If (x+2) and (x-3) are factors of f(x)= x^4 -4x^3 -7x^2 +22x +24, write f(x) as a product of linear factors and find all roots. Having issues factoring this one. Please help.

Question

If (x+2) and (x-3) are factors of f(x)= x^4 -4x^3 -7x^2 +22x +24, write f(x) as a product of linear factors and find all roots.  Having issues factoring this one. Please help.

hero · Answer

1. Multiply (x + 2)(x - 3).
2. Divide f(x)/((x + 2)(x - 3))
3. Factor the remaining quadratic

anonymous · Answer

I don't understand. If I multiply (x+2)(x-3) I get x^2-x-6'  then what?

hero · Answer

Then divide f(x)/(x^2 - x - 6) :

$$\frac{f(x)}{x^2 - x - 6} = \frac{x^4 - 4x^3 - 7x^2 + 22x + 24}{x^2 - x - 6}$$

anonymous · Answer

Can I use synthetic division?

hero · Answer

It would probably be best to use polynomial division in this case. You could use synthetic division if you divided the factors one at a time.

hero · Answer

There's another method you could use, but it's not a popular method

anonymous · Answer

Let me see what I get using long division

hero · Answer

The method I refer to takes multiples of $x^2 - x - 6$:


$$ \frac{x^4 - 4x^3 - 7x^2 + 22x + 24}{x^2 - x - 6}  $$
$$ = \frac{x^4 - x^3 - 6x^2 -3x^3 -x^2 + 22x + 24}{x^2 - x - 6}$$
$$ = \frac{x^4 - x^2 - 6x^2}{x^2 - x - 6} - \frac{3x^3 + x^2 - 22x - 24}{x^2 - x - 6}$$
$$  = x^2 - \frac{3x^3 -3x^2 -18x + 4x^2 - 4x - 24}{x^2 - x - 6}$$
$$=x^2 - \frac{3x^3 - 3x^2 - 18x}{x^2 - x - 6} - \frac{4x^2 - 4x - 24}{x^2 - x - 6}$$
$$=x^2 - 3x - 1$$

hero · Answer

oops

$$x^2 - 3x - 4$$

hero · Answer

Should factor to (x - 4)(x + 1) which would be the last remaining factors of f(x)

anonymous · Answer

I can 't get the long division to work out right and not sure what. I am doing wrong  I get x ^2 then is have -3x^3-x^2+ 22x+ 24' then I get 3x with 2x^2 - 18x + 24 and. I'm stuck

hero · Answer

I'm not a big fan of polynomial division because anyone can easily get signs confused and such. You should probably stick with the synthetic division. Or you can try to learn my alternative division methodology.

anonymous · Answer

I'll go back and rework with synthetic division. Thanks for your help!

hero · Answer

Good luck @Nancyo

anonymous · Answer

I think I finally got it. You guys are a great help!!!! Happy Easter!

hero · Answer

I didn't help you with the synthetic division. You did that on your own. Great job