Factor 4x^4+35x^2-9=0

$4 x^{4} + 35 x^{2} - 9 = 0$

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

$4 {(x^{2})}^{2} + 35 x^{2} - 9 = 0$

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

$4 u^{2} + 35 u - 9 = 0$

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 $b = 35$ .

Factor $35$ out of $35 u$ .

$4 u^{2} + 35 (u) - 9 = 0$

Rewrite $35$ as $- 1$ plus $36$

$4 u^{2} + (- 1 + 36) u - 9 = 0$

Apply the distributive property.

$4 u^{2} - 1 u + 36 u - 9 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) + 36 u - 9 = 0$

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

$u (4 u - 1) + 9 (4 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u + 9) = 0$

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

$(4 x^{2} - 1) (x^{2} + 9) = 0$

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

$({(2 x)}^{2} - 1) (x^{2} + 9) = 0$

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

$({(2 x)}^{2} - 1^{2}) (x^{2} + 9) = 0$

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 $b = 1$ .

$(2 x + 1) (2 x - 1) (x^{2} + 9) = 0$

Factor 4x^4+35x^2-9=0

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