Question 1170466: Factors of Polynomial Equations of Degree greater than 2
Give the factored form of the polynomial equations.
1.x^3-2x^2-5x+6=0
2.x^3-x^2-10x-8=0
give the factored for of
1.2x^4+5x^3-5x-2=0, given that one root is 1
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Once again tutor @MathLover1 has shown a method for factoring polynomials without showing HOW the factorization is done. It looks like magic and thus teaches the student nothing.
Perhaps one day she will provide a response to a similar question in which she explains how her method is actually done....
I will solve these three similar problems by two very different methods. The first is a formal process which is what will be taught in most resources; the second uses some powerful mathematical ideas but also involves some playing with numbers.
Here is the general process for the standard formal method:
(1) Use the rational roots theorem to find the possible rational roots;
(2) use substitution and/or synthetic division to find the actual roots;
(3) when the remaining polynomial has been reduced to second degree, solve by factoring, or by using the quadratic formula.
example 1: 
The possible rational roots are (plus or minus) 1, 2, 3, and 6.
It's always easy to check roots 1 and -1 by substituting. In this case, x=1 satisfies the equation, so 1 is a root and (x-1) is a factor of the polynomial. Remove that factor using synthetic division.
1 | 1 -2 -5 6
| 1 -1 -6
----------------
1 -1 -6 0
So
Then factoring the quadratic gives us the final factored form of the equation:
example 2: 
The possible rational roots are (plus or minus) 1, 2, 4, and 8.
In this case, direct substitution shows that x=1 does not satisfy the equation, but x=-1 does. So again, with x=-1 a root, remove the factor (x+1) using synthetic division.
-1 | 1 -1 -10 -8
| -1 2 8
----------------
1 -2 -8 0
So
And factoring the quadratic gives us the final factored form of the equation:
example 3: 
we are given that x=1 is a root, so remove the factor (x-1) using synthetic division.
1 | 2 5 0 -5 -2
| 2 7 7 2
-------------------
2 7 7 2 0
So
The possible rational roots are now (plus or minus) 1, 2, 1/2.
Direct substitution shows that x=-1 is a root, so remove the factor of (x+1) using synthetic division.
-1 | 2 7 7 2
| -2 -5 -2
------------
2 5 2 0
So
Factoring the quadratic then gives us the final factored form of the equation:
There are the solutions of the three examples using the standard process.
Now here are solutions using Vieta's Theorem, which tells us that, in a cubic polynomial equation
The sum of the three roots is -b/a and the product of the three roots is -d/a.
We can use this along with the rational roots theorem to play around with the possible rational roots to find the factored forms of the three example equations.
example 1:
The possible rational roots are (plus or minus) 1, 2, 3, and 6.
The sum of the three roots is 2 and the product is -6.
Playing around with the possible rational roots we can find the roots are 1, -2, and 3: (1)+(-2)+(3) = 2; (1)(-2)(3) = -6. Then with those roots the factored form of the equation is (as before,of course!)
example 2:
The possible rational roots are (plus or minus) 1, 2, 4, and 8.
The sum of the roots is 1 and the product of the three roots is 8. Again playing around with the possible rational roots, we can find the roots are -1, -2, and 4: (-1)+(-2)+(4) = 1; (-1)(-2)(4) = 8. Then with those roots the factored form, agreeing with the result above, is
example 3:
For this one, we will not repeat the process of removing the (x-1) factor corresponding to the given root of x=1. We will start with the cubic polynomial equation
The possible rational roots are again (plus or minus) 1, 2, 1/2.
The sum of the three roots is -7/2 and the product of the three roots is -1. This one is harder because of the fractions; but again playing around with the possible rational roots gives us the roots -1, -2, and -1/2. And that again gives us a factored form of the equation that agrees with our earlier result:
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