Precalculus Examples

${(a^{2} + 4 a)}^{2} - (a^{2} + 4 a) - 20$

Step 1

Let $u = a^{2} + 4 a$ . Substitute $u$ for all occurrences of $a^{2} + 4 a$ .

$u^{2} - u - 20$

Step 2

Factor $u^{2} - u - 20$ using the AC method.

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Step 2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 2.2

Write the factored form using these integers.

$(u - 5) (u + 4)$

Step 3

Replace all occurrences of $u$ with $a^{2} + 4 a$ .

$(a^{2} + 4 a - 5) (a^{2} + 4 a + 4)$

Step 4

Factor $a^{2} + 4 a - 5$ using the AC method.

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Step 4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 4.2

Write the factored form using these integers.

$(a - 1) (a + 5) (a^{2} + 4 a + 4)$

Step 5

Factor using the perfect square rule.

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Step 5.1

Rewrite $4$ as $2^{2}$ .

$(a - 1) (a + 5) (a^{2} + 4 a + 2^{2})$

Step 5.2

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

$4 a = 2 \cdot a \cdot 2$

Step 5.3

Rewrite the polynomial.

$(a - 1) (a + 5) (a^{2} + 2 \cdot a \cdot 2 + 2^{2})$

Step 5.4

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

$(a - 1) (a + 5) {(a + 2)}^{2}$