How do you simplify #\frac { - 5y ^ { 2} + 20} { y ^ { 2} - 5y - 14}#?

1 Answer
Somebody N.
Sep 20, 2017

See below.

Explanation:

First factor the denominator. Notice that we are looking for 2 numbers whose product is -14 and whose sum is -5. This will be -7 and 2 .i.e 2 x -7 = -14 and 2 + (-7) = -5.

So we have:

#(y +2 )(y - 7)#

Factor out -5 from the numerator:

#-5(y^2 - 4)# this can be expressed as #-5(y^2 - 2^2)#

#(y^2 - 2^2)# is the difference of 2 squares. This is equal to the product of sum and difference of the 2 numbers i.e.

#(y^2 - 2^2) -= (y + 2 )(y - 2 )#

So our numerator is now:

#-5(y + 2 )( y - 2 )#

So we get:

#(-5(y + 2 )( y - 2 ))/((y + 2 )( y - 7 ))#

Cancelling like factors:

#(-5cancel((y + 2 ))( y - 2 ))/(cancel((y + 2 ))( y - 7 ))#

#(-5(y - 2 ))/(y - 7 ) -> (-5y + 2 )/(y - 7 )#

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