Algebra Examples

$4 x^{4} + 7 x^{2} - 2 = 0$

Step 1

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

$4 {(x^{2})}^{2} + 7 x^{2} - 2 = 0$

Step 2

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

$4 u^{2} + 7 u - 2 = 0$

Step 3

Factor by grouping.

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Step 3.1

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 $a \cdot c = 4 \cdot - 2 = - 8$ and whose sum is $b = 7$ .

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Step 3.1.1

Factor $7$ out of $7 u$ .

$4 u^{2} + 7 (u) - 2 = 0$

Step 3.1.2

Rewrite $7$ as $- 1$ plus $8$

$4 u^{2} + (- 1 + 8) u - 2 = 0$

Step 3.1.3

Apply the distributive property.

$4 u^{2} - 1 u + 8 u - 2 = 0$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) + 8 u - 2 = 0$

Step 3.2.2

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

$u (4 u - 1) + 2 (4 u - 1) = 0$

Step 3.3

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

$(4 u - 1) (u + 2) = 0$

Step 4

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

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

Step 5

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

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

Step 6

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

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

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 $b = 1$ .

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