Factor 4x^4-5x^2-9

$4 x^{4} - 5 x^{2} - 9$

Step 1

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

$4 {(x^{2})}^{2} - 5 x^{2} - 9$

Step 2

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

$4 u^{2} - 5 u - 9$

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 $a \cdot c = 4 \cdot - 9 = - 36$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 u$ .

$4 u^{2} - 5 (u) - 9$

Rewrite $- 5$ as $4$ plus $- 9$

$4 u^{2} + (4 - 9) u - 9$

Apply the distributive property.

$4 u^{2} + 4 u - 9 u - 9$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} + 4 u) - 9 u - 9$

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

$4 u (u + 1) - 9 (u + 1)$

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

$(u + 1) (4 u - 9)$

Step 4

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

$(x^{2} + 1) (4 x^{2} - 9)$

Step 5

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

$(x^{2} + 1) ({(2 x)}^{2} - 9)$

Step 6

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

$(x^{2} + 1) ({(2 x)}^{2} - 3^{2})$

Step 7

Factor.

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 $b = 3$ .

$(x^{2} + 1) ((2 x + 3) (2 x - 3))$

Remove unnecessary parentheses.

$(x^{2} + 1) (2 x + 3) (2 x - 3)$

Factor 4x^4-5x^2-9

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