How do you factor #(x^2-14x+49)/(x^2-2x-35)#?

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Answer 1 · 2016-06-30T12:53:05

#(x-7)/(x+5)#

The numerator is a squared bynomial:

#x^2-14x+49=(x-7)^2#

the denominator is similar to a trynomial of the type

#x^2+sx+p#

where you can find two numbers whose sum is s and whose product is p, in this case the numbers are -7 and 5 because

-7+5=-2 and -7*5=-35

so

#x^2-2x-35=(x-7)(x+5)#

now you can rewrite the fraction:

#(x-7)^cancel2/(cancel(x-7)(x+5))#