# Dividing polynomials

• Dec 6th 2011, 04:30 AM
bubbletea999
Dividing polynomials
I'm learning how to divide polynomials by expressions like a + 1 and I'm stuck on a particular question:

4x^4 - 3x^2 + x + 2 divided by 2x + 3

The answer must be expressed in the form (2x+3)*Quotient+Remainder.

Here's how I've been going about it.

4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx + c) + R

Then after expanding and collecting like terms, I have:

2ax^4 + 3ax^3 + 2bx^2 + (2b+2c)x + 3c + R

And it's here that I get stuck. The problem here is that I don't know how to go on. What for example do I do with the x^3 bit?

Many thanks
• Dec 6th 2011, 04:42 AM
CaptainBlack
Re: Dividing polynomials
Quote:

Originally Posted by bubbletea999
I'm learning how to divide polynomials by expressions like a + 1 and I'm stuck on a particular question:

4x^4 - 3x^2 + x + 2 divided by 2x + 3

The answer must be expressed in the form (2x+3)*Quotient+Remainder.

Here's how I've been going about it.

4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx + c) + R

Then after expanding and collecting like terms, I have:

2ax^4 + 3ax^3 + 2bx^2 + (2b+2c)x + 3c + R

And it's here that I get stuck. The problem here is that I don't know how to go on. What for example do I do with the x^3 bit?

Many thanks

Have you done synthetic division? If so use it.

If not try equating coeficients of like powers in each of your expressions, to get you started (assuming you have no algebraic mistakes) for the fourth power term:

4=2a

CB
• Dec 6th 2011, 04:54 AM
sbhatnagar
Re: Dividing polynomials
You have:
$\displaystyle 4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx^2 + {\color{red}cx}+d) + R$

Open the brackets:

\displaystyle \begin{align*} 4x^4 - 3x^2 + x + 2 &=(2a)x^4+(2b)x^3+(2c)x^2+(2d)x+(3a)x^3+(3b)x^2+(3 c)x+(3d)+R \end{align*}

\displaystyle \begin{align*} \implies 4x^4 - 3x^2 + x + 2 &= (2a)x^4+(2b+3a)x^3+(2c+3b)x^2+(3c+2d)x+(3d+R) \end{align*}

Equating the coefficients:

$\displaystyle 2a=4$

$\displaystyle 2b+3a=-3$

$\displaystyle 2c+3b=0$

$\displaystyle 3c+2d=1$

$\displaystyle 3d+R=2$

The correct solutions to these equations will give you the value of $\displaystyle a,b,c,d \ \textrm{and} \ R$.
• Dec 23rd 2011, 12:45 AM
bubbletea999
Re: Dividing polynomials

I've now learned how to do synthetic division, and in the process of doing so also learned about the general form, which ended up solving my problem with this! I was going wrong because I wasn't accounting for the "missing power" in the original equation. It jumps from x^4 to x^2. I've solved the problem by adding 0x^3 in between them -- does this sound right? It's produced the right answer, at any rate.

Thanks again.
• Dec 24th 2011, 01:25 PM
HallsofIvy
Re: Dividing polynomials
Another way to divide $\displaystyle 4x^4 - 3x^2 + x + 2$ by $\displaystyle 2x + 3$, more like "standard" numerical long division:

$\displaystyle 2x$ divides into $\displaystyle 4x^4$ $\displaystyle 2x^3$ times. Multiplying that by $\displaystyle 2x+ 3$ gives [tex](2x+3)(2x^3)= 4x^4+ 6x^3. Subtract [tex](4x^4+ 0x^3)- (4x^4+ 6x^3)= -6x^3. That leaves us $\displaystyle -6x^3- 3x^2+ x+ 2$. Now, $\displaystyle 2x$ divides into $\displaystyle -6x^3$ $\displaystyle -3x^2$ times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle -6x^3- 9x^2$. Subtract $\displaystyle (-6x^3- 3x^2)- (-6x^3- 9x^2)= 6x^2$. That leaves $\displaystyle 6x^2+ x+ 2$. $\displaystyle 2x$ divides into $\displaystyle 6x^2$ 3x times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle 6x^2+ 9x$. Subtracting $\displaystyle (6x^2+ x)-(6x^2+ 9x)= -8x$. That leaves us $\displaystyle -8x+ 2$. $\displaystyle 2x$ divides into $\displaystyle -8x$ -4 times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle -8x- 12$. Subtracting $\displaystyle (-8x+ 2)- (-8x- 12)= 14$.

That is, the quotient is $\displaystyle 2x^3- 3x^2+ 3x- 4$ with remainder 14.

(Captain Black, I was under the impression that "synthetic division" could only be used to divide by something of the form x+ a.)