Polynomials questions are a useful source for the students of Class 9 and Class 10 since the chapter Polynomial is one of the important concepts for these classes. Here, we have provided polynomials questions with detailed explanations. In this article, you will find questions and answers on polynomials, along with some extra practice questions.
Learn: Polynomials
Polynomial Definition
A polynomial p(x) in one variable x is an algebraic expression in x of the form:
p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0,
where a0, a1 , a2, . . ., an are constants and an ≠0.
a0, a1, a2, . . ., an are respectively the coefficients of x0, x1, x2 , . . ., xn.
Each of anxn, an–1 xn–1, …, a0, with an ≠0, is called a term of the polynomial p(x).
The highest power of the variable in a polynomial is called the degree of that polynomial.
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Polynomial with – |
Name of the polynomial |
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One term |
Monomial |
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Two terms |
Binomial |
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Three terms |
Trinomial |
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Degree 1 |
Linear polynomial |
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Degree 2 |
Quadratic polynomial |
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Degree 3 |
Cubic polynomial |
Factor Theorem: x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor of p(x), then p(a) = 0.
Remainder Theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
Polynomials Questions with Answers
1. For the polynomial (x3 + 2x + 1)/5 – (7/2)x2 – x6, write
(i) the degree of the polynomial
(ii) the coefficient of x3
(iii) the coefficient of x6
(iv) the constant term
Solution:
Given polynomial is:
(x3 + 2x + 1)/5 – (7/2)x2 – x6
Or
(1/5)x3 + (2/5)x + (1/5) – (7/2)x2 – x6
(i) Degree of the polynomial = 6 {since the highest power of variable x is 6}
(ii) Coefficient of x3 = (1/5)
(iii) Coefficient of x6 = -1
(iv) Constant term = 1/5
2. Find the value of a, if x – a is a factor of x3 – ax2 + 2x + a – 1.
Solution:
Let p(x) = x3 – ax2 + 2x + a – 1
Given that x – a is a factor of p(x).
⇒ p(a) = 0
i.e., a3 – a(a)2 + 2a + a – 1 = 0
a3 – a3 + 2a + a – 1 = 0
3a – 1 = 0
3a = 1
a = 1/3
Therefore, a = 1/3.
3. If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution:
Given,
x + y = 12
xy = 27
x3 + y3 = (x + y) (x2 – xy + y2)
= (x + y) [(x + y)2 – 3xy]
= 12 × (122 – 3 × 27)
= 12 × (144 – 81)
= 12 × 63
= 756
4. Factorise: 12x2 – 7x + 1
Solution:
Let the given polynomial be:
p(x) = 12x2 – 7x + 1
= 12x2 – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(4x – 1)
Hence, the factors of 12x2 – 7x + 1 are (3x – 1) and (4x – 1).
5. Factorise the polynomial p(x) = x3 – 3x2 – 9x – 5.
Solution:
Given,
p(x) = x3 – 3x2 – 9x – 5
By trial and error method, substitute x = -1 in p(x).
p(-1) = (-1)3 – 3(-1)2 – 9(-1) – 5
= -1 – 3 + 9 – 5
= -9 + 9 = 0
Since p(-1) = 0 and by the factor theorem we can say x + 1 is a factor of p(x).
By dividing p(x) by x + 1 we can get the remaining factors.
Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5)
Consider quotient, x2 – 4x – 5
= x2 – 5x + x – 5
= x(x – 5) + 1(x – 5)
= (x + 1)(x – 5)
Therefore, p(x) = x3 – 3x2 – 9x – 5 = (x + 1)(x + 1)(x – 5) = (x + 1)2(x – 5).
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If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then; Sum of zeroes = α + β = -b/a Product of zeroes = αβ = c/a If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then α +β + γ = -b/a, αβ + β γ + γ α = c/a αβγ = -d/a The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) Here, r(x) = 0 or degree r(x) < degree g(x) |
6. Find the value of x³ + y³ – 12xy + 64, when x + y = – 4.
Solution:
Given,
x + y = -4
Or
x + y + 4 = 0….(i)
Let the given polynomial be:
p(x) = x³ + y³ – 12xy + 64
= x³ + y³ + 4³ – 3(4xy) {since 64 = 4³}
We know that is a + b + c = 0 then a³ + b³ + c³ = 3abc
That means if x + y + 4 = 0, x³ + y³ + 4³ = 3(x)(y)(4) {from (i)}
So, p(x) = 3(x)(y)(4) – 3(4xy)
= 12xy – 12xy
= 0
7. Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2. [Hint: Factorise x2 – 3x + 2]
Solution:
Let p(x) = 2x4 – 5x3 + 2x2 – x + 2
Let us factorise x2 – 3x + 2.
x2 – 3x + 2 = x2 – 2x – x + 2
= x(x – 2) -1(x – 2)
= (x – 1)(x – 2)
Hence, 1 and 2 are the zeroes of x2 – 3x + 2.
