# Thread: Polynomial Problem and not a clue how to proceed

1. ## Polynomial Problem and not a clue how to proceed

I'm on my last problem for my lesson and I'm a little stumped and need a push in the right direction.

Problem: "One factor of $\displaystyle 4x^3+15x^2-31x-30$ is x-2. Find the other factors."

I don't want the answer but if someone could explain to me how they even got x-2 as a factor I should be able to figure the rest out or if someone knows of a lesson that explains this that would be helpful as well.

Thanks for looking!

2. "How they got" that factor isn't (per se) relevant to how you should proceed, but I'll tell you anyway
There's a result known as the "rational root theorem" which gives us a finite amount of possible linear factors to check.
There's also calculators, which can crank out solutions (factors) in no time.

The big picture is:
a) you know how to factor/solve quadratics
b) this is a cubic
c) dividing a cubic by a (given) linear factor will produce a quadratic, which you know how to factor!

So we proceed with synthetic division. Set x - 2 = 0 to get x = 2 as the number "in the box"
[2] 4 15 -31 -30
......... 8_ 46_ 30
....4... 23.. 15 ...|0|

Now your quadratic is $\displaystyle 4x^2 + 23x + 15 = (4x + 3)(x + 5)$

3. Originally Posted by Substince
I'm on my last problem for my lesson and I'm a little stumped and need a push in the right direction.

Problem: "One factor of $\displaystyle 4x^3+15x^2-31x-30$ is x-2. Find the other factors."

I don't want the answer but if someone could explain to me how they even got x-2 as a factor I should be able to figure the rest out or if someone knows of a lesson that explains this that would be helpful as well.

Thanks for looking!
By the Factor theorem. It states that "a polynomial f(x) has a factor (x − k) if and only if f(k) = 0." Therefore given a polynomial you can switch some values and check if the value of f(x) is equal to zero.

In this problem, $\displaystyle f(x)=4x^3+15x^2-31x-30$

$\displaystyle f(2)=4.2^3+15.2^2-31.2-30=32+60-62-30=0$

Hence by the Factor theorem, (x-2) is a factor. If you want more details refer, Factor theorem - Wikipedia, the free encyclopedia.

4. Well, I suspect that "they" know that x- 2 is a factor because they created the polynomial by multiplying a quadratic by x- 2! Of course, you could tell that x- 2 is a factor by noting that $\displaystyle 4(2)^3+ 15(2)^2- 31(2)- 30= 32+ 60- 62- 30= 92- 92= 0$. That tells you that x- 2 evenly divides the polynoial- if it did not we would have $\displaystyle P(x)= (x- 2)Q(x)+ r$ where r is the remainder. Taking x= 2 in that would give $\displaystyle P(2)= (0)Q(x)+ r= r$. Since P(2)= 0, there is no remainder.

But the fact that you were told that x- 2 is a factor is all you really needed to know. Now, divide the given polynomial by x- 2 (or use "synthetic division" since the divisor is of the form x- a).

5. lol wow do I ever feel stupid right now :P

Thanks for all the comments guys and clearing things up for me

6. Originally Posted by Substince
lol wow do I ever feel stupid right now :P
WHY? You should feel "good" as you know you'll never have that problem again !

7. But "feeling good" and "feeling stupid" are, fortuately, not mutually exclusive! I learned that a long time ago!