SOLUTION: There are four answers using the quartic equation to this question p(x)=x^5-x^4-13x^3+13x^2+36x-36 has five real roots. One of them is x=1.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: There are four answers using the quartic equation to this question p(x)=x^5-x^4-13x^3+13x^2+36x-36 has five real roots. One of them is x=1.       Log On


   

Question 18439: There are four answers using the quartic equation to this question p(x)=x^5-x^4-13x^3+13x^2+36x-36 has five real roots. One of them is x=1.
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Factor
+x%5E5-x%5E4-13x%5E3%2B13x%5E2%2B36x-36+
= +x%5E4+%28x-1%29+-+13x%5E2%28x-1%29+%2B+36%28x-1%29
= %28x-1%29+%28x%5E4+-+13x%5E2+%2B+36%29
= %28x-1%29+%28x%5E2+-4%29%28x%5E2+-9%29
= +%28x-1%29+%28x-2%29%28x%2B2%29+%28x-3%29%28x%2B3%29+
So, the other four roots are
....
Kenny