Precalculus Examples

$f (x) = 4 x^{4} - 24 x^{3} + 35 x^{2} + 6 x - 9$

Step 1

Regroup terms.

$f (x) = - 24 x^{3} + 6 x + 4 x^{4} + 35 x^{2} - 9$

Step 2

Factor $- 6 x$ out of $- 24 x^{3} + 6 x$ .

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Step 2.1

Factor $- 6 x$ out of $- 24 x^{3}$ .

$f (x) = - 6 x (4 x^{2}) + 6 x + 4 x^{4} + 35 x^{2} - 9$

Step 2.2

Factor $- 6 x$ out of $6 x$ .

$f (x) = - 6 x (4 x^{2}) - 6 x \cdot - 1 + 4 x^{4} + 35 x^{2} - 9$

Step 2.3

Factor $- 6 x$ out of $- 6 x (4 x^{2}) - 6 x (- 1)$ .

$f (x) = - 6 x (4 x^{2} - 1) + 4 x^{4} + 35 x^{2} - 9$

Step 3

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

$f (x) = - 6 x ({(2 x)}^{2} - 1) + 4 x^{4} + 35 x^{2} - 9$

Step 4

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

$f (x) = - 6 x ({(2 x)}^{2} - 1^{2}) + 4 x^{4} + 35 x^{2} - 9$

Step 5

Factor.

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Step 5.1

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 $b = 1$ .

$f (x) = - 6 x ((2 x + 1) (2 x - 1)) + 4 x^{4} + 35 x^{2} - 9$

Step 5.2

Remove unnecessary parentheses.

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 x^{4} + 35 x^{2} - 9$

Step 6

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 {(x^{2})}^{2} + 35 x^{2} - 9$

Step 7

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 u^{2} + 35 u - 9$

Step 8

Factor by grouping.

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Step 8.1

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 $a \cdot c = 4 \cdot - 9 = - 36$ and whose sum is $b = 35$ .

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Step 8.1.1

Factor $35$ out of $35 u$ .

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 u^{2} + 35 (u) - 9$

Step 8.1.2

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 u^{2} + (- 1 + 36) u - 9$

Step 8.1.3

Apply the distributive property.

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u + 36 u - 9$

Step 8.2

Factor out the greatest common factor from each group.

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Step 8.2.1

Group the first two terms and the last two terms.

$f (x) = - 6 x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u + 36 u - 9$

Step 8.2.2

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + u (4 u - 1) + 9 (4 u - 1)$

Step 8.3

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + (4 u - 1) (u + 9)$

Step 9

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + (4 x^{2} - 1) (x^{2} + 9)$

Step 10

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + ({(2 x)}^{2} - 1) (x^{2} + 9)$

Step 11

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

$f (x) = - 6 x (2 x + 1) (2 x - 1) + ({(2 x)}^{2} - 1^{2}) (x^{2} + 9)$

Step 12

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 $b = 1$ .

$f (x) = - 6 x (2 x + 1) (2 x - 1) + (2 x + 1) (2 x - 1) (x^{2} + 9)$

Step 13

Factor $(2 x + 1) (2 x - 1)$ out of $- 6 x (2 x + 1) (2 x - 1) + (2 x + 1) (2 x - 1) (x^{2} + 9)$ .

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Step 13.1

Factor $(2 x + 1) (2 x - 1)$ out of $- 6 x (2 x + 1) (2 x - 1)$ .

$f (x) = (2 x + 1) (2 x - 1) (- 6 x) + (2 x + 1) (2 x - 1) (x^{2} + 9)$

Step 13.2

Factor $(2 x + 1) (2 x - 1)$ out of $(2 x + 1) (2 x - 1) (- 6 x) + (2 x + 1) (2 x - 1) (x^{2} + 9)$ .

$f (x) = (2 x + 1) (2 x - 1) (- 6 x + x^{2} + 9)$

Step 14

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$f (x) = (2 x + 1) (2 x - 1) (- 6 u + u^{2} + 9)$

Step 15

Factor using the perfect square rule.

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Step 15.1

Rearrange terms.

$f (x) = (2 x + 1) (2 x - 1) (u^{2} - 6 u + 9)$

Step 15.2

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

$f (x) = (2 x + 1) (2 x - 1) (u^{2} - 6 u + 3^{2})$

Step 15.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Step 15.4

Rewrite the polynomial.

$f (x) = (2 x + 1) (2 x - 1) (u^{2} - 2 \cdot u \cdot 3 + 3^{2})$

Step 15.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

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

Step 16

Replace all occurrences of $u$ with $x$ .

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