Finite Math Examples

$a^{4} b^{4} + 4 a^{2} b^{2} - 96$

Step 1

Rewrite $a^{2} b^{2}$ as ${(a b)}^{2}$ .

${(a b)}^{4} + 4 {(a b)}^{2} - 96$

Step 2

Rewrite ${(a b)}^{4}$ as ${({(a b)}^{2})}^{2}$ .

${({(a b)}^{2})}^{2} + 4 {(a b)}^{2} - 96$

Step 3

Let $u = {(a b)}^{2}$ . Substitute $u$ for all occurrences of ${(a b)}^{2}$ .

$u^{2} + 4 u - 96$

Step 4

Factor $u^{2} + 4 u - 96$ 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 $- 96$ and whose sum is $4$ .

$- 8, 12$

Step 4.2

Write the factored form using these integers.

$(u - 8) (u + 12)$

Step 5

Replace all occurrences of $u$ with ${(a b)}^{2}$ .

$({(a b)}^{2} - 8) ({(a b)}^{2} + 12)$

Step 6

Simplify.

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Step 6.1

Apply the product rule to $a b$ .

$(a^{2} b^{2} - 8) ({(a b)}^{2} + 12)$

Step 6.2

Apply the product rule to $a b$ .

$(a^{2} b^{2} - 8) (a^{2} b^{2} + 12)$