How do you multiply and simplify $\frac{1}{x + 2} \cdot \frac{x^{2} - 9 x + 20}{4 - x}$ $\frac { 1} { x + 2} \cdot \frac { x ^ { 2} - 9x + 20} { 4- x }$ ?

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Answer 1 · 2017-03-18T15:20:36

$\frac{5 - x}{x + 2}$ $(5-x)/(x+2)$

You would factor $x^{2} - 9 x + 20 = (x - 5) (x - 4)$ $x^2-9x+20=(x-5)(x-4)$ and substitute it in the expression to get:

$\frac{1}{x + 2} \cdot \frac{(x - 5) (x - 4)}{4 - x}$ $1/(x+2)*((x-5)(x-4))/(4-x)$

Since $\frac{x - 4}{4 - x} = - 1$ $(x-4)/(4-x)=-1$ , you get:

$1/(x+2)*((x-5)cancel((x-4))^-1)/cancel((4-x))$

$=(5-x)/(x+2)$

How do you multiply and simplify \frac { 1} { x + 2} \cdot \frac { x ^ { 2} - 9x + 20} { 4- x }1x+2⋅x2−9x+204−x\frac { 1} { x + 2} \cdot \frac { x ^ { 2} - 9x + 20} { 4- x }?