relating unkown coefficients with a known factor of a polynomial

Hi, I think this is in the right place. I think it's also very simple but the text book I'm working from just doesn't explain it.

The question is:

If x + p is a factor of ax^{2} + bx + 3, show that p(b - ap) = 3

According to the book, I think what they want one to do is long division and then, knowing x + p is a factor and therefore remainder = 0, equate the unknown coefficients. But the example in the text is sufficiently different to not explain how to do it with this particular expression. I've got so far with it, and then get totally stuck. I know this is probably very simple but I just can't figure it out.

I can't believe I'm getting stuck so early on. I'm on section 3 of chapter 1. I've had no problems with the first two sections I couldn't work out myself but I'm really stuck on this one! Please help.

Re: relating unkown coefficients with a known factor of a polynomial

Hey indymogul.

Have you done the long division part first? If so show us what you tried and if not show us what is troubling you about long division for polynomials.

Re: relating unkown coefficients with a known factor of a polynomial

Hi, thanks for the quick reply.

Well, yes I did try to do the long division but I got stuck.ax

**____________**

x + p **)** ax^{2}+ bx + 3ax^{2 }+ p(ax)

(Pardon my crude long division formatting, I don't know how else to do it.) It's about at this point I get lost. Obviously the ax^{2}s cancel out, but I can't figure out how to subtract p(ax) from b(x) or if they are even, and I can't figure out what the next part of the quotient is.

Does p(ax) = bx +3? Cause that would make it a bit easier. Or am I way off there.

I also don't know whether to bring the 3 down at this line, or at the next after subtracting. Basically I think I'm totally lost. I know that somewhere the 2 lines will be equal, because x + p is a factor of the quadratic expression and therefore there's no remainder, but even from that I can't see how to prove that p(b - ap) = 3.

The example in the textbook is completely different. That's why I'm stuck on this one. There's a few others like this one, 3 more to be exact. I just skipped them because I feel that I just don't get it. I did all the rest of the 18 questions on factoring and got them pretty much all correct. I think if I understand this one I would be able to do the others.

Thanks so much for your help. I'm determined to learn this and I really appreciate it.

Re: relating unkown coefficients with a known factor of a polynomial

It's a lot easier to use the long division method.

If you divide (x + p) into ax^2 + bx + c then you get ax + (b-ap) as your quotient with 3 - p(b - ap) as your remainder but if (x+p) is a factor then the remainder must be zero.

So if the remainder is zero, what can you say about that term?

Re: relating unkown coefficients with a known factor of a polynomial

Hello, indymogul!

Quote:

$\displaystyle \text{If }x+p\text{ is a factor of }\:f(x) \:=\:ax^2+bx+3,\,\text{ show that }\,p(b - ap)\:=\:3$

$\displaystyle \text{If }x+p\text{ is a factor of }f(x),\,\text{ then: }\,f(\text{-}p) \:=\:0$

$\displaystyle \begin{array}{cc}\text{So we have:} & a(\text{-}p)^2 + b(\text{-}p) + 3 \:=\:0 \\ & ap^2 - bp + 3 \:=\:0 \\ & bp-ap^2 \:=\:3 \\ & p(b - ap) \:=\:3 \end{array}$

Re: relating unkown coefficients with a known factor of a polynomial

Thanks chiro. I get that 3 - p(b - ap) = 0 or -p(b - ap) = -3 or p(b - ap) = 3.The problem I'm having is how to do the long division part. The example in the textbook is ax^3 + bx^2 +1 divided by x^2 - 2x +1, which I could do easily enough. It's all these extra variables in this one, a, p and b, that are throwing me. I don't get how to do the long division with these in there. Can someone walk me through it? It's about 15 years since I did my grade 12 and I haven't studied a scrap of maths since then. It's in there somewhere though! Just needs a bit of refreshing.

Soroban, yours looks like a much easier method, I've seen this elsewhere while researching the problem, but the textbook is doing it with long division so I think I'm going to try to work out how to to do it this way. I think your method is covered later in the textbook. Thanks everyone for your help.

Re: relating unkown coefficients with a known factor of a polynomial

I was able to get the first part of the quotient, ax. It's the (b - ap) I could not figure out how to arrive at. This was as far as I got:

ax

____________

x + p ) ax^2+ bx + 3

ax^2 + p(ax)

So the two ax^2s cancel out. Does the 3 carry down at the last line? And how do you divide x + p into p(ax) to get (b - ap)? This completely throws me. I don't have a clue.

Re: relating unkown coefficients with a known factor of a polynomial

Just getting into the factor theorem now in the next section. I'd really like to understand how that long division is done though, if anyone can help there. Thanks.

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Re: relating unkown coefficients with a known factor of a polynomial

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Since the remainder must be zero, we have:

$\displaystyle 3-(b-ap)p=0$

$\displaystyle p(b-ap)=3$

Re: relating unkown coefficients with a known factor of a polynomial

OK I think I get it now! Thank you so much everyone for your help. After looking at the examples of how the long division is done, I think I understand how to do it. Moving on to the factor theorem now anyway. Thanks again you guys. Very much appreciated.

Re: relating unkown coefficients with a known factor of a polynomial

Quote:

Originally Posted by

**indymogul** The question is:

If x + p is a factor of ax^{2} + bx + 3, show that p(b - ap) = 3

If $\displaystyle (x+p)$ is a factor then $\displaystyle -p$ is a root.

So $\displaystyle ap^2-bp+3=0$ or $\displaystyle p(b-ap)=3$.

Re: relating unkown coefficients with a known factor of a polynomial

Yes that is correct but this was before the section on the Factor Theorem where one is expected to show the proof using long division only. Of course with the Factor Theorem one knows that -p is a root. Well, I know that now because I've completed the next section. Thanks for your help.