Factor 25x^6+10x^5+x^4

$25 x^{6} + 10 x^{5} + x^{4}$

Step 1

Factor $x^{4}$ out of $25 x^{6} + 10 x^{5} + x^{4}$ .

Factor $x^{4}$ out of $25 x^{6}$ .

$x^{4} (25 x^{2}) + 10 x^{5} + x^{4}$

Factor $x^{4}$ out of $10 x^{5}$ .

$x^{4} (25 x^{2}) + x^{4} (10 x) + x^{4}$

Multiply by $1$ .

$x^{4} (25 x^{2}) + x^{4} (10 x) + x^{4} \cdot 1$

Factor $x^{4}$ out of $x^{4} (25 x^{2}) + x^{4} (10 x)$ .

$x^{4} (25 x^{2} + 10 x) + x^{4} \cdot 1$

Factor $x^{4}$ out of $x^{4} (25 x^{2} + 10 x) + x^{4} \cdot 1$ .

$x^{4} (25 x^{2} + 10 x + 1)$

Step 2

Factor using the perfect square rule.

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

$x^{4} ({(5 x)}^{2} + 10 x + 1)$

Rewrite $1$ as $1^{2}$ .

$x^{4} ({(5 x)}^{2} + 10 x + 1^{2})$

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

$10 x = 2 \cdot (5 x) \cdot 1$

Rewrite the polynomial.

$x^{4} ({(5 x)}^{2} + 2 \cdot (5 x) \cdot 1 + 1^{2})$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 5 x$ and $b = 1$ .

$x^{4} {(5 x + 1)}^{2}$