Welcome to our community

Be a part of something great, join today!

kkittiee's question at Yahoo! Answers involving factoring a cubic polynomial

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello kkittiee,

We are given to factor:

$\displaystyle f(x)=x^3+11x^2+51x+41$

The rational roots theorem tells us that if this polynomial has any rational roots, they will come from the list:

$\displaystyle \pm(1,41)$

and in fact, we find:

$\displaystyle f(-1)=(-1)^3+11(-1)^2+51(-1)+41=0$

So, we know $\displaystyle x+1$ is a factor. Performing synthetic division, we find:

kkittiee.jpg

And so we know:

$\displaystyle f(x)=x^3+11x^2+51x+41=(x+1)(x^2+10x+41)$

The quadratic factor is prime, so we are done.