c) x² – 3x – 10
(d) x² + 3x + 10
Solution:
(c) Sum of zeros = 3
Product of zeros = -10
Polynomial : x² – (Sum of zeros) x + Product of zeros
= x² – 3x – 10

Question 9.
A quadratic polynomial whose zeros are 5 and -3, is
(a) x² + 2x – 15
(b) x² – 2x + 15
(c) x² – 2x – 15
(d) none of these
Solution:
(c) Zeros are 5 and -3
Sum of zeros = 5 – 3 = 2
Product of zeros = 5 x (-3) = -15
Polynomial: x² – (Sum of zeros) x + Product of zeros
= x² – 2x – 15

Question 10.
A quadratic polynomial whose zeros are and , is
(a) 10x² + x + 3
(b) 10x² + x – 3
(c) 10x² – x + 3
(d) 10x² – x – 3
Solution:
(d)

Question 11.
The zeros of the quadratic polynomial x² + 88x + 125 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
Solution:
(b) Let f(x) = x² + 88x +125
Here, sum of roots = = -88
and product = = 125
Product is positive,
Both zeros can be both positive or both negative.
Sum is negative.
Both zeros are negative.

Question 12.
If α and β are the zeros of x² + 5x + 8 then the value of (α + β) is
(a) 5
(b) -5
(c) 8
(d) -8
Solution:
(b) α and β are the zeros of x² + 5x + 8
Then sum of zeros (α + β) = = = -5

Question 13.
If α and β are the zeros of 2x² + 5x – 9 then the value of αβ is
(a)
(b)
(c)
(d)
Solution:
(c) α and β are the zeros of 2x² + 5x – 9
Product of zeros (αβ) = =

Question 14.
If one zero of the quadratic polynomial kx² + 3x + k is 2 then the value of k is
(a)
(b)
(c)
(d)
Solution:
(d) 2 is a zero of kx² + 3x + k
It will satisfy the quadratic equation kx² + 3x + k = 0
k(2)² + 3x² + 1 = 0
4k + 6 + k = 0
=> 5k = -6
k =

Question 15.
If one zero of the quadratic polynomial (k – 1) x² + kx + 1 is -4 then the value of k is
(a)
(b)
(c)
(d)
Solution:
(b) -4 is a zero of (k – 1) x² + 4x + 1
-4 will satisfy the equation (k – 1) x² + kx + 1 = 0
=> (k – 1)(-4)² + k(-4) + 1 =0
=> 16k – 16 – 4k + 1 = 0
=> 12k – 15 = 0
=> 12k = 15
=> k = =

Question 16.
If -2 and 3 are the zeros of the quadratic polynomial x² + (a + 1)x + b then
(a) a = -2, b = 6
(b) a = 2, b = -6
(c) a = -2,b = -6
(d) a = 2, b = 6
Solution:
(c)

Question 17.
If one zero of 3x² + 8x + k be the reciprocal of the other then k = ?
(a) 3
(b) -3
(c)
(d)
Solution:
(a)

Question 18.
If the sum of the zeros of the quadratic polynomial kx² + 2x + 3k is equal to the product of its zeros then k = ?
(a)
(b)
(c)
(d)
Solution:
(d) Polynomial: kx² + 2x + 3k

Question 19.
If α, β are the zeros of the polynomial x² + 6x + 2 then
(a) 3
(b) -3
(c) 12
(d) -12
Solution:
(b) α, β are the zeros of the polynomial x² + 6x + 2

Question 20.
If α, β, γ are the zeros of the polynomial x³ – 6x² – x + 30 then (αβ + βγ + γα) = ?
(a) -1
(b) 1
(c) -5
(d) 30
Solution:
(a) α, β, γ are the zeros of x³ – 6x² – x + 30
Then αβ + βγ + γα = = = -1

Question 21.
If α, β, γ are the zeros of the polynomial 2x³ + x² – 13x + 6 then αβγ = ?
(a) -3
(b) 3
(c)
(d)
Solution:
(a) α, β, γ are the zeros of 2x³ + x² – 13x + 6, then
αβγ = = = -3

