How do you factor completely $15 a^{2} + 55 a - 20$ $15a^2 + 55a - 20$ ?

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How do you factor completely $15 a^{2} + 55 a - 20$ $15a^2 + 55a - 20$ ?

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Answer 1 · 2016-10-26T06:18:06

$15 a^{2} + 55 a - 20 = 5 (3 a - 1) (a + 4)$ $15a^2+55a-20 = 5(3a-1)(a+4)$

Explanation:

First separate out the common scalar factor $5$ $5$ to find:

$15 a^{2} + 55 a - 20 = 5 (3 a^{2} + 11 a - 4)$ $15a^2+55a-20 = 5(3a^2+11a-4)$

Next use an AC Method to factor $3 a^{2} + 11 a - 4$ $3a^2+11a-4$ :

Find a pair of factors of $A C = 3 \cdot 4 = 12$ $AC = 3*4 = 12$ which differ by $B = 11$ $B=11$ .

The pair $12, 1$ $12, 1$ works, in that $12 \cdot 1 = 12$ $12*1 = 12$ and $12 - 1 = 11$ $12-1 = 11$ .

Use this pair to split the middle term and factor by grouping:

$3 a^{2} + 11 a - 4 = 3 a^{2} + 12 a - a - 4$ $3a^2+11a-4 = 3a^2+12a-a-4$

$3 a^{2} + 11 a - 4 = (3 a^{2} + 12 a) - (a + 4)$ $color(white)(3a^2+11a-4) = (3a^2+12a)-(a+4)$

$3 a^{2} + 11 a - 4 = 3 a (a + 4) - 1 (a + 4)$ $color(white)(3a^2+11a-4) = 3a(a+4)-1(a+4)$

$3 a^{2} + 11 a - 4 = (3 a - 1) (a + 4)$ $color(white)(3a^2+11a-4) = (3a-1)(a+4)$

So:

$15 a^{2} + 55 a - 20 = 5 (3 a - 1) (a + 4)$ $15a^2+55a-20 = 5(3a-1)(a+4)$

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