# Thread: Nonlinear factor polynomial divisions + remainder theorem question

1. ## Nonlinear factor polynomial divisions + remainder theorem question

Hi,

I'm abit confused with these two things. Firstly, I was taught this quick shortcut for dividing polynomials (although I struggle with using it for dividing by non-linear factors). I'll try explain it and see if anyone can help me with it.

To do it, using this example: $5x^5 + 13x^4 - 2x^2 - 6 / x + 1$

I write out the (x + 1) five times (since highest power is 5) and to make it like this:

(x + 1) (x + 1) (x + 1) (x + 1) (x + 1)

Then I'd go: $5x^4$ is the first number, since that's what needed to do the first power. That would leave $5x^4$ left. Then I'd need $+ 8x^4$ to get to $13x^4$ so my equation would now look like:

$5x^4(x + 1) + 8x^3 (x + 1) (x + 1) (x + 1) (x + 1)$

Then so on until i get to:

$5x^4(x + 1) + 8x^3 (x + 1) - 8x^2(x + 1) + 6x(x + 1) - 6(x + 1)$

and then I can factor to find solution is $5x^4 + 8x^3 - 8x^2 + 6x - 6$

Dunno if anyone else knows that method. But if so, how would I do it if there was a remainder? Also, how could I use this to divide by non linear factors (ie: $(x^2 + 1)$ which has a remainder when it is $x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1$ is divided by it)

Also, I was wondering. with that above question (find remainder when $x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1$ is divided by $(x^2 + 1)$), how could I use the remainder thereom with it? Is it possible?

Thanks!

2. ## Hi

Iīm a bit confused. Partially because you speak of division and but the equation looks

y = aX^5+ bX^4 + cX^3 +dX^2 + 1/x. or you mean:
( aX^5+ bX^4 + cX^3 +dX^2)/(x+1) ??

If there is a ramainder that canīt be devided, you donīt devide it, for example say the reminder is the constant 7, gives you
ax^3 + bx^2 + 7/(x+1) <--------the reminder cant be devided.

If you wanīt some more explaining search youtube. Some videos on this topic are great.

3. Originally Posted by Rudey
Hi,

I'm abit confused with these two things. Firstly, I was taught this quick shortcut for dividing polynomials (although I struggle with using it for dividing by non-linear factors). I'll try explain it and see if anyone can help me with it.

To do it, using this example: $5x^5 + 13x^4 - 2x^2 - 6 / x + 1$

I write out the (x + 1) five times (since highest power is 5) and to make it like this:

(x + 1) (x + 1) (x + 1) (x + 1) (x + 1)

Then I'd go: $5x^4$ is the first number, since that's what needed to do the first power. That would leave $5x^4$ left. Then I'd need $+ 8x^4$ to get to $13x^4$ so my equation would now look like:

$5x^4(x + 1) + 8x^3 (x + 1) (x + 1) (x + 1) (x + 1)$

Then so on until i get to:

$5x^4(x + 1) + 8x^3 (x + 1) - 8x^2(x + 1) + 6x(x + 1) - 6(x + 1)$

and then I can factor to find solution is $5x^4 + 8x^3 - 8x^2 + 6x - 6$

Dunno if anyone else knows that method.
I've seen it but I find it awkward to use. Perhaps just because I learned 'regular division' first.

But if so, how would I do it if there was a remainder?
Nothing new. you would just arrive a constant, not 0 term in the last step.

Also, how could I use this to divide by non linear factors (ie: $(x^2 + 1)$ which has a remainder when it is $x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1$ is divided by it)
You wouldn't. At least not in the form you give here which consists of writing powers of your divisor.

Also, I was wondering. with that above question (find remainder when $x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1$ is divided by $(x^2 + 1)$), how could I use the remainder thereom with it? Is it possible?
The "remainder theorem" says that the remainder when polynomial p(x) is divided by x-a is just p(a). You can't use that here because $x^2+ 1$ is not of the form "x- a".

Of course, the "Euclidean algorithm" tells you that you can divide and that the remainder will be either a constant or a first degree polyomial.

Thanks!

4. how could I use the remainder thereom with it? Is it possible?
Well in fact you can, but you must understand polynomial division first:

If you divide a polynomial $p(x)$ by $x^2+1$ we can find a polynomial $q(x)$ such that $p(x) = q(x)(x^2+1)+r$

so filling in a zero of $(x^2+1)$, that is $x_0 = \pm i$ gives $p(x_0) = r$

(observe that $(x^2+1)= (x-i)(x+1)$ wich gives zero's $\pm i$. Do you know what complex numbers are?))

5. So how could I solve it? Could someone please work through it, so I can see how it's done and apply it to similar questions. Here is the question:

Let $P(x)=x^5-3x^4+2x^3-2x^2+3x+1$
a) Show that neither (x-1) nor (x+1) is a factor of P(x)

This I did by using the factor theorem and showing that P(1) and P(-1) do not equal 0

b) Given that P(x) can be written in the form
$(x^2 - 1)Q(x) + ax + b$ where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by $(x^2-1)$
The solution is: 6x - 4

6. ## The first one.

Given $P(x) = x^5 - 3x^4 + 2x^3 - 2x^2 +3x +1$

Showing that P(1) is not 0
$P(1) = 1^5 - 3*1^4 +2*1^3 -2*1^2 +3*1 +1 = 2$
so its not zero. That shows you that when x = 1; y = 2 so the curve does not intersect X-axis when X is 1 correct: (x-1) can therefor not be a factor.

If you wanīt to show that neither x-1; x+1 are factors just divide.

$(1^5 - 3*1^4 +2*1^3 -2*1^2 +3*1 +1)/(x+1)$

IF there are no remainders then it factors but if there is a remainder it can not factor. If you donīt know how to devide I can follow you through. A tip is to check out the youtube videos on polynomial division, there are som very good once. You will be solving this in notime yourself, when u get how to do it.

7. Hi, thanks for reply. But I managed to get that, what I was having trouble with was how to solve the second part (part b)

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# Dividing a polynomial with a non factor

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