How do you factor $3 {(x - 4)}^{2} + 16 (x - 4) - 12$ $3(x-4)^2+16(x-4) -12$ ?

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How do you factor $3 {(x - 4)}^{2} + 16 (x - 4) - 12$ $3(x-4)^2+16(x-4) -12$ ?

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Answer 1 · 2015-11-28T23:56:31

Factor as a quadratic in $(x - 4)$ $(x-4)$ then simplify at the last stage to find:

$3 {(x - 4)}^{2} + 16 (x - 4) - 12 = (x + 2) (3 x - 14)$ $3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)$

Explanation:

One way is to leave the $(x - 4)$ $(x-4)$ 's intact until the end and look for a pair of factors of $3 \times 12 = 36$ $3 xx 12 = 36$ , whose difference is $16$ $16$ . The pair $2$ $2$ , $18$ $18$ works, so use that to split the middle term and factor by grouping:

$3 {(x - 4)}^{2} + 16 (x - 4) - 12$ $3(x-4)^2+16(x-4)-12$

$= 3 {(x - 4)}^{2} - 2 (x - 4) + 18 (x - 4) - 12$ $=3(x-4)^2-2(x-4)+18(x-4)-12$

$= (3 {(x - 4)}^{2} - 2 (x - 4)) + (18 (x - 4) - 12)$ $=(3(x-4)^2-2(x-4))+(18(x-4)-12)$

$= (x - 4) (3 (x - 4) - 2) + 6 (3 (x - 4) - 2)$ $=(x-4)(3(x-4)-2)+6(3(x-4)-2)$

$= ((x - 4) + 6) (3 (x - 4) - 2)$ $=((x-4)+6)(3(x-4)-2)$

$= (x + 2) (3 x - 14)$ $=(x+2)(3x-14)$

Another way of expressing this solution is to substitute $t = x - 4$ $t = x-4$ and proceed as follows:

$3 {(x - 4)}^{2} + 16 (x - 4) - 12$ $3(x-4)^2+16(x-4)-12$

$= 3 t^{2} + 16 t - 12$ $=3t^2+16t-12$

$= 3 t^{2} - 2 t + 18 t - 12$ $=3t^2-2t+18t-12$

$= (3 t^{2} - 2 t) + (18 t - 12)$ $=(3t^2-2t)+(18t-12)$

$= t (3 t - 2) + 6 (3 t - 2)$ $=t(3t-2)+6(3t-2)$

$= (t + 6) (3 t - 2)$ $=(t+6)(3t-2)$

$= ((x - 4) + 6) (3 (x - 4) - 2)$ $=((x-4)+6)(3(x-4)-2)$

$= (x + 2) (3 x - 14)$ $=(x+2)(3x-14)$

Answer 2 · 2015-11-29T00:07:06

Alternatively, multiply out the original quadratic and simplify before factoring to find:

$3 {(x - 4)}^{2} + 16 (x - 4) - 12 = (x + 2) (3 x - 14)$ $3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)$

Explanation:

$3 {(x - 4)}^{2} + 16 (x - 4) - 12$ $3(x-4)^2+16(x-4)-12$

$= 3 (x^{2} - 8 x + 16) + 16 (x - 4) - 12$ $=3(x^2-8x+16)+16(x-4)-12$

$= 3 x^{2} - 24 x + 48 + 16 x - 64 - 12$ $=3x^2-24x+48+16x-64-12$

$= 3 x^{2} - 8 x - 28$ $=3x^2-8x-28$

Look for a pair of factors of $3 \times 28 = 84$ $3 xx 28 = 84$ with difference $8$ $8$ .

The pair $6$ $6$ , $14$ $14$ works.

So:

$3 x^{2} - 8 x - 28$ $3x^2-8x-28$

$= 3 x^{2} - 14 x + 6 x - 28$ $=3x^2-14x+6x-28$

$= (3 x^{2} - 14 x) + (6 x - 28)$ $=(3x^2-14x)+(6x-28)$

$= x (3 x - 14) + 2 (3 x - 14)$ $=x(3x-14)+2(3x-14)$

$= (x + 2) (3 x - 14)$ $=(x+2)(3x-14)$

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