Factor 81p^4-36p^3+4p^2

$81 p^{4} - 36 p^{3} + 4 p^{2}$

Step 1

Factor $p^{2}$ out of $81 p^{4} - 36 p^{3} + 4 p^{2}$ .

Factor $p^{2}$ out of $81 p^{4}$ .

$p^{2} (81 p^{2}) - 36 p^{3} + 4 p^{2}$

Factor $p^{2}$ out of $- 36 p^{3}$ .

$p^{2} (81 p^{2}) + p^{2} (- 36 p) + 4 p^{2}$

Factor $p^{2}$ out of $4 p^{2}$ .

$p^{2} (81 p^{2}) + p^{2} (- 36 p) + p^{2} \cdot 4$

Factor $p^{2}$ out of $p^{2} (81 p^{2}) + p^{2} (- 36 p)$ .

$p^{2} (81 p^{2} - 36 p) + p^{2} \cdot 4$

Factor $p^{2}$ out of $p^{2} (81 p^{2} - 36 p) + p^{2} \cdot 4$ .

$p^{2} (81 p^{2} - 36 p + 4)$

Step 2

Factor using the perfect square rule.

Rewrite $81 p^{2}$ as ${(9 p)}^{2}$ .

$p^{2} ({(9 p)}^{2} - 36 p + 4)$

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

$p^{2} ({(9 p)}^{2} - 36 p + 2^{2})$

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

$36 p = 2 \cdot (9 p) \cdot 2$

Rewrite the polynomial.

$p^{2} ({(9 p)}^{2} - 2 \cdot (9 p) \cdot 2 + 2^{2})$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 9 p$ and $b = 2$ .

$p^{2} {(9 p - 2)}^{2}$