Welcome to our community

Be a part of something great, join today!

jesusluvsponies's question at Yahoo! Answers (Real and rational roots)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

I have no idea how to do these, I missed lecture because I had the flu.
Can you please explain? I have an exam and this is part of the material covered. Thanks!!!

Find all rational roots of the polynomial
6x^3 + 7x^2 + 2x -10

Find all real roots of the polynomial
x^3 + x^2 -8x -8

Factor the polynomial as a product of linear factors and a factor g(x) such that g(x) is either a constant or a polynomial that has no rational roots.

x^5 -2^4 +2x^3 -3x +2


Thank you so much!!!!!
Here is a link to the question:

Polynomials, please help 10 points!!!!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello jesusluvsponies,

We'll use the following theorem:

Rational root theorem - Wikipedia, the free encyclopedia

$(a)\;p(x)=6x^3 + 7x^2 + 2x -10$.

In this case, $p=\pm 1,\pm 2,\pm 5$ and $q=\pm 1,\pm 2,\pm3,\pm 6$. Substituting we get $p(5/6)=0$. Using the algorithm of Ruffini we get

$p(x)=(6x-5)(x^2+2x-2)$

But $x^2+2x-2$ has no real roots, so the only rational root of $p(x)$ is

$\boxed{\;x=5/6\;}$

$(b)\;q(x)=x^3 + x^2 -8x -8$.

In this case, $q(-1)=0$. Using the algorithm of Ruffini we get

$q(x)=(x+1)(x^2-8)$

and the real roots of $x^2-8$ are $\pm 2\sqrt{2}$, so the real roots of $q(x)$ are

$\boxed{\;x=-1,x=\pm 2\sqrt{2}\;}$

$(c)\;r(x)=x^5 -2x^4 +2x^3 -3x +2 $.

In this case, $r(1)=0$. Using the algorithm of Ruffini we get

$r(x)=(x-1)(x^4-x^3+x^2+x-2)$

But $x=1$ is a real root of $x^4-x^3+x^2+x-2$ wich implies (again Ruffini)

$x^4-x^3+x^2+x-2=(x-1)(x^3+x+2)$

But $x=-1$ is a root of $x^3+x+2$ wich implies (again Ruffini)

$x^3+x+2=(x+1)(x^2-x+2)$

But $x^2-x+2$ has no real roots, so

$\boxed{\;r(x)=(x-1)^2(x+1)(x^2-x+2)\;}$