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factoring a 5th degree polynomial **10pts**?
I am trying to factor this polynomial but can't get the right answer.
(x^5)-(3x^4)-(4x^3)+(28x^2)-(37x)+15
I thought (x-5)*(x+1)*(x-1)*(x-1)*(x+3) was the correct answer, but then when I multiplied it out it was not right. I am multiplying wrong or do I have the wrong answer?
Help please!! Anyone who can tell me what the correct answers are will get 10pts!!!
1 Answer
- PaulaLv 71 decade agoFavorite Answer
OK, the only way I know how to do these is to try some simle numbers:
x=1 => 1 - 3 - 4 + 28 - 37 + 15 = 0
So that's a good start, (x-1) is a factor.
=> f(x) = (x-1)(x^4 - 2x^3 - 6x^2 + 22x - 15)
Now look at (x^4 - 2x^3 - 6x^2 + 22x - 15)
x=1 is a factor again, so...
f(x) = (x-1)(x-1)(x^3 - x^2 - 7x + 15)
Now look at (x^3 - x^2 - 7x + 15)
Is x=1 a factor... no.
Is x=-1 a factor... no
Is x=2 a factor... no, in fact integer factors need to be odd
Is x=3 a factor...no
is x=-3 a factor... yes!
(x^3 - x^2 - 7x + 15) = (x+3)(x^2 - 4x + 5)
and now checking the quadratic equation (which you must know if you're doing 5th degree polynomials), (x^2 - 4x + 5) has no real factors, so we can't go any further (unless we do imaginary factors).
=> f(x) = (x-1)(x-1)(x+3)(x^2 - 4x + 5)