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Solve for y y^2-16y+64=0

$y^{2} - 16 y + 64 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$y^{2} - 16 y + 8^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 y = 2 \cdot y \cdot 8$

Rewrite the polynomial.

$y^{2} - 2 \cdot y \cdot 8 + 8^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 8$ .

${(y - 8)}^{2} = 0$

${(y - 8)}^{2} = 0$

Step 2

Set the $y - 8$ equal to $0$ .

$y - 8 = 0$

Step 3

Add $8$ to both sides of the equation.

$y = 8$

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