SOLUTION: x^3-8x^2+17x-4=0
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Question 756034: x^3-8x^2+17x-4=0
Answer by tommyt3rd(5050) (Show Source): You can put this solution on YOUR website!
x^3-8x^2+17x-4=
(x^3-8x^2+16x)+(x-4)=
x(x^2-8x+16)+(x-4)=
x(x-4)^2+(x-4)=
(x-4)[x(x-4)+1]=
(x-4)(x^2-4x+1)=
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