Can #y= x^2+7x-30 # be factored? If so what are the factors ?

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Answer 1 · 2016-01-04T00:16:39

#x^2+7x-30 = (x+10)(x-3)#

Find a pair of factors of #30# which multiply to give #30# and differ by #7#. The pair #10#, #3# works, hence:

#x^2+7x-30 = (x+10)(x-3)#

#x^2+7x-30# is in the form #ax^2+bx+c# with #a=1#, #b=7# and #c=-30#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 7^2-(4xx1xx-30) = 49+120 = 169 = 13^2#

Since this is a perfect square, the quadratic has two linear factors with rational coefficients.

Rather than search for a suitable pair of factors of #30# we could use the quadratic formula to find the zeros of #x^2+7x-30# and hence its factors:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)#

#=(-7+-13)/2#

That is #x=-10# or #x=3#

Hence factors #(x+10)# and #(x-3)#