My Math Forum Synthetic Division Problem

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 August 6th, 2011, 08:28 AM #1 Newbie   Joined: Jul 2011 Posts: 3 Thanks: 0 Synthetic Division Problem Hi Can anyone be kind enough to explain to me what "synthetic division"? and how I can use that to solve the following: The question says,"use synthetic division to determine if (x + 1) is a factor of P(x) = x^4 + 3x^3 - 2x^2 -12x - 8.If so ,write P(x)as (x+1)times the reduced polynomial.Then find the zeros of P(x)(with multiplicity) Graph the polynomial ,clearly labeling key parts of the graph. How do I approach this question? Thanks
 August 6th, 2011, 08:37 AM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Synthetic Division Problem Synthetically divide by (x + 1). If you don't have any idea how to do that, it would be better for you to do a web search. Set x + 1 = 0 to find that x = -1 So -1 is the number "in the box". The remainder will either be 0, or not 0.
August 7th, 2011, 01:56 PM   #3
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Joined: Jul 2011

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Re: Synthetic Division Problem

Quote:
 Originally Posted by Kodwo Hi Can anyone be kind enough to explain to me what "synthetic division"? and how I can use that to solve the following: The question says,"use synthetic division to determine if (x + 1) is a factor of P(x) = x^4 + 3x^3 - 2x^2 -12x - 8.If so ,write P(x)as (x+1)times the reduced polynomial.Then find the zeros of P(x)(with multiplicity) Graph the polynomial ,clearly labeling key parts of the graph. How do I approach this question? Thanks
Synthetic divison is the extension of divison to algebraic expressions.

\begin{align}
&x^3
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
\end{align}

Mulitply $(x+1)$ by $x^3$ and subtract it from the polynomial. Thus:

\begin{align}
&x^3
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
&-(x^4+x^3)
&\text{---------------}
&2x^3-2x^2-12x-8
\end{align}

Simple enough, right? Just like basic division.

Continuing on:

\begin{align}
&x^3+2x^2
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
&-(x^4+x^3)
&\text{---------------}
&2x^3-2x^2-12x-8
\end{align}

Subtract $2x^2$ times $(x+1)$ from our polynomial, yielding:

\begin{align}
&x^3+2x^2
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
&-(x^4+x^3)
&\text{---------------}
&2x^3-2x^2-12x-8
&-(2x^3+2x^2)
&\text{---------------}
&-4x^2-12x-8
\end{align}

If you are confused, please note the fact that the subtraction sign is being distributed onto the thing which we're subtracting. So, it's actually subtracting the positive values rather than adding them.

Continuing on, multiply $(x+1)$ by $-4x$ and subtract it, which yields us:

\begin{align}
&x^3+2x^2-4x
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
&-(x^4+x^3)
&\text{---------------}
&2x^3-2x^2-12x-8
&-(2x^3+2x^2)
&\text{---------------}
&-4x^2-12x-8
&-(-4x^2-4x)
&\text{---------------}
&-8x-8
\end{align}

To finally complete what may feel like a long problem, multiply $(x+1)$ by $8$ and subtract it:

\begin{align}
&x^3+2x^2-4x+8
&\text{---------------}
{x+1}&){x^4+3x^3-2x^2-12x-8}
&-(x^4+x^3)
&\text{---------------}
&2x^3-2x^2-12x-8
&-(2x^3+2x^2)
&\text{---------------}
&-4x^2-12x-8
&-(-4x^2-4x)
&\text{---------------}
&-8x-8
&-(8x+

&\text{---------------}
&0
\end{align}" />

This means:

$(x+1)(x^3+2x^2-4x+=x^4+3x^3-2x^2-12x-8" />
i.e.
$(x+1)(x^3+2x^2-4x+=P(x)" />
Hence, $(x+1)$ is a factor of $P(x)$

I want to continue this problem on to find a simpler factorization:

$P(x)=(x+1)(x^3+2x^2-4x-
=(x+1)(x(x^2+2x-4)-
=(x+1)(x(x(x+2)-4)-
=(x+1)(x^2(x+2)-4x-
=(x+1)(x^2(x+2)-(x+2)4)
=(x+1)(x+2)(x^2-4)
=(x+1)(x+2)(x+2)(x-2)
=(x+1)(x-2)(x+2)^2" />

Thus, by multiplicity:

$P(x)=0 \text{ if and only if } \{x=(-1,-2, \vee 2)\}$

You can do the graphing. I [expletive] hate graphing.

 August 7th, 2011, 02:51 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Synthetic Division Problem That's nice, but it's not synthetic division!
August 7th, 2011, 03:19 PM   #5
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Synthetic Division Problem

Hello, Kodwo!

If you're not familiar with Synthetic Division, I recommend a search.

Quote:
 (a) Use synthetic division to determine if $(x + 1)$ is a factor of $P(x) \:=\: x^4\,+\,3x^3\,-\,2x^2\,-\,2x\,-\,8.$ (b) If so, write $P(x)$ as $(x+1)$ times the reduced polynomial. (c) Then find the zeros of $P(x)$ (with multiplicity) (d) Graph the polynomial, clearly labeling key parts of the graph.

$(a)\;\;\begin{array}{cccccccc}-1 & | & 1 & 3 & -2 & -12 & -8 \\ \\
& | && -1 & -2 & +4 & +8 \\ && -- & -- & -- & -- & -- \\ && 1 & 2 & -4 & -8 & \;0 \end{array}$

$(b)\;\;P(x) \:=\:(x\,+\,1)(x^3\,+\,2x^2\,-\,4x\,-\,8)$

$\text{(c) The cubic factors.}$
[color=beige]. . . [/color]$x^3\,+\,2x^2\,-\,4x\,-\,8 \;=\;x^2(x\,+\,2)\,-\,4(x\,+\,2) \;=\;(x\,+\,2)(x^2\,-\,4)$
[color=beige]. = . . . . . . . . . . . . . . . [/color]$=\;(x\,+\,2)(x\,-\,2)(x\,+\,2) \;=\;(x\,-\,2)(x\,+\,2)^2$

[color=beige]. . . [/color]$P(x) \;=\;(x\,+\,1)(x\,-\,2)(x\,+\,2)^2$

[color=beige]. . . [/color]The zeros are:[color=beige] .[/color] -1, 2, and -2 (multiplicity 2).

Code:
                           |
*                    |
|          *
*        *          |
*    *     *       |         *
-------o---------o-----+--------o----
-2        -1*    |      * 2
*  |   *
*
|

August 7th, 2011, 03:26 PM   #6
Senior Member

Joined: Jul 2011

Posts: 245
Thanks: 0

Re: Synthetic Division Problem

Quote:
 Originally Posted by The Chaz That's nice, but it's not synthetic division!
Wow, wtf. My textbook said they were the same thing. . .

o_O.

Oh well, I hope I helped in some form or another. . .

 August 7th, 2011, 03:59 PM #7 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Synthetic Division Problem You did polynomial long division... nothing wrong with that! Synthetic (as the abacus showed) takes up a lot less space
August 7th, 2011, 04:01 PM   #8
Senior Member

Joined: Jul 2011

Posts: 245
Thanks: 0

Re: Synthetic Division Problem

Quote:
 Originally Posted by The Chaz You did polynomial long division... nothing wrong with that! Synthetic (as the abacus showed) takes up a lot less space
It's like augmented matrices all over again.

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