How do you factor given that $f (- 2) = 0$ $f(-2)=0$ and $f (x) = x^{3} - 5 x^{2} - 2 x + 24$ $f(x)=x^3-5x^2-2x+24$ ?

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Answer 1 · 2016-09-20T20:20:08

$f (x) = (x + 2) (x - 4) (x - 3)$ $f(x)=(x+2)(x-4)(x-3)$ .

Having verified $f (- 2) = 0, (x + 2)$ $f(-2)=0, (x+2)$ is a factor of $f (x) .$ $f(x).$

Long Division is well-known Method to factorise, but we proceed as under :-

$f (x) = x^{3} - 5 x^{2} - 2 x + 24$ $f(x)=x^3-5x^2-2x+24$

$=ul(x^3+2x^2)-ul(7x^2-14x)+ul(12x+24)$

$=x^2(x+2)-7x(x+2)+12(x+2)$

$=(x+2)(x^2-7x+12)$

$=(x+2){ul(x^2-4x)-ul(3x+12)}$

$=(x+2){x(x-4)-3(x-4)}$

$=(x+2)(x-4)(x-3)$ .

Enjoy Maths.!

How do you factor given that f(-2)=0f(−2)=0f(-2)=0 and f(x)=x^3-5x^2-2x+24f(x)=x3−5x2−2x+24f(x)=x^3-5x^2-2x+24?