SOLUTION: How do I factor x^3-4x^2+x+6 to find out the solutions? I was given that x=-2 is one of the solutions. The factoring is what is giving me trouble.

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Question 1190368: How do I factor x^3-4x^2+x+6 to find out the solutions?
I was given that x=-2 is one of the solutions.
The factoring is what is giving me trouble.
Found 3 solutions by MathLover1, greenestamps, josgarithmetic:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!


if given that is one of the solutions, means = is one factor
so, use long division to see if you can factor it completely

---------(
|
----------.........subtract
-------------....bring down
-------------
-------------........subtract
-------------------...bring down
-------------------
-------------------.......subtract
----------------------->reminder




Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The other tutor went through the motions without looking at what the problem says. She found that dividing the given polynomial by x+2 leaves a remainder, which means x=-2 is NOT a zero (root) of the polynomial.

So the problem as posted is defective.

Re-post showing the correct information....

And, as you are requested to do when submitting your question, show the work you have tried to do on the problem, or at least tell us exactly what it is about the problem that you are having trouble with. Saying "The factoring is what is giving me trouble." tells us nothing.
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
You want the roots or the zeros?
x^3-4x^2+x+6
x^3+x-4x^2+6
x^3+x-(4x^2-6)
x^3+x-2(2x^2-3)
No, this did not help.

Look at Rational Roots Theorem.
Try -2 and +2. 
2   |   1   -4   1   6
    |
    |       2   -4   -6
    |-------------------------
        1   -2   -3   0

The result means one factor is and other factor
which is composed of and .

Your factorization is .




(testing for -2 did not give a zero remainder.)
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