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August 6th, 2011, 08:28 AM   #1
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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
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August 6th, 2011, 08:37 AM   #2
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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.
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August 7th, 2011, 01:56 PM   #3
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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.



Mulitply by and subtract it from the polynomial. Thus:



Simple enough, right? Just like basic division.

Continuing on:



Subtract times from our polynomial, yielding:



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 by and subtract it, which yields us:



To finally complete what may feel like a long problem, multiply by and subtract it:


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

This means:

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

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


=(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:



You can do the graphing. I [expletive] hate graphing.
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August 7th, 2011, 02:51 PM   #4
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Re: Synthetic Division Problem

That's nice, but it's not synthetic division!
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August 7th, 2011, 03:19 PM   #5
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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 is a factor of
(b) If so, write as times the reduced polynomial.
(c) Then find the zeros of (with multiplicity)
(d) Graph the polynomial, clearly labeling key parts of the graph.








[color=beige]. . . [/color]
[color=beige]. = . . . . . . . . . . . . . . . [/color]

[color=beige]. . . [/color]

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



Code:
                           |
      *                    |
                           |          *
       *        *          |
        *    *     *       |         *
    -------o---------o-----+--------o----
          -2        -1*    |      * 2
                        *  |   *
                           *
                           |
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August 7th, 2011, 03:26 PM   #6
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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. . .
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August 7th, 2011, 03:59 PM   #7
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Re: Synthetic Division Problem

You did polynomial long division... nothing wrong with that!
Synthetic (as the abacus showed) takes up a lot less space
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August 7th, 2011, 04:01 PM   #8
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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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