How do you multiply and simplify \frac { x ^ { 2} - 7x - 30} { x ^ { 2} - 4x - 21} \cdot \frac { 14- 2x } { x ^ { 2} - 19x + 90}x2−7x−30x2−4x−21⋅14−2xx2−19x+90?

2 Answers
Bill Jorgensen
Jun 22, 2018

(2(7-x))/((x-7)(x-9))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}x2−7x−30x2−4x−21⋅14−2xx2−19x+90

first factor each term:

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

Jacobi J.
Jun 22, 2018

-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)

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