Factor 2x^6+13x^4+6x^2

$2 x^{6} + 13 x^{4} + 6 x^{2}$

Step 1

Factor $x^{2}$ out of $2 x^{6} + 13 x^{4} + 6 x^{2}$ .

Factor $x^{2}$ out of $2 x^{6}$ .

$x^{2} (2 x^{4}) + 13 x^{4} + 6 x^{2}$

Factor $x^{2}$ out of $13 x^{4}$ .

$x^{2} (2 x^{4}) + x^{2} (13 x^{2}) + 6 x^{2}$

Factor $x^{2}$ out of $6 x^{2}$ .

$x^{2} (2 x^{4}) + x^{2} (13 x^{2}) + x^{2} \cdot 6$

Factor $x^{2}$ out of $x^{2} (2 x^{4}) + x^{2} (13 x^{2})$ .

$x^{2} (2 x^{4} + 13 x^{2}) + x^{2} \cdot 6$

Factor $x^{2}$ out of $x^{2} (2 x^{4} + 13 x^{2}) + x^{2} \cdot 6$ .

$x^{2} (2 x^{4} + 13 x^{2} + 6)$

Step 2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x^{2} (2 {(x^{2})}^{2} + 13 x^{2} + 6)$

Step 3

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

$x^{2} (2 u^{2} + 13 u + 6)$

Step 4

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = 13$ .

Factor $13$ out of $13 u$ .

$x^{2} (2 u^{2} + 13 (u) + 6)$

Rewrite $13$ as $1$ plus $12$

$x^{2} (2 u^{2} + (1 + 12) u + 6)$

Apply the distributive property.

$x^{2} (2 u^{2} + 1 u + 12 u + 6)$

Multiply $u$ by $1$ .

$x^{2} (2 u^{2} + u + 12 u + 6)$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{2} ((2 u^{2} + u) + 12 u + 6)$

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

$x^{2} (u (2 u + 1) + 6 (2 u + 1))$

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$x^{2} ((2 u + 1) (u + 6))$

Step 5

Factor.

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

$x^{2} ((2 x^{2} + 1) (x^{2} + 6))$

Remove unnecessary parentheses.

$x^{2} (2 x^{2} + 1) (x^{2} + 6)$

Factor 2x^6+13x^4+6x^2

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