Now, substitute x = 1 and x = 2 in p(x).
p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2
= 2 – 5 + 2 – 1 + 2
= 6 – 6
= 0
p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2
= 2(16) – 5(8) + 2(4)
= 32 – 40 + 8
= 40 – 40
= 0
Therefore, p(x) = 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2.
8. Find the zeroes of the polynomial 4x2 – 3x – 1 by the factorisation method and verify the relation between the zeroes and the coefficients of the polynomial.
Solution:
Let the given polynomial be:
p(x) = 4x2 – 3x – 1
= 4x2 -4x + x – 1
= 4x(x – 1) + 1(x – 1)
= (4x + 1)(x – 1)
Thus, x = -1/4 and x = 1 are the zeroes of the given polynomial.
Now,
Sum of zeroes = (-1/4) + 1 = (-1 + 4)/4 = 3/4 = -coefficient of x/coefficient of x2
Product of zeroes = (-1/4)(1) = -1/4 = constant term/coefficient of x2
Hence, the relation between the zeroes and the coefficients of the polynomial is verified.
9. Find a quadratic polynomial whose sum and product respectively of the zeroes are 21/8 and 5/16. Also, find the zeroes of the polynomial by factorisation.
Solution:
Given,
Sum of zeroes = 21/8
Product of zeroes = 5/16
A quadratic polynomial with the sum and product of zeroes is given by:
p(x) = x2 – (sum of zeros) x + (product of roots)
= x2 – (21/8)x + (5/16)
Consider p(x) = 0 to solve for the zeroes.
⇒ x2 – (21/8)x + (5/16) = 0
⇒ 16x2 – 42x + 5 = 0
⇒ 16x2 – 40x – 2x + 5 = 0
⇒ 8x(2x – 5) – 1(2x – 5) = 0
⇒ (2x – 5) (8x – 1) = 0
⇒ (2x – 5) = 0, (8x – 1) = 0
⇒ 2x = 5, 8x = 1
⇒ x = 5/2, x = 1/8
Therefore, the two zeroes of the polynomial are 5/2 and 1/8.
10. For which values of a and b, are the zeroes of q(x) = x3 + 2x2 + a, also the zeroes of the polynomial p(x) = x5 – x4 – 4x3 + 3x2 + 3x + b? Which zeroes of p(x) are not the zeroes of q(x)?
Solution:
Given,
p(x) = x5 – x4 – 4x3 + 3x2 + 3x + b
q(x) = x3 + 2x2 + a
By division algorithm of polynomials,
Dividend = (divisor) (quotient) + remainder
⇒ p(x) = g(x).q(x) + r(x)
r(x) = -(a + 1)x2 + 3(1 + a)x + b – 2a = 0
and
g(x) = x2 – 3x + 2
By comparing the coefficients,
– (a + 1) = 0 , (1 + a) = 0
⇒ a = – 1
Also, b – 2a = 0
b = 2a = 2(- 1) = – 2
⇒ b = – 2
q(x) = x3 + 2x2 – 1
Equating q(x) = 0 to get the zeroes.
x3 + 2x2 – 1 = 0
x3 + x2 + x2 – 1 = 0
x2(x + 1) + (x2 – 1) = 0
x2(x + 1) + (x + 1)(x – 1) = 0
(x + 1)(x2 + x – 1) = 0
g(x) = x2 – 3x + 2
Equating g(x) = 0 to get the zeroes.
x2 – 3x + 2 = 0
x2 – (x + 2x) + 2 = 0
x(x – 1) – 2(x – 1) = 0
(x – 1)(x – 2) = 0
Dividend = (divisor) (quotient) + remainder
⇒ p(x) = g(x).q(x) + r(x) P(x) = x5 – x4 – 4x3 + 3x – 2 = q(x).g(x) + 0
= (x3 + 2x2 – 1) (x2 – 3x + 2)
= (x + 1) (x2 + x – 1) (x – 1) (x – 2)
Therefore, 1 and 2 are the zeroes of the polynomial p(x) that are not in q(x).
Practice Questions on Polynomials
Solve the following polynomials questions for better practice.
- If the remainder on division of x3 + 2x2 + kx +3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 + 2x2 + kx – 18.
- If the polynomials az3 + 4z2 + 3z – 4 and z3 – 4z + a leave the same remainder when divided by z – 3, find the value of a.
- If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.
- Verify that 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
- Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.
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