Class 8 Maths Chapter 2 Question Answers - Polynomials

Q1: If the polynomial (x⁴ + 2x³ + 8x² + 12x + 18) is divided by another polynomial (x² + 5), the remainder comes out to be (px + q), find the values of p and q.
Ans:

Remainder = 2x + 3
px + q = 2x + 3
p = 2 and q = 3.

Q2: If a polynomial 3x⁴ – 4x³ – 16x² + 15x + 14 is divided by another polynomial x² – 4, the remainder comes out to be px + q. Find the value of p and q.
Ans:


Q3: Find the values of a and b so that x⁴ + x³ + 8x² +ax – b is divisible by x² + 1.
Ans: 

If x⁴ + x³ + 8x² + ax – b is divisible by x² + 1
Remainder = 0
(a – 1)x – b – 7 = 0
(a – 1)x + (-b – 7) = 0 . x + 0
a – 1 = 0, -b – 7 = 0
a = 1, b = -7
a = 1, b = -7

Q4: What must be subtracted from p(x) = 8x⁴ + 14x³ – 2x² + 8x – 12 so that 4x² + 3x – 2 is factor of p(x)? This question was given to group of students for working together.
Ans: 

Polynomial to be subtracted by (15x – 14).

Q5: If α and β are the zeroes of the polynomial p(x) = 2x² + 5x + k, satisfying the relation, α² + β² + αβ = 21/4 then find the value of k.
Ans: Given polynomial is p(x) = 2x² + 5x + k
Here a = 2, b = 5, c = k


Q6: If α and β are zeroes of p(x) = kx² + 4x + 4, such that α² + β² = 24, find k.
Ans: We have, p(x) = kx² + 4x + 4
Here a = k, b = 4, c = 4

⇒ 24k² = 16 – 8k
⇒ 24k² + 8k – 16 = 0
⇒ 3k² + k – 2 = 0 …[Dividing both sides by 8]
⇒ 3k² + 3k – 2k – 2 = 0
⇒ 3k(k + 1) – 2(k + 1) = 0
⇒ (k + 1)(3k – 2) = 0
⇒ k + 1 = 0 or 3k – 2 = 0
⇒ k = -1 or k = 2/3

Q7: If p(x) = x³ – 2x² + kx + 5 is divided by (x – 2), the remainder is 11. Find k. Hence find all the zeroes of x³ + kx² + 3x + 1.
Ans:
p(x) = x³ – 2x² + kx + 5,
When x – 2,
p(2) = (2)³ – 2(2)² + k(2) + 5
⇒ 11 = 8 – 8 + 2k + 5
⇒ 11 – 5 = 2k
⇒ 6 = 2k
⇒ k = 3
Let q(x) = x³ + kx² + 3x + 1
= x³ + 3x² + 3x + 1
= x³ + 1 + 3x² + 3x
= (x)³ + (1)³ + 3x(x + 1)
= (x + 1)³
= (x + 1) (x + 1) (x + 1) …[∵ a³ + b³ + 3ab (a + b) = (a + b)³]
All zeroes are:
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
Hence zeroes are -1, -1 and -1.

Q8: Find all the zeroes of the polynomial 8x⁴ + 8x³ – 18x² – 20x – 5, if it is given that two of its zeroes are  and
Ans:

Q9: If a polynomial x⁴ + 5x³ + 4x² – 10x – 12 has two zeroes as -2 and -3, then find the other zeroes.
Ans:
Since two zeroes are -2 and -3.
(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6
Dividing the given equation with x² + 5x + 6, we get 

x⁴ + 5x³ + 4x² – 10x – 12
= (x² + 5x + 6)(x² – 2)
= (x + 2)(x + 3)(x – √2 )(x + √2 )
Other zeroes are:
x – √2 = 0 or x + √2 = 0
x = √2 or x = -√2 

Q10: Given that x – √5 is a factor of the polynomial x³ – 3√5 x² – 5x + 15√5, find all the zeroes of the polynomial.
Ans:
Let P(x) = x³ – 3√5 x² – 5x + 15√5
x – √5 is a factor of the given polynomial.
Put x = -√5, 

Other zero:
x – 3√5 = 0 ⇒ x = 3√5
All the zeroes of P(x) are -√5, √5 and 3√5.