1. Factor Theorum (algebra)

1. Use the factor theorem to factorise completely f(x)= x³+x²-4x-4.Hence solve the equation x³+x²-4x-4=0

f(x) = x³ + x² - 4x - 4

f(x) = (x³ + x²) - (4x + 4)

f(x) = x²(x + 1) - 4(x + 1)

f(x) = (x + 1)(x² - 4)

f(x) = (x + 1)(x² - 2²)

f(x) = (x + 1)(x + 2)(x - 2)

x³ + x² - 4x - 4 = 0

(x + 1)(x + 2)(x - 2) = 0

Hi I was wondering if someone can help me with this assignment question, I did it and handed it in to my tutor and this is the feedback he gave me. He said I haven't applied the factor theorum correctly and factorised it with out using factor theorum where have I gone wrong?

(These are my tutors comment)The start.Initially you had f(x) =x³+x²-4x-4.Then checking you find f(-1) = 0.Hence f(x)=(x+1)g(x) and work out what g is.And repeat for g(x).It's not the easiest way to do it. And your method would probably be the best way to go about it but the task says "Use the factor theorem".

2. Re: Factor Theorum (algebra)

You did not actually go wrong, but your method is not the one you are tested on. I'm not sure you understood what your tutor is saying. You need to use the factor theorem as demonstrated below:

Basically you should look for a root in your original polynomial. x³ + x² - 4x - 4

For example, $\displaystyle f(-1)=0$ as your tutor has noted, so $\displaystyle (x+1)$ is a factor of $\displaystyle f(x)$.

Now dividing $\displaystyle f(x)$ by $\displaystyle (x+1)$, you get

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

Set $\displaystyle g(x)=x^2-4$ and observe that $\displaystyle g(2)=0$. Therefore $\displaystyle (x-2)$ is a factor of $\displaystyle g(x)$. Dividing $\displaystyle g(x)$ by $\displaystyle x-2$ gives $\displaystyle x+2$. That is, $\displaystyle g(x)=(x-2)(x+2)$. Hence

$\displaystyle f(x)=(x+1)g(x)=(x+1)(x-2)(x+2)$

3. Re: Factor Theorum (algebra)

Originally Posted by Gusbob
You did not actually go wrong, but your method is not the one you are tested on. I'm not sure you understood what your tutor is saying. You need to use the factor theorem as demonstrated below:

Basically you should look for a root in your original polynomial. x³ + x² - 4x - 4

For example, $\displaystyle f(-1)=0$ as your tutor has noted, so $\displaystyle (x+1)$ is a factor of $\displaystyle f(x)$.

Now dividing $\displaystyle f(x)$ by $\displaystyle (x+1)$, you get

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

Set $\displaystyle g(x)=x^2-4$ and observe that $\displaystyle g(2)=0$. Therefore $\displaystyle (x-2)$ is a factor of $\displaystyle g(x)$. Dividing $\displaystyle g(x)$ by $\displaystyle x-2$ gives $\displaystyle x+2$. That is, $\displaystyle g(x)=(x-2)(x+2)$. Hence

$\displaystyle f(x)=(x+1)g(x)=(x+1)(x-2)(x+2)$
I still can't understand what my tutor wants me to do, Have I wrote my equations in the wrong format? How should the simplified expressions look?

4. Re: Factor Theorum (algebra)

Originally Posted by Andrew187
I still can't understand what my tutor wants me to do, Have I wrote my equations in the wrong format? How should the simplified expressions look?
Do you know what the factor theorem is? What your tutor wants you to do is exactly what I wrote down.

5. Re: Factor Theorum (algebra)

I'm no expert on factor theorum, but if you can kindly show me how the answer is meant to be laid out I would be very grateful.

6. Re: Factor Theorum (algebra)

Originally Posted by Andrew187
I'm no expert on factor theorum, but if you can kindly show me how the answer is meant to be laid out I would be very grateful.
I did. It was in my first reply.