Polynomial Problem and not a clue how to proceed

**Substince** · March 11th 2011, 06:13 AM

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 $4x^3+15x^2-31x-30$ 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!

**TheChaz** · March 11th 2011, 06:31 AM

"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 $4x^2 + 23x + 15 = (4x + 3)(x + 5)$

**Sudharaka** · March 11th 2011, 06:32 AM

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 $4x^3+15x^2-31x-30$ 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, $f(x)=4x^3+15x^2-31x-30$

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

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

**HallsofIvy** · March 11th 2011, 06:41 AM

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 $4(2)^3+ 15(2)^2- 31(2)- 30= 32+ 60- 62- 30= 92- 92= 0$ . That tells you that x- 2 evenly divides the polynoial- if it did not we would have $P(x)= (x- 2)Q(x)+ r$ where r is the remainder. Taking x= 2 in that would give $P(2)= (0)Q(x)+ r= r$ . 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).

**Substince** · March 11th 2011, 08:34 AM

lol wow do I ever feel stupid right now :P

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

**Wilmer** · March 11th 2011, 08:38 AM

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 !

**HallsofIvy** · March 11th 2011, 08:50 AM

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

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