Write as a Set of Linear Factors 8x^4-11x^2+3=0

$8 x^{4} - 11 x^{2} + 3 = 0$

Step 1

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

$8 {(x^{2})}^{2} - 11 x^{2} + 3 = 0$

Step 2

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

$8 u^{2} - 11 u + 3 = 0$

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 $a \cdot c = 8 \cdot 3 = 24$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 u$ .

$8 u^{2} - 11 u + 3 = 0$

Rewrite $- 11$ as $- 3$ plus $- 8$

$8 u^{2} + (- 3 - 8) u + 3 = 0$

Apply the distributive property.

$8 u^{2} - 3 u - 8 u + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(8 u^{2} - 3 u) - 8 u + 3 = 0$

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

$u (8 u - 3) - (8 u - 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $8 u - 3$ .

$(8 u - 3) (u - 1) = 0$

Step 4

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

$(8 x^{2} - 3) (x^{2} - 1) = 0$

Step 5

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

$(8 x^{2} - 3) (x^{2} - 1^{2}) = 0$

Step 6

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 $b = 1$ .

$(8 x^{2} - 3) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(8 x^{2} - 3) (x + 1) (x - 1) = 0$

Write as a Set of Linear Factors 8x^4-11x^2+3=0

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