How do you factor $4 x^{2} - 32 x + 60$ $4x^2-32x+60$ ?

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How do you factor $4 x^{2} - 32 x + 60$ $4x^2-32x+60$ ?

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Answer 1 · 2018-03-03T09:39:01

factor out 4

$4 (x^{2} - 8 x + 15)$ $4(x^2-8x+15)$

what numbers add up to -8 and multiply to 15?
-3 and -5

$4 (x - 3) (x - 5)$ $4(x-3)(x-5)$

Answer 2 · 2018-03-03T09:44:06

$4 x^{2} - 32 x + 60 = (2 x - 6) (2 x - 10)$ $4x^2-32x +60 = (2x-6)(2x-10)$

Explanation:

We can write that trinomial as

$4 x^{2} - 32 x + 60 = 4 x^{2} - 32 x + 64 - 4$ $4x^2 - 32x + 60 = 4x^2 - 32x + 64 - 4$

Now,
$4 x^{2} - 32 x + 64 = {(2 x)}^{2} - 2 \cdot 2 x \cdot 8 + 8^{2}$ $4x^2 - 32x + 64 = (color(red)(2x))^2 - 2*color(red)(2x)*color(red)8 + color(red)8^2$

Knowing that
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2 = a^2-2ab+b^2$

We can deduce $4 x^{2} - 32 x + 64$ $4x^2 - 32x + 64$ to be a perfect square, precisely ${(2 x - 8)}^{2}$ $color(red)((2x-8)^2)$ .

Substitute it into the original expression :

$4 x^{2} - 32 x + 64 - 4 = {(2 x - 8)}^{2} - 2^{2}$ $4x^2 - 32x + 64 - 4 = (2x-8)^2 - 2^2$

That is a difference of squares, so we can factor it out as:

$(2 x - 8 + 2) (2 x - 8 - 2)$ $(2x-8color(red)+ 2)(2x-8color(red)-2)$ = $(2 x - 6) (2 x - 10)$ $color(blue)((2x-6)(2x-10)$

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How do you factor $4 x^{2} - 32 x + 60$ $4x^2-32x+60$ ?

2 Answers

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