Question

Operations with rational expressions

Answer 1

`$(50-2 w^(2))/(3 w^(2)+9 w-30) \cdot (w^(2)+5 w-14)/(6 w-30)=-(w+7)/(9)`

Solution:

Given expression:

`$(50-2 w^(2))/(3 w^(2)+9 w-30) \cdot (w^(2)+5 w-14)/(6 w-30)`

To solve the given expression:

First simplify:
`(50-2 w^(2))/(3 w^(2)+9 w-30)`

`$(50-2 w^(2))/(3 w^(2)+9 w-30)=-(2(w+5)(w-5))/(3(w-2)(w+5))`

Cancel the common factor (w + 5).

`$=-(2(w-5))/(3(w-2))`

Now substitute this in the given expression.

`$(50-2 w^(2))/(3 w^(2)+9 w-30) \cdot (w^(2)+5 w-14)/(6 w-30)=-(2(w-5))/(3(w-2)) \cdot (w^(2)+5 w-14)/(6 w-30)`

Multiply the fractions
`(a)/(b) \cdot (c)/(d)=(a \cdot c)/(b \cdot d)`

`$=-(2(w-5)\left(w^(2)+5 w-14\right))/(3(w-2)(6 w-30))`

Factor the denominator
`3(w-2)(6 w-30) =18(w-2)(w-5)`

`$=-(2(w-5)\left(w^(2)+5 w-14\right))/(18(w-2)(w-5))`

Cancel the common factor 2(w – 5).

`$=-(w^(2)+5 w-14)/(9(w-2))`

Factor the numerator
`w^(2)+5 w-14=(w-2)(w+7)`

`$=-((w-2)(w+7))/(9(w-2))`

Cancel the common factor (w – 2).

`$=-(w+7)/(9)`

`$(50-2 w^(2))/(3 w^(2)+9 w-30) \cdot (w^(2)+5 w-14)/(6 w-30)=-(w+7)/(9)`