# Help with factoring cubic polynomials and find the zeros of the function

• May 2nd 2007, 05:37 AM
lilrhino
Help with factoring cubic polynomials and find the zeros of the function
Yikes! I spent hours on this problem w/o any result.

f(x) = x^3 - x^2 - 2x + 2 = 0

My first step was to plug in some numbers and see if any of them would result in a zero.

f(x) = 0

I plugged in 1 and it resulted in a zero. So I figured that (x+1) is one of my factors. I proceeded to do synthetic division and that's when things got crazy.

• May 2nd 2007, 06:46 AM
CaptainBlack
Quote:

Originally Posted by lilrhino
Yikes! I spent hours on this problem w/o any result.

f(x) = x^3 - x^2 - 2x + 2 = 0

My first step was to plug in some numbers and see if any of them would result in a zero.

f(x) = 0

I plugged in 1 and it resulted in a zero. So I figured that (x+1) is one of my factors. I proceeded to do synthetic division and that's when things got crazy.

(x-1) is a factor, so try:

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

The linear term on the right is then -a-2x, which should be equal to that
on the left, so a=0.

Now check by multiplying out the following:

(x-1)(x^2-2)

Now (x^2-2)=(x-sqrt(2))(x+sqrt(2), so:

x^3 - x^2 - 2x + 2 = (x-1)(x-sqrt(2))(x+sqrt(2)).

RonL
• May 2nd 2007, 07:16 AM
lilrhino
Quote:

Originally Posted by CaptainBlack
(x-1) is a factor, so try:

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

The linear term on the right is then -a-2x, which should be equal to that
on the left, so a=0.

Now check by multiplying out the following:

(x-1)(x^2-2)

Now (x^2-2)=(x-sqrt(2))(x+sqrt(2), so:

x^3 - x^2 - 2x + 2 = (x-1)(x-sqrt(2))(x+sqrt(2)).

RonL

Thanks RonL, but I'm still trying to comprehend the logic around "the linear term -a-2x should equal the term on the left, so a=0".

Can you provide a little more detail as to why this is used to solve and why the terms should equal one another on both sides?

It appears as if you found terms to eliminate the others so you could end with:

(x-1)(x^2-2)

I know you're a math whiz, what how would a novice like myself know to do that? What is the trigger to know that I should perform these particular steps? Sorry for all the questions, but I really want to understand this. Thanks again!
• May 2nd 2007, 07:59 AM
CaptainBlack
Quote:

Originally Posted by lilrhino
Thanks RonL, but I'm still trying to comprehend the logic around "the linear term -a-2x should equal the term on the left, so a=0".

Can you provide a little more detail as to why this is used to solve and why the terms should equal one another on both sides?

It appears as if you found terms to eliminate the others so you could end with:

(x-1)(x^2-2)

I know you're a math whiz, what how would a novice like myself know to do that? What is the trigger to know that I should perform these particular steps? Sorry for all the questions, but I really want to understand this. Thanks again!

Just multiply out the right hand side of:

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

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

Now the coefficients of like powers of x on both sides of this equation
must be equal, so in particular the coeficients of the linear terms are equal
so:

-2 = -2-a

and so a=0.

RonL
• May 2nd 2007, 08:41 AM
lilrhino
Quote:

Originally Posted by CaptainBlack
Just multiply out the right hand side of:

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

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

Now the coefficients of like powers of x on both sides of this equation
must be equal, so in particular the coeficients of the linear terms are equal
so:

-2 = -2-a

and so a=0.

RonL

I'll print this out and study it. Thank you so much for the detailed explanation.