Factor by Grouping 4y^(4a)+12y^(2a)+10y^(2a)+30

$4 y^{4 a} + 12 y^{2 a} + 10 y^{2 a} + 30$

Factor out the GCF of $2$ from $4 y^{4 a} + 12 y^{2 a} + 10 y^{2 a} + 30$ .

Factor out the GCF of $2$ from each term in the polynomial.

Factor out the GCF of $2$ from the expression $4 y^{4 a}$ .

$2 (2 y^{4 a}) + 12 y^{2 a} + 10 y^{2 a} + 30$

Factor out the GCF of $2$ from the expression $12 y^{2 a}$ .

$2 (2 y^{4 a}) + 2 (6 y^{2 a}) + 10 y^{2 a} + 30$

Factor out the GCF of $2$ from the expression $10 y^{2 a}$ .

$2 (2 y^{4 a}) + 2 (6 y^{2 a}) + 2 (5 y^{2 a}) + 30$

Factor out the GCF of $2$ from the expression $30$ .

$2 (2 y^{4 a}) + 2 (6 y^{2 a}) + 2 (5 y^{2 a}) + 2 (15)$

Since all the terms share a common factor of $2$ , it can be factored out of each term.

$2 (2 y^{4 a} + 6 y^{2 a} + 5 y^{2 a} + 15)$

Add $6 y^{2 a}$ and $5 y^{2 a}$ .

$2 (2 y^{4 a} + 11 y^{2 a} + 15)$

Rewrite $y^{4 a}$ as ${(y^{2 a})}^{2}$ .

$2 (2 {(y^{2 a})}^{2} + 11 y^{2 a} + 15)$

Let $u = y^{2 a}$ . Substitute $u$ for all occurrences of $y^{2 a}$ .

$2 (2 u^{2} + 11 u + 15)$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = 11$ .

Factor $11$ out of $11 u$ .

$2 (2 u^{2} + 11 (u) + 15)$

Rewrite $11$ as $5$ plus $6$

$2 (2 u^{2} + (5 + 6) u + 15)$

Apply the distributive property.

$2 (2 u^{2} + 5 u + 6 u + 15)$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((2 u^{2} + 5 u) + 6 u + 15)$

Factor out the greatest common factor (GCF) from each group.

$2 (u (2 u + 5) + 3 (2 u + 5))$

Factor the polynomial by factoring out the greatest common factor, $2 u + 5$ .

$2 ((2 u + 5) (u + 3))$

Replace all occurrences of $u$ with $y^{2 a}$ .

$2 ((2 y^{2 a} + 5) (y^{2 a} + 3))$

Factor by Grouping 4y^(4a)+12y^(2a)+10y^(2a)+30

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