Factor 24ab-16a^2-9b^2

$24 a b - 16 a^{2} - 9 b^{2}$

Step 1

Add parentheses.

$24 (a b) - 16 a^{2} - 9 b^{2}$

Step 2

Let $u = a b$ . Substitute $u$ for all occurrences of $a b$ .

$24 u - 16 a^{2} - 9 b^{2}$

Step 3

Factor $- 1$ out of $24 u - 16 a^{2} - 9 b^{2}$ .

Move $24 u$ .

$- 16 a^{2} - 9 b^{2} + 24 u$

Factor $- 1$ out of $- 16 a^{2}$ .

$- (16 a^{2}) - 9 b^{2} + 24 u$

Factor $- 1$ out of $- 9 b^{2}$ .

$- (16 a^{2}) - (9 b^{2}) + 24 u$

Factor $- 1$ out of $24 u$ .

$- (16 a^{2}) - (9 b^{2}) - (- 24 u)$

Factor $- 1$ out of $- (16 a^{2}) - (9 b^{2})$ .

$- (16 a^{2} + 9 b^{2}) - (- 24 u)$

Factor $- 1$ out of $- (16 a^{2} + 9 b^{2}) - (- 24 u)$ .

$- (16 a^{2} + 9 b^{2} - 24 u)$

Step 4

Replace all occurrences of $u$ with $a b$ .

$- (16 a^{2} + 9 b^{2} - 24 (a b))$

Step 5

Remove parentheses.

$- (16 a^{2} + 9 b^{2} - 24 a b)$

Step 6

Factor using the perfect square rule.

Rearrange terms.

$- (16 a^{2} - 24 a b + 9 b^{2})$

Rewrite $16 a^{2}$ as ${(4 a)}^{2}$ .

$- ({(4 a)}^{2} - 24 a b + 9 b^{2})$

Rewrite $9 b^{2}$ as ${(3 b)}^{2}$ .

$- ({(4 a)}^{2} - 24 a b + {(3 b)}^{2})$

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

$24 a b = 2 \cdot (4 a) \cdot (3 b)$

Rewrite the polynomial.

$- ({(4 a)}^{2} - 2 \cdot (4 a) \cdot (3 b) + {(3 b)}^{2})$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 4 a$ and $b = 3 b$ .

$- {(4 a - 3 b)}^{2}$