How do you factor $y^{2} - 16 + 64$ $y^2-16+64$ ?

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Answer 1 · 2016-01-24T21:57:13

This trinomial is a perfect square trinomial, so can be factored as ${(a - b)}^{2}$ $(a-b)^2$

The square root of 64 is 8 and the square root of $y^{2}$ $y^2$ is y.

The middle term should be of the form 2ab. In this case, a=y and b= 8

So, 2(y)(8), or 16y, which is the middle term. This short proof justifies that it is indeed a perfect square trinomial.

$y^{2}$ $y^2$ - 16y + 64
= (y - 8)(y - 8)
= ${(y - 8)}^{2}$ $(y - 8)^2$

Your answer is ${(y - 8)}^{2}$ $(y -8)^2$

Exercises:

a). $y^{2}$ $y^2$ - 16

B) $y^{2}$ $y^2$ + 8y + 16

C) $y^{2}$ $y^2$ + 16

D) $y^{2}$ $y^2$ - 8y + 16

$4 x^{2}$ $4x^2$ + 28x + 49

Hopefully you understand now!

How do you factor y2−16+64y^2-16+64?