Question

17. Factorise using factor theorem:
x^3-3x^2-4x+12​

Answer 1

(x - 2)(x - 3)(x + 2).

Explanation:

The Factor Theorem states that if x - a is a factor of f(x) the f(a) = 0.

f(x) = x^3 - 3x^2 - 4x + 12​

We try the factors of 12:

f(2) = 2^3 - 3*4 - 8 +12

= 8 - 12 - 8 + 12 = 0

- so (x - 2) is a factor of f(x)

Dividing:

x^2 - x - 6 <-----------Quotient

-----------------------------

x - 2 )x^3 - 3x^2 - 4x + 12​

x^3 - 2x^2

-x^2 - 4x

-x^2 + 2x

-6x + 12

-6x + 12

So we factor x^2 - x - 6:

= (x - 3)(x + 2)

Answer 2

(x - 3)(x - 2)(x + 2)

Explanation:

Given

x³ - 3x² - 4x + 12 ( factor the first/second and third/fourth terms )

= x²(x - 3) - 4(x - 3) ← factor out (x - 3) from each term

= (x - 3)(x² - 4) ← x² - 4 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 4

= x² - 2²

= (x - 2)(x + 2)

Thus

x³ - 3x² - 4x + 12 = (x - 3)(x - 2)(x + 2) ← in factored form