SOLUTION: find the remaining factors of this polynomial. factor completely. 6x^3-25x^2+2x+8;(2x+1) is a factor

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find the remaining factors of this polynomial. factor completely. 6x^3-25x^2+2x+8;(2x+1) is a factor      Log On


   

Question 725566: find the remaining factors of this polynomial. factor completely. 6x^3-25x^2+2x+8;(2x+1) is a factor
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If 2x+1 is a factor of the polynomial then it will divide evenly into the polynomial. Not only that but the quotient will be the other factor. So we'll start by dividing the polynomial by 2x+1.

We could use long division. And since synthetic division is designed for division by an expression of the form x-r it would seem that we must use long division. But I'm going to show you how to use synthetic division. First we must transform 2x+1 to fit the x-r pattern/ Factoring out a 2 we get:
2%28x%2B1%2F2%29
And then rewriting it as a subtraction:
2%28x-%28-1%2F2%29%29
In this form, we can see that if 2x+1 is a factor, so will 2 and (x-(-1/2)). We can divide by the (x-(-1/2)) factor using synthetic division:
-1/2 |   6   -25   2   8
------        -3  14  -8
        -----------------
         6   -28  16   0
The quotient, "6 -28 16", translates into 6x%5E2-28x%2B16. So far we have:
6x%5E3-25x%5E2%2B2x%2B8+=+%28x-%28-1%2F2%29%29%286x%5E2-28x%2B16%29
The second factor has the factor of 2 we already knew about. Now we have:
6x%5E3-25x%5E2%2B2x%2B8+=+%28x-%28-1%2F2%29%29%282%29%283x%5E2-14x%2B8%29
"Re-combining" the first two factors we get:
6x%5E3-25x%5E2%2B2x%2B8+=+%282x%2B1%29%283x%5E2-14x%2B8%29
Now we finish factoring:
6x%5E3-25x%5E2%2B2x%2B8+=+%282x%2B1%29%283x%5E2-14x%2B8%29+=+%282x%2B1%29%283x-2%29%28x-4%29