Precalculus Examples

$16 y^{4} {(y + 8)}^{3} + y^{5} {(y + 8)}^{4}$

Step 1

Factor $y^{4} {(y + 8)}^{3}$ out of $16 y^{4} {(y + 8)}^{3} + y^{5} {(y + 8)}^{4}$ .

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Step 1.1

Factor $y^{4} {(y + 8)}^{3}$ out of $16 y^{4} {(y + 8)}^{3}$ .

$y^{4} {(y + 8)}^{3} (16) + y^{5} {(y + 8)}^{4}$

Step 1.2

Factor $y^{4} {(y + 8)}^{3}$ out of $y^{5} {(y + 8)}^{4}$ .

$y^{4} {(y + 8)}^{3} (16) + y^{4} {(y + 8)}^{3} (y (y + 8))$

Step 1.3

Factor $y^{4} {(y + 8)}^{3}$ out of $y^{4} {(y + 8)}^{3} (16) + y^{4} {(y + 8)}^{3} (y (y + 8))$ .

$y^{4} {(y + 8)}^{3} (16 + y (y + 8))$

Step 2

Apply the distributive property.

$y^{4} {(y + 8)}^{3} (16 + y \cdot y + y \cdot 8)$

Step 3

Multiply $y$ by $y$ .

$y^{4} {(y + 8)}^{3} (16 + y^{2} + y \cdot 8)$

Step 4

Move $8$ to the left of $y$ .

$y^{4} {(y + 8)}^{3} (16 + y^{2} + 8 y)$

Step 5

Reorder terms.

$y^{4} {(y + 8)}^{3} (y^{2} + 8 y + 16)$

Step 6

Factor using the perfect square rule.

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Step 6.1

Rewrite $16$ as $4^{2}$ .

$y^{4} {(y + 8)}^{3} (y^{2} + 8 y + 4^{2})$

Step 6.2

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

$8 y = 2 \cdot y \cdot 4$

Step 6.3

Rewrite the polynomial.

$y^{4} {(y + 8)}^{3} (y^{2} + 2 \cdot y \cdot 4 + 4^{2})$

Step 6.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 4$ .

$y^{4} {(y + 8)}^{3} {(y + 4)}^{2}$