Is it possible to factor #y=x^2-x-56 #? If so, what are the factors?

1 Answer
Mar 5, 2016

Answer:

#(x+7)(x-8)#

Explanation:

For finding whether or not #ax^2+bx+c=0# can be factorized or not, we find the value #b^2-4ac#.

As the equation is #x^2−x−56#, #b^2-4ac=1^2-4xx1xx(-56)=1+224=225#. As it is a perfect square, it should be possible to factorize it.

Hence for factorizing, we split #ac# (the product of coefficient of #x^2# and independent term into two factors, whose sum is #b#, the coefficient of #x#.

In the polynomial #x^2−x−56#, the product is #-56# and hence factors whose sum is #-1#, these would be #-8# and #7#. Hence splitting middle term this way, we get

#x^2−x−56=x^2−8x+7x−56# i.e.

#x(x-8)+7(x-8)# or

#(x+7)(x-8)#