How do you factor $9 (x^{2} + 8 x + 16) + 12 (x + 4) + 4$ $9(x^2+8x+16)+12(x+4)+4$ ?

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How do you factor $9 (x^{2} + 8 x + 16) + 12 (x + 4) + 4$ $9(x^2+8x+16)+12(x+4)+4$ ?

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Answer 1 · 2015-05-15T22:00:00

Since you have the quadratic in this form, first note that

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

So we are looking for factors of $9 {(x + 4)}^{2} + 12 (x + 4) + 4$ $9(x+4)^2+12(x+4)+4$

Note further that $9$ $9$ and $4$ $4$ are both square numbers, being $3^{2}$ $3^2$ and $2^{2}$ $2^2$ . Note also that $12$ $12$ is twice $3 \cdot 2$ $3*2$ , so

$9 {(x + 4)}^{2} + 12 (x + 4) + 4 = {(3 (x + 4) + 2)}^{2}$ $9(x+4)^2+12(x+4)+4 = (3(x+4)+2)^2$

$3 (x + 4) + 2 = 3 x + 12 + 2 = 3 x + 14$ $3(x+4)+2 = 3x+12+2 = 3x+14$

Putting this all together:

$9 (x^{2} + 8 x + 16) + 12 (x + 4) + 4 = {(3 x + 14)}^{2}$ $9(x^2+8x+16)+12(x+4)+4 = (3x+14)^2$

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How do you factor $9 (x^{2} + 8 x + 16) + 12 (x + 4) + 4$ $9(x^2+8x+16)+12(x+4)+4$ ?

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