Question 22.
If α, β, γ be the zeros of the polynomial p(x) such that (α + β + γ) = 3, (αβ + βγ + γα) = -10 and αβγ = -24 then p(x) = ?
(a) x³ + 3x² – 10x + 24
(b) x³ + 3x² + 10x – 24
(c) x³ – 3x² – 10x + 24
(d) None of these
Solution:
(c) α, β, γ are the zeros of p(x) such that

Question 23.
If two of the zeros of the cubic polynomial ax³ + bx² + cx + d are 0 then the third zero is
(a)
(b)
(c)
(d)
Solution:
(a)

Question 24.
If one of the zeros of the cubic polynomial ax³ + bx² + cx + d is 0 then the product of the other two zeros is
(a)
(b)
(c) 0
(d)
Solution:
(b) If one zero of cubic polynomial ax³ + bx² + cx + d = 0
Let a be zero, then

Question 25.
If one of the zeros of the cubic polynomial x³ + ax² + bx + c is -1 then the product of the other two zeros is
(a) a – b – 1
(b) b – a – 1
(c) 1 – a + b
(d) 1 + a – b
Solution:
(c)

Question 26.

(a) 3
(b) -3
(c) -2
(d) 2
Solution:
(d)

Question 27.
On dividing a polynomial p(x) by a nonzero polynomial q(x), let g(x) be the quotient and r(x) be the remainder then p(x) = q(x) x g(x) + r(x), where
(a) r(x) = 0 always
(b) deg r(x) < deg g(x) always
(c) either r(x) = 0 or deg r(x) < deg g(x)
(d) r(x) = g(x)
Solution:
(c) p(x) is divided by q(x), then
p(x) = q(x) x g(x) + r(x)
Either r(x) = 0
Degree of r(x) < deg of g(x)

Question 28.
Which of the following is a true statement?
(a) x² + 5x – 3 is a linear polynomial.
(b) x² + 4x – 1 is a binomial.
(c) x + 1 is a monomial.
(d) 5x² is a monomial.
Solution:
(d) (a) is not a linear polynomial.
(b) is trinomial not binomial.
(c) is not a monomial.
(d) 5x² is monomial is true.

TEST YOURSELF

Question 1.
Zeros of p(x) = x² – 2x – 3 are
(a) 1, -3
(b) 3, -1
(c) -3, -1
(d) 1, 3
Solution:
(b) Polynomial is x² – 2x – 3
=> x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
Either x – 3 = 0, then x = 3
or x + 1 = 0, then x = -1
Zeros are 3, -1

Question 2.
If α, β, γ are the zeros of the polynomial x³ – 6x² – x + 30 then the value of (αβ + βγ + γα) is
(a) -1
(b) 1
(c) -5
(d) 30
Solution:
(a) α, β, γ are the zeros of Polynomial is x³ – 6x² – x + 30
Here, a = 1, b = -6, c = -1, d = 30
αβ + βγ + γα = = = -1

Question 3.
If α, β are the zeros of kx² – 2x + 3k such that α + β = αβ then k = ?
(a)
(b)
(c)
(d)
Solution:
(c)

Question 4.
It is given that the difference between the zeros of 4x² – 8kx + 9 is 4 and k > 0. Then, k = ?
(a)
(b)
(c)
(d)
Solution:
(c) Let α and β be the zeros of the polynomial 4x² – 8kx + 9

Short-Answer Questions
Question 5.
Find the zeros of the polynomial x² + 2x – 195.
Solution:

Either x + 15 = 0, then x = -15
or x – 13 = 0, then x = 13
Zeros are 13, -15

Question 6.
If one zero of the polynomial (a² + 9) x² + 13x + 6a is the reciprocal of the other, find the value of a.
Solution:
The polynomial is (a² + 9) x² + 13x + 6a

Question 7.
Find a quadratic polynomial whose zeros are 2 and -5.
Solution:
Zeros are 2 and -5
Sum of zeros = 2 + (-5) = 2 – 5 = -3
and product of zeros = 2 x (-5) = -10
Now polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-3)x + (-10)
= x² + 3x – 10

