Question

Using factor theorem, factorize each of the following polynomials:
(i) x3 – 6x2 + 3x + 10

Answer 1

Let f(x) = x3 – 6x2 + 3x + 10

Constant term = 10

Factors of 10 are ±1, ±2, ±5, ±10

Let x + 1 = 0 or x = -1

f(-1) = (-1)3 – 6(-1)2 + 3(-1) + 10

= 10 – 10 = 0

Answer 2

Step-by-step explanation:

We have to factorise x³ - 6x² + 3x + 10.

→ x³ - 6x² + 3x + 10

Take x = 2

→ (2)³ - 6(2)² + 3(2) + 10

→ 8 - 6(4) + 6 + 10

→ 8 - 24 + 6 + 10

→ 24 - 24

→ 0

Therefore, (x - 2) is a factor.

x - 2 ) x³ - 6x² + 3x + 10 ( x² - 4x - 5

x³ - 2x² (change the signs)

_____________

- 4x² + 3x

- 4x² - 2x (change the signs)

________________

5x + 10

5x + 10 (change the signs)

__________________

0

Quotient is x² - 4x - 5

Split the middle term in x² - 4x - 5 in such a way that it's sum is -4 and product is -5.

→ x² - 4x - 5

→ x² + x - 5x - 5

→ x(x + 1) -5(x + 1)

→ (x + 1) (x - 5)

Therefore, the factors are (x + 1) (x - 2) (x - 5)