# How do you simplify (x^2-9)/(x^2-16)*(x^2-8x+16)/(x^2+6x+9)?

Jul 23, 2015

Factor and cancel out common factors to find:

$\frac{{x}^{2} - 9}{{x}^{2} - 16} \cdot \frac{{x}^{2} - 8 x + 16}{{x}^{2} + 6 x + 9}$

$= \frac{\left(x - 3\right) \left(x - 4\right)}{\left(x + 3\right) \left(x + 4\right)}$

$= 1 - \frac{14 x}{\left(x + 3\right) \left(x + 4\right)}$

with exclusion $x \ne 4$

#### Explanation:

Use difference of squares identity:

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

Use perfect square trinomial identity:

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

$\frac{{x}^{2} - 9}{{x}^{2} - 16} \cdot \frac{{x}^{2} - 8 x + 16}{{x}^{2} + 6 x + 9}$

$= \frac{\left({x}^{2} - {3}^{2}\right) \left({x}^{2} - 2 \cdot 4 x + {4}^{2}\right)}{\left({x}^{2} - {4}^{2}\right) \left({x}^{2} + 2 \cdot 3 x + {3}^{2}\right)}$

$= \frac{\left(x - 3\right) \left(x + 3\right) {\left(x - 4\right)}^{2}}{\left(x - 4\right) \left(x + 4\right) {\left(x + 3\right)}^{2}}$

$= \frac{\left(x - 3\right) \left(x - 4\right)}{\left(x + 3\right) \left(x + 4\right)}$

$= \frac{{x}^{2} - 7 x + 12}{{x}^{2} + 7 x + 12}$

$= \frac{\left({x}^{2} + 7 x + 12\right) - 14 x}{{x}^{2} + 7 x + 12}$

$= 1 - \frac{14 x}{\left(x + 3\right) \left(x + 4\right)}$

with exclusion $x \ne 4$

Jul 23, 2015

$\frac{\left(x - 3\right) \left(x - 4\right)}{\left(x + 3\right) \left(x + 4\right)}$

#### Explanation:

$\frac{{x}^{2} - 9}{{x}^{2} - 16} . \frac{{x}^{2} - 8 x + 16}{{x}^{2} + 6 x + 9}$
$= \frac{\left(x + 3\right) \left(x - 3\right)}{\left(x + 4\right) \left(x - 4\right)} . {\left(x - 4\right)}^{2} / {\left(x + 3\right)}^{2}$
$= \frac{\left(x - 3\right) \left(x - 4\right)}{\left(x + 3\right) \left(x + 4\right)}$

Ideas:

1. ${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$
Thus, in the numerator in the fraction on the right, put $y = 4$ and in the denominator, put $y = 3$.

2. ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$
Again, like in the previous point, in the numerator in the fraction on the left, put $y = 3$ and in the denominator, put $y = 4$.

3. In the 3rd step, cancel out common terms, namely $\left(x - 4\right)$ and $\left(x + 3\right)$.