Question 8.
If the zeros of the polynomial x³ – 3x² + x + 1 are (o -b), a and (a + b), find the values of a and b.
Solution:
(a – b), a, (a + b) are the zeros of the polynomial x³ – 3x² + x + 1
Here, a = 1, b = -3, c = 1, d = 1
Now, sum of zeros = = = 3
a – b + a + a + b = 3

Question 9.
Verify that 2 is a zero of the polynomial x³ + 4x² – 3x – 18.
Solution:
Let f(x) = x³ + 4x² – 3x – 18
If 2 is its zero, then it will satisfy it
Now, (x – 2) is a factor Dividing by (x – 2)
Hence, x = 2 is a zero of f(x)

Question 10.
Find the quadratic polynomial, the sum of whose zeros is -5 and their product is 6.
Solution:
Sum of zeros = -5
and product of zeros = 6
Quadratic polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-5) x + 6
= x² + 5x + 6

Question 11.
Find a cubic polynomial whose zeros are 3, 5 and -2.
Solution:

Question 12.
Using remainder theorem, find the remainder when p(x) = x³ + 3x² – 5x + 4 is divided by (x – 2).
Solution:
p(x) = x³ + 3x² – 5x + 4
g(x) = x – 2
Let x – 2 = 0, then x = 2
Remainder = p(2) = (2)³ + 3(2)² – 5(2) + 4 = 8 + 12 – 10 + 4 = 14

Question 13.
Show that (x + 2) is a factor of f(x) = x³ + 4x² + x – 6.
Solution:
f(x) = x³ + 4x² + x – 6
and g(x) = x + 2
Let x + 2 = 0, then x = -2
f(-2) = (-2)³ + 4(-2)² + (-2) – 6 = -8 + 16 – 2 – 6 = 0
Remainder is zero, x + 2 is a factor of f(x)

Question 14.
If α, β, γ are the zeros of the polynomial p(x) = 6x³ + 3x² – 5x + 1, find the value of
Solution:

Question 15.
If α, β are the zeros of the polynomial f(x) = x² – 5x + k such that α – β = 1, find the value of k.
Solution:

Question 16.
Show that the polynomial f(x) = x⁴ + 4x² + 6 has no zero.
Solution:
f(x) = x⁴ + 4x² + 6
=> (x²)² + 4(x²) + 6 = y² + 4y + 6 (Let x² = y)
Let α, β be the zeros of y² + 4y + 6
Sum of zeros = -4
and product of zeros = 6
But there are no factors of 6 whose sum is -4 {Factors of 6 = 1 x 6 and 2 x 3}
Hence, f(x) Has no zero (real).

Long-Answer Questions
Question 17.
If one zero of the polynomial p(x) = x³ – 6x² + 11x – 6 is 3, find the other two zeros.
Solution:
3 is one zero of p(x) = x³ – 6x² + 11x – 6
(x – 3) is a factor of p(x)
Dividing, we get

Question 18.
If two zeros of the polynomial p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2 are √2 and – √2 , find its other two zeros.
Solution:

Question 19.
Find the quotient when p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2 is divided by (x² + 3x + 1).
Solution:
p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2
Dividing by x² + 3x + 1,
we get,
Quotient = 3x² – 4x + 2

Question 20.
Use remainder theorem to find the value of k, it being given that when x³ + 2x² + kx + 3 is divided by (x – 3), then the remainder is 21.
Solution:
Let p(x) = x³ + 2x² + kx + 3
g(x) = x – 3
and r(x) = 21
Dividing p(x) by g(x), we get

But remainder = 21
3 + 3k + 45 = 21
3k = 21 – 45 – 3
=> 3k = 21 – 48 = -27
k = -9
Second method:
x – 3 is a factor of p(x) : x = 3
Substituting the value of x in p(x)
p(3) = 3³ + 2 x 3² + k x 3 + 3
= 27 + 18 + 3k + 3
48 + 3k = 21
=> 3k = -48 + 21 = -27
k = -9
Hence, k = -9