Factor P(x)=4x^4-45x^2+81

$P (x) = 4 x^{4} - 45 x^{2} + 81$

Step 1

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

$P (x) = 4 {(x^{2})}^{2} - 45 x^{2} + 81$

Step 2

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

$P (x) = 4 u^{2} - 45 u + 81$

Step 3

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 $a \cdot c = 4 \cdot 81 = 324$ and whose sum is $b = - 45$ .

Factor $- 45$ out of $- 45 u$ .

$P (x) = 4 u^{2} - 45 u + 81$

Rewrite $- 45$ as $- 9$ plus $- 36$

$P (x) = 4 u^{2} + (- 9 - 36) u + 81$

Apply the distributive property.

$P (x) = 4 u^{2} - 9 u - 36 u + 81$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$P (x) = (4 u^{2} - 9 u) - 36 u + 81$

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

$P (x) = u (4 u - 9) - 9 (4 u - 9)$

Factor the polynomial by factoring out the greatest common factor, $4 u - 9$ .

$P (x) = (4 u - 9) (u - 9)$

Step 4

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

$P (x) = (4 x^{2} - 9) (x^{2} - 9)$

Step 5

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

$P (x) = ({(2 x)}^{2} - 9) (x^{2} - 9)$

Step 6

Rewrite $9$ as $3^{2}$ .

$P (x) = ({(2 x)}^{2} - 3^{2}) (x^{2} - 9)$

Step 7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 3$ .

$P (x) = (2 x + 3) (2 x - 3) (x^{2} - 9)$

Step 8

Rewrite $9$ as $3^{2}$ .

$P (x) = (2 x + 3) (2 x - 3) (x^{2} - 3^{2})$

Step 9

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$P (x) = (2 x + 3) (2 x - 3) ((x + 3) (x - 3))$

Remove unnecessary parentheses.

$P (x) = (2 x + 3) (2 x - 3) (x + 3) (x - 3)$

Factor P(x)=4x^4-45x^2+81

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