# 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

I´m 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 can´t be devided, you don´t 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 wan´t 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 wan´t 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 don´t 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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### Division of polynomial by non linear polynomial

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