# Thread: How do you factor x^3+7x^2+8x-16?

1. ## How do you factor x^3+7x^2+8x-16?

How do you factor x^3+7x^2+8x-16?

Thanks.

2. ## Re: How do you factor x^3+7x^2+8x-16?

First thing to do is find factors of -16. Can you list them?

3. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by pickslides
First thing to do is find factors of -16. Can you list them?
-16 has the factors of the following: 2, 4, 8, 16.

4. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by david213
-16 has the factors of the following: 2, 4, 8, 16.
Also, -1,1,-2,-4,-8 and -16.

Now substitute these values into the equation like this

$x=2: 2^3+7(2)^2+8(2)-16 = 36$

what are the other results?

5. ## Re: How do you factor x^3+7x^2+8x-16?

I kinda have a clue right now. So you are trying to get the number that will make the equation equals to zero? I plugged all these numbers into my calculator, and it turns out that 1 will make the equation equals to zero.

6. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by david213
I kinda have a clue right now. So you are trying to get the number that will make the equation equals to zero? I plugged all these numbers into my calculator, and it turns out that 1 will make the equation equals to zero.
Yes, so since f(1) = 0, that means (x - 1) is a factor. So now long divide...

7. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by david213
I kinda have a clue right now. So you are trying to get the number that will make the equation equals to zero?
Yep!

The factor theorem says: For $f(x)$, if $f(a)=0$ then $x-a$ is a factor.

Originally Posted by david213
I plugged all these numbers into my calculator, and it turns out that 1 will make the equation equals to zero.
Great, so therefore $(x-1)$ is a factor.

Now find $\displaystyle \frac{x^3+7x^2+8x-16}{x-1}= \dots$ and you are done!

8. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by pickslides
Yep!

The factor theorem says: For $f(x)$, if $f(a)=0$ then $x-a$ is a factor.

Great, so therefore $(x-1)$ is a factor.

Now find $\displaystyle \frac{x^3+7x^2+8x-16}{x-1}= \dots$ and you are done!
ALMOST done, you probably still have to factorise the resulting quadratic factor...

9. ## Re: How do you factor x^3+7x^2+8x-16?

Okay, so, after I did the long division, I got x^2+8x+16. Thus, the the factored form of x^3+7x^2+8x-16 is (x-1)(x+4)^2?

10. ## Re: How do you factor x^3+7x^2+8x-16?

Well done, you can expand it back out to check.

11. ## Re: How do you factor x^3+7x^2+8x-16?

Originally Posted by pickslides
First thing to do is find factors of -16. Can you list them?
Youn should provide the reason for this, though perhaps asking the OP to tell us what they have tried should also be considered?

- "The rational root theorem tells you that if this has any "nice" roots/factor that the roots are factors of -16"

CB