Factor 4x^4+7x^2-36

$4 x^{4} + 7 x^{2} - 36$

Step 1

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

$4 {(x^{2})}^{2} + 7 x^{2} - 36$

Step 2

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

$4 u^{2} + 7 u - 36$

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 $a \cdot c = 4 \cdot - 36 = - 144$ and whose sum is $b = 7$ .

Factor $7$ out of $7 u$ .

$4 u^{2} + 7 (u) - 36$

Rewrite $7$ as $- 9$ plus $16$

$4 u^{2} + (- 9 + 16) u - 36$

Apply the distributive property.

$4 u^{2} - 9 u + 16 u - 36$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} - 9 u) + 16 u - 36$

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

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

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

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

Step 4

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

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

Step 5

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

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

Step 6

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

$({(2 x)}^{2} - 3^{2}) (x^{2} + 4)$

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 $b = 3$ .

$(2 x + 3) (2 x - 3) (x^{2} + 4)$

Factor 4x^4+7x^2-36

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