Question

How do you multiply and simplify `\frac { x ^ { 2} - 7x - 30} { x ^ { 2} - 4x - 21} \cdot \frac { 14- 2x } { x ^ { 2} - 19x + 90}`?

Answer 1

`(2(7-x))/((x-7)(x-9))`

Explanation:

`\frac { x ^ { 2} - 7x - 30} { x ^ { 2} - 4x - 21} \cdot \frac { 14- 2x } { x ^ { 2} - 19x + 90}`

first factor each term:

`((x+3)(x-10))/((x+3)(x-7))*(2(7-x))/((x-9)(x-10))`

Cancel terms:

`(cancel((x+3))cancel((x-10)))/(cancel((x+3))(x-7))*(2(7-x))/((x-9)cancel((x-10)))`

`(2(7-x))/((x-7)(x-9))`

Answer 2

`-2/(x-9)`

Explanation:

For the second-degree terms, we need to think of two numbers that sum up to the middle term and have a product of the last term. These will become our factors.

`((color(blue)(x^2-7x-30))/color(darkviolet)(x^2-4x-21))*((14-2x)/(color(lime)(x^2-19x+90)))`

`color(blue)((x-10)(x+3))/(color(darkviolet)((x-7)(x+3)))*(14-2x)/(color(lime)((x-10)(x-9)))`

We can factor a `2` out of the black term, which will leave us with

`((x-10)(x+3))/((x-7)(x+3))*(2(7-x))/((x-10)(x-9))`

Like terms on the top and bottom will cancel. We're left with

`(cancel((x-10))cancel((x+3)))/((x-7)cancel((x+3)))*(2(7-x))/(cancel((x-10))(x-9))`

`1/(x-7)*(2(7-x))/(x-9)`

Which simplifies to

`(2color(orange)((7-x)))/((x-7)(x-9))`

`(2*color(orange)(-1(x-7)))/((x-7)(x-9))`

`(2*color(orange)(-1cancel((x-7))))/(cancel((x-7))(x-9))`

`=>-2/(x-9)`