How do you factor the trinomial $4 x^{2} + 2 x - 20$ $4x^2+2x-20$ ?

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Answer 1 · 2015-11-26T03:21:44

$4 x^{2} + 2 x - 20 = 2 (2 x + 5) (x - 2)$ $4x^2+2x-20=2(2x+5)(x-2)$

$4 x^{2} + 2 x - 20$ $4x^2+2x-20$

Factor out the GCF $2$ $2$ .

$2 (2 x^{2} + x - 10)$ $2(2x^2+x-10)$

Factor the remaining trinomial using the AC method, where $a = 2, b = 1, and c = - 10$ $a=2, b=1, and c=-10$ .

Multiply $a$ $a$ times $c$ $c$ .

$2 \times - 10 = - 20$ $2xx-10=-20$

Find two numbers that when added equal $1$ $1$ and when multiplied equal $- 20$ $-20$ . The numbers $5$ $5$ and $- 4$ $-4$ meet the criteria.

Rewrite the trinomial substituting $5 x$ $5x$ and $- 4 x$ $-4x$ in place of $x$ $x$ .

$2 (2 x^{2} + 5 x - 4 x - 10)$ $2(2x^2+5x-4x-10)$

Separate the two pairs of terms.

$2 (2 x^{2} + 5 x) - (4 x - 10)$ $2(2x^2+5x)-(4x-10)$

Factor $x$ $x$ out of the first pair of terms.

$2 (x (2 x + 5) - (4 x - 10)$ $2(x(2x+5)-(4x-10)$

Factor $- 2$ $-2$ out of the second pair of terms.

$2 (x (2 x + 5) - 2 (2 x + 5)$ $2(x(2x+5)-2(2x+5)$

Factor out $(2 x + 5)$ $(2x+5)$ .

$2 (2 x + 5) (x - 2)$ $2(2x+5)(x-2)$

How do you factor the trinomial 4x2+2x−204x^2+2x-20?