How do I factor out a constant before factoring out a quadratic? Factor completely: 3w^2-3w-90

1 answer:

3 0

$3w^2-3w-90 =0 \ \ \:3 \\ \\w^2- w-30 =0 \\ \\a=1 , \ b= -1 , c= -30 \\ \\\Delta =b^2-4ac = (-1)^2 -4\cdot1\cdot (-30) =1+120=121 \\ \\w_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{1-\sqrt{121}}{2 }=\frac{ 1-11}{2}=\frac{-10}{2}=-5$

$w_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{1+\sqrt{121}}{2 }=\frac{ 1+11}{2}=\frac{12}{2}=6 \\ \\ Answer:\\ \\ w^2- w-30 =(x+5)(x-6)$

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