[SOLVED] Algebraic Long Division

**Quacky** · January 9th 2010, 07:40 AM

I have a few questions about algebraic division:

a)As an example, how would I work out $2x^3-4x^2+6x-3$ divided by $x+1$ ?
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 $2x^3-4x^2+6x-3$ divided by $2x+1$ , how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

Thanks in advance.

**earboth** · January 9th 2010, 08:16 AM

Originally Posted by Quacky

I have a few questions about algebraic division:

a)As an example, how would I work out $2x^3-4x^2+6x-3$ divided by $x+1$ ?
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 $2x^3-4x^2+6x-3$ divided by $2x+1$ , how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

Thanks in advance.

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.

**Quacky** · January 9th 2010, 08:28 AM

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.

**Dinkydoe** · January 9th 2010, 09:19 AM

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

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

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

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

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

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

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

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

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

**Quacky** · January 9th 2010, 09:30 AM

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

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