Asked by Vic
Factor f(x)=x^4+x^3-8x^2+6x+36 completely
Answers
Answered by
drwls
If there is a rational root, it would be +or- 1,2,3,4,6,8 or 12.
x = -2 works, so x+2 is a factor.
So, after one factoring step, and a bit of polynomial long division, you have
(x+2)(x^3 -x^2 -6x +18) = 0
Next, use the rational root theorm on the cubic factor. You will find that -3 is a root, making (x+3) a factor.
So now you have
x^4+x^3-8x^2+6x+36
(x+2)(x+3)(x^2 -4x +6)
The remaining two factors are complex conjugates.
x = -2 works, so x+2 is a factor.
So, after one factoring step, and a bit of polynomial long division, you have
(x+2)(x^3 -x^2 -6x +18) = 0
Next, use the rational root theorm on the cubic factor. You will find that -3 is a root, making (x+3) a factor.
So now you have
x^4+x^3-8x^2+6x+36
(x+2)(x+3)(x^2 -4x +6)
The remaining two factors are complex conjugates.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.