Question #4c1b0

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Question #4c1b0

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Answer 1 · 2017-10-01T10:50:39

$f (x) = (x + 2) (x + 4)$ $f(x)=(x+2)(x+4)$

Explanation:

$expand f (x) = {(x + 3)}^{2} - 1$ $"expand "f(x)=(x+3)^2-1$

$\Rightarrow f (x) = x^{2} + 6 x + 9 - 1 = x^{2} + 6 x + 8$ $rArrf(x)=x^2+6x+9-1=x^2+6x+8$

$the factors of + 8 which sum to + 6 are + 2 and + 4$ $"the factors of + 8 which sum to + 6 are + 2 and + 4"$

$\Rightarrow x^{2} + 6 x + 8 = (x + 2) (x + 4)$ $rArrx^2+6x+8=(x+2)(x+4)$

$f (x) = {(x + 3)}^{2} - 1 = (x + 2) (x + 4) in intercept form$ $f(x)=(x+3)^2-1=(x+2)(x+4)color(blue)" in intercept form"$

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