Factorisation of Polynomials Questions
Chapter – Polynomials HP Board
Topic – Factorisation of Polynomials
Question 1. Factorise: x² + 7x + 12.
Solution:
Given:
x² + 7x + 12
Find two numbers whose sum is 7 and product is 12.
The numbers are 3 and 4.
Split the middle term:
= x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Answer: (x + 3)(x + 4)
Question 2. Factorise: x² − 9x + 20.
Solution:
Given:
x² − 9x + 20
Find two numbers whose sum is −9 and product is 20.
The numbers are −4 and −5.
Split the middle term:
= x² − 4x − 5x + 20
= x(x − 4) − 5(x − 4)
= (x − 4)(x − 5)
Answer: (x − 4)(x − 5)
Question 3. Factorise: 4x² − 25.
Solution:
Given:
4x² − 25
Using the identity:
a² − b² = (a − b)(a + b)
We can write:
4x² − 25 = (2x)² − 5²
= (2x − 5)(2x + 5)
Answer: (2x − 5)(2x + 5)
Question 4. Factorise: x² + 10x + 25.
Solution:
Given:
x² + 10x + 25
Using the identity:
a² + 2ab + b² = (a + b)²
Here:
a = x and b = 5
Therefore:
x² + 10x + 25
= x² + 2(x)(5) + 5²
= (x + 5)²
Answer: (x + 5)²
Question 5. Factorise: 9x² − 12x + 4.
Solution:
Given:
9x² − 12x + 4
Using the identity:
a² − 2ab + b² = (a − b)²
Here:
a = 3x and b = 2
Therefore:
9x² − 12x + 4
= (3x)² − 2(3x)(2) + 2²
= (3x − 2)²
Answer: (3x − 2)²
Question 6. Factorise: 2x² + 7x + 3.
Solution:
Given:
2x² + 7x + 3
Multiply the coefficient of x² by the constant term:
2 × 3 = 6
Find two numbers whose sum is 7 and product is 6.
The numbers are 6 and 1.
Split the middle term:
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)
Answer: (2x + 1)(x + 3)
Question 7. Factorise: x³ − 8.
Solution:
Given:
x³ − 8
We know that:
8 = 2³
Using the identity:
a³ − b³ = (a − b)(a² + ab + b²)
Therefore:
x³ − 8 = x³ − 2³
= (x − 2)(x² + 2x + 4)
Answer: (x − 2)(x² + 2x + 4)
Question 8. Factorise: 8x³ + 27.
Solution:
Given:
8x³ + 27
We can write:
8x³ = (2x)³
27 = 3³
Using the identity:
a³ + b³ = (a + b)(a² − ab + b²)
Therefore:
8x³ + 27
= (2x)³ + 3³
= (2x + 3)(4x² − 6x + 9)
Answer: (2x + 3)(4x² − 6x + 9)
Question 9. Factorise: x³ − 6x² + 11x − 6.
Solution:
Let:
p(x) = x³ − 6x² + 11x − 6
First, check x = 1.
p(1) = 1 − 6 + 11 − 6
= 0
Therefore, by the Factor Theorem, (x − 1) is a factor.
Dividing the polynomial by (x − 1), we get:
x² − 5x + 6
Now factorise:
x² − 5x + 6
= x² − 2x − 3x + 6
= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3)
Therefore:
x³ − 6x² + 11x − 6
= (x − 1)(x − 2)(x − 3)
Answer: (x − 1)(x − 2)(x − 3)
Question 10. Factorise: x³ + 3x² − 4x − 12.
Solution:
Given:
x³ + 3x² − 4x − 12
Group the terms:
= (x³ + 3x²) − (4x + 12)
Take common factors:
= x²(x + 3) − 4(x + 3)
Take (x + 3) common:
= (x + 3)(x² − 4)
Using the identity:
a² − b² = (a − b)(a + b)
We get:
x² − 4 = (x − 2)(x + 2)
Therefore:
x³ + 3x² − 4x − 12
= (x + 3)(x − 2)(x + 2)
Answer: (x + 3)(x − 2)(x + 2)