SOLUTION: Factor the Polynomial: x^3+7x^2+15x+9 I don't even know where to start, however I made an attempt, by trying to use the factor by grouping method which all I got is: (x^3+7x^2)+

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the Polynomial: x^3+7x^2+15x+9 I don't even know where to start, however I made an attempt, by trying to use the factor by grouping method which all I got is: (x^3+7x^2)+      Log On


   



Question 684000: Factor the Polynomial:
x^3+7x^2+15x+9
I don't even know where to start, however I made an attempt, by trying to use the factor by grouping method which all I got is:
(x^3+7x^2)+(15x+9)
x^2(x+7)+3(5x+3)
and that's were I stopped because I know that two of them should be the same
-thank you in advance :D

Found 2 solutions by rapaljer, MathLover1:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This one does NOT work by grouping! You need to use Synthetic Division. Start by looking at the LAST term, which is 9. You need to find a number that is a factor of 9, and do synthetic division by that number until you get a remainder of 0. I can see that one that works here is -1, maybe others as well.

I don't know exactly how to lay out a synthetic division problem on this website, but I have quite a few examples on my website in Intermediate Algebra or College Algebra.

To see my website go to www.mathinlivingcolor.com, named after my theme "Math in Living Color." Look for a single link near the bottom of this page that will take you to my Homepage. Once on my Homepage, for explanation of Sythetic Division, look for "Basic, Intermediate and College Algebra: One Step at a Time", and select "Intermediate Algebra". Look in "Chapter 2" for "Section 2.10" page 232! College Algebra Chapter 3 has additional explanation about using Synthetic Division in order to factor polynomials. Send me an email if you need additional help with this one.

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida

www.mathinlivingcolor.com.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3%2B7x%5E2%2B15x%2B9.......write 7x%5E2 as 6x%5E2%2Bx%5E2 and 15x as 9x%2B6x

x%5E3%2B6x%5E2%2B9x%2Bx%5E2%2B6x%2B9...group
%28x%5E3%2Bx%5E2%29%2B%286x%5E2%2B6x%29%2B%289x%2B9%29
x%5E2%28x%2B1%29%2B6x%28x%2B1%29%2B9%28x%2B1%29
%28x%2B1%29+%28x%5E2%2B6x%2B9%29
%28x%2B1%29+%28x%2B3%29%5E2
%28x%2B1%29+%28x%2B3%29+%28x%2B3%29