How do you factor the trinomial $3 x^{2} - 6 x - 9$ $3x^2-6x-9$ ?

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How do you factor the trinomial $3 x^{2} - 6 x - 9$ $3x^2-6x-9$ ?

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Answer 1 · 2016-08-13T14:29:54

$= 3 (x - 3) (x + 1)$ $=3(x-3)(x+1)$

Explanation:

$3 x^{2} - 6 x - 9$ $3x^2-6x-9$

$= 3 (x^{2} - 2 x - 3)$ $=3(x^2-2x-3)$

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

$= 3 (x (x - 3) + 1 (x - 3))$ $=3(x(x-3)+1(x-3))$

$= 3 (x - 3) (x + 1)$ $=3(x-3)(x+1)$

Answer 2 · 2016-08-13T14:34:40

$3 (x - 3) (x + 1)$ $3(x-3)(x+1)$

Explanation:

The first step is to take out the $common factor$ $color(blue)"common factor"$ of 3

$\Rightarrow 3 x^{2} - 6 x - 9 = 3 (x^{2} - 2 x - 3)$ $rArr3x^2-6x-9=3(x^2-2x-3)$

Now we require to factorise the quadratic inside the bracket.

Need 2 factors which multiply to give - 3 and sum to - 2

These are -3 and +1

$\Rightarrow x^{2} - 2 x - 3 = (x - 3) (x + 1)$ $rArrx^2-2x-3=(x-3)(x+1)$

$\Rightarrow 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ $rArr3x^2-6x-9=3(x-3)(x+1)$

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How do you factor the trinomial $3 x^{2} - 6 x - 9$ $3x^2-6x-9$ ?

2 Answers

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How do you factor the trinomial $3 x^{2} - 6 x - 9$ $3x^2-6x-9$ ?