Is it possible to factor #y= x^2 + 4x -21 #? If so, what are the factors?

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Answer 1 · 2018-06-18T09:39:37

#(x+7)(x-3)#

#"the factors of - 21 which sum to + 4 are + 7 and - 3"#

#x^2+4x-21=(x+7)(x-3)#

Answer 2 · 2018-06-18T09:55:05

Yes: #(x+7)(x-3)#

You can do it solving: #x^2+4x-21=0#

#x^2+4x-21=0#

#a=1, b=4, c=-21#

#x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}#

#x=\frac{-(4)\pm \sqrt{(4)^2-4(1)(-21)}}{2(1)}#

#x=\frac{-4\pm \sqrt{16+84}}{2(1)}#

#x=\frac{-4\pm \sqrt{100}}{2}#

#x=\frac{-4\pm 10}{2}#
then
#x_1=\frac{-4+10}{2}=\frac{6}{2}=3#
#x_2=\frac{-4-10}{2}=\frac{-14}{2}=-7#
And

#x^2+bx+c=(x-x_1)(x-x_2)#
Or
#x^2+4x-21=(x-3)(x-(-7))#

#x^2+4x-21=(x-3)(x+7)#
Q.E.D.