# Thread: [SOLVED] Algebraic Long Division

1. ## [SOLVED] Algebraic Long Division

I have a few questions about algebraic division:

a)As an example, how would I work out $\displaystyle 2x^3-4x^2+6x-3$ divided by $\displaystyle x+1$?
I've never understood algebraic long division so a step by step guide (or link) would be greatly appreciated.

b) If I have something like $\displaystyle 2x^3-4x^2+6x-3$ divided by $\displaystyle 2x+1$, how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

2. Originally Posted by Quacky
I have a few questions about algebraic division:

a)As an example, how would I work out $\displaystyle 2x^3-4x^2+6x-3$ divided by $\displaystyle x+1$?
I've never understood algebraic long division so a step by step guide (or link) would be greatly appreciated.

b) If I have something like $\displaystyle 2x^3-4x^2+6x-3$ divided by $\displaystyle 2x+1$, how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

To a) You use the first summand of the divisor as divisor. To find the term which has to be subtracted you use the complete divisor. (Please read this sentence three times before you decide that you didn't understand it )
Code:
(2x^2 - 4x^2 + 6x - 3) ÷ (x + 1) = 2x^2 - 6x + 12, R = -15
-(2x^3 + 2x^2)
---------------
-6x^2 + 6x
-(-6x^2 - 6x)
-------------
12x - 3
-(12x + 12)
-----------------
-15
to b) Yes.

3. You use the first summand of the divisor as divisor. To find the term which has to be subtracted you use the complete divisor.
Wow, that's certainly a phrase to get my head around , but thanks for the explanations, I think I can follow your working when I try on paper now, I'll try some more examples by myself to see if I've understood.

Thanks again.

4. (a)
In general: Let $\displaystyle f(x) = c_nx^n +c_{n-1}x^{n-1}+\cdots c_1x + c_0$ be a polynomial of degree n. Say we want to divide this by a linear factor $\displaystyle (ax+b)$. We will seek to try to find a polynomial q(x) of degree n-1 such that $\displaystyle f(x) = q(x)(ax+b)+ r$ where $\displaystyle r$ is the remainder of degree 0.(a constant)

We reduce the polynomial step by step:
Let $\displaystyle g_1(x) = f(x) - \frac{c_n}{a}x^{n-1}\cdot (ax+b)= (c_{n-1}-\frac{c_nb}{a})x^{n-1}+c_{n-2}x^{n-2}+\cdots c_1x+c_0$
Let $\displaystyle q_1(x) = \frac{c_n}{a}x^{n-1}$

We now reduced the polynomial with one degree and we have $\displaystyle f(x) = q_1(x)(ax+b)+g_1(x)$

Continue in the same fashion with the new polynomial $\displaystyle g_1(x)$ by repeatedly removing the head-coefficient.

Your example: $\displaystyle f(x) = 2x^3-4x^2+6x-3$. Divide by $\displaystyle x+1$:

So we let $\displaystyle q_1(x) = 2x^2$
and $\displaystyle g_1(x) = f(x) - 2x^2(x+1) = -6x^2+6x-3$
$\displaystyle q_2(x) = -6x$
$\displaystyle g_2(x) = g_1(x)-(-6x(x+1)) = 12x - 3$
$\displaystyle q_3(x) = 12$
$\displaystyle g_3(x) = g_2(x) - 12(x+1) = -15$

Now $\displaystyle q(x) = q_1(x)+q_2(x)+q_3(x) = 2x^2-6x+12$ and $\displaystyle f(x) = (2x^2-6x+12)(x+1) - 15$.

See now we got f into the form $\displaystyle f(x) = q(x)(ax+b)+r$ with $\displaystyle r=-15$ as our remainder.

(b) yes. The idea is that you got your polynomial into the form: $\displaystyle f(x)= q(x)(ax+b)+r$ so if you plug in a zero of the linear factor we get $\displaystyle f(\frac{-b}{a}) = r$

5. Thanks for the explanation, I think between the two of you, everything I needed to know has been covered.