Solve a^2-b^2+6b-9=

$$a ^ { 2 } - b ^ { 2 } + 6 b - 9 =$$

Factor

$$\left(a+b-3\right)\left(a-b+3\right)$$

Solution Steps

Rewrite as difference of two terms.

$$a^{2}-\left(b^{2}-6b+9\right)$$

Rewrite the terms as squares.

$$a^{2}-\left(3-b\right)^{2}$$

The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-3+b\right)\left(a+3-b\right)$$

Evaluate

$$-\left(b-3\right)^{2}+a^{2}$$