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How does synthetic division produce the results it does?

  • #1
808
37

Homework Statement


f(x) = x^3
Find f(-2)
f(-2) = -8
The text says that by dividing a function [x^3] by [x minus a given input (-2)] using synthetic division, I'll be able to produce the correct output (-8). I want to know why this happens.

Homework Equations


(x^3)/(x-(-2)) = -8


The Attempt at a Solution


I'm trying to prove that x^3 divided by x minus the input will produce the output. I made it simple enough, but all I manage to do is prove that synthetic division does what it does.

(x^3)/(x+2) = -8 [multiply both sides by (x+2)]
x^3 = -8x - 16 [move right side to left]
x^3 + 8x + 16 [divide whole equation by (x+2)]
(x^3 + 8x + 16)/(x+2) [synthetic division]

-2 1 0 8 16
-2 4 -24
1 -2 4 -8

I still don't know why synthetic division produces an output for a given input when subtracted from x, and becomes the divisor for the original equation. It still ends up with me having to use synthetic division to obtain my result.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
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Homework Statement


f(x) = x^3
Find f(-2)
f(-2) = -8
The text says that by dividing a function [x^3] by [x minus a given input (-2)] using synthetic division, I'll be able to produce the correct output (-8). I want to know why this happens.

Homework Equations


(x^3)/(x-(-2)) = -8


The Attempt at a Solution


I'm trying to prove that x^3 divided by x minus the input will produce the output. I made it simple enough, but all I manage to do is prove that synthetic division does what it does.

(x^3)/(x+2) = -8 [multiply both sides by (x+2)]
x^3 = -8x - 16 [move right side to left]
x^3 + 8x + 16 [divide whole equation by (x+2)]
(x^3 + 8x + 16)/(x+2) [synthetic division]

-2 1 0 8 16
-2 4 -24
1 -2 4 -8

I still don't know why synthetic division produces an output for a given input when subtracted from x, and becomes the divisor for the original equation. It still ends up with me having to use synthetic division to obtain my result.
For any polynomial ##p(x)##, ##p(a)## is the remainder when ##p(x)## is divided by ##(x-a)##. The reason is simple: division produces a lower-degree polynomial ##q(x)## such that for all ##x## we have
[tex] p(x) = (x-a) q(x) + r[/tex]
The remainder ##r## has lower degree than the divisor ##x-a## (which is already of degree 1); thus, the remainder is just a number and does not conain ##x## at all---that is, it is a constant. Since the equation holds for all ##x##, just look at what it gives you when you set ##x = a##.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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933
Synthetic division is simply "division" with non-necessary parts dropped. For example, to divide [itex]x^3- 5x^2+ 7x- 2[/itex] by x- 2, I would, as a "trial divisor" note that x will divide into [itex]x^3[/itex] [itex]x^2[/itex] times. Multiplying x- 2 by [itex]x^2[/itex] gives [itex]x^3- 2x^2[/itex]. Subtracting that from [itex]x^3- 5x^2+ 7x- 2[/itex] leaves a remainder of [itex]-3x^2+ 7x- 2[/itex]. x will divide into [itex]-3x^2[/itex] [itex]-3x[/itex] times. Multiplying x- 2 by [itex]-3x[/itex] gives [itex]-3x^2+ 6x[/itex]. Subtracting that from [itex]-3x^2+ 7x- 2[/itex] leaves [itex]x- 2[/itex] and x- 2 obviously divides into that exactly once. [itex]x- 2[/itex] divides into [itex]x^3- 5x^2+ 7x- 2[/itex] exactly [itex]x^2- 3x+ 1[/itex] times.

Now, for synthetic division, since the powers of x are always in the same order, we don't really need to write those- write the dividend as [itex]\begin{array}{cccc}1 & -5 & 7 & -2\end{array}[/itex]. And we can change those subtractions into additions by changing the sign- instead of "x- 2" we just write the divisor as "2". We have [itex]\begin{array} {cccccc} & || 1 & -5 & 7 & -2\end{array}[/itex]. Bring down that first 1: and multiply the 2 by that: [tex]\begin{array} {cccccc}2 & || & 1 & -5 & 7 & -2 \\ & & 2 & & & \\ & & 1 & -3 & & \end{array}[/tex]. 2 times that -3 is -6 we would next have [tex]\begin{array} {cccccc}2 & || & 1 & -5 & 7 & -2 \\ & & & 2 & -6 & \\ & & 1 & -3 & 1 & \end{array}[/tex]. And 2 times that 1 is 2 so [tex]\begin{array} {cccccc}2 & || & 1 & -5 & 7 & -2 \\ & & & 2 & -6 & 2\\ & & 1 & -3 & 1 & 0 \end{array}[/tex]. That bottom row, "[itex]\begin{array}{cccc}1 & -3 & 1 & 0 \end{array}[/itex]" we interpret as [itex]x^2- 3x+ 1[/itex].
 
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  • #4
808
37
The reason is simple: division produces a lower-degree polynomial q(x)q(x) such that for all xx we have
p(x)=(xa)q(x)+r
p(x) = (x-a) q(x) + r
The remainder rr has lower degree than the divisor xax-a (which is already of degree 1);
I'm sorry; I don't quite understand what you mean by 'lower-degree'. Do you mean like exponents?
 
  • #5
33,285
4,989
I'm sorry; I don't quite understand what you mean by 'lower-degree'. Do you mean like exponents?
Yes. For example the degree of x3 - 2x + 1 is 3, which is the degree of the x3 term.
 
  • #6
808
37
So what you're saying is: if a coefficient in a polynomial equation possesses a degree of 0 (that is, x to the power of zero), it is a number all its own; given that anything to the power of zero is one, and anything multiplied by one will not change?

For any polynomial p(x)p(x), p(a)p(a) is the remainder when p(x)p(x) is divided by (xa)(x-a). The reason is simple: division produces a lower-degree polynomial q(x)q(x) such that for all xx we have
p(x)=(xa)q(x)+r
So is this a basic law, or something?
 
  • #7
Ray Vickson
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So what you're saying is: if a coefficient in a polynomial equation possesses a degree of 0 (that is, x to the power of zero), it is a number all its own; given that anything to the power of zero is one, and anything multiplied by one will not change?



So is this a basic law, or something?
No. It is the definition of what is meant by "division".

It involves a bit of work to show that the operation of division actually exists (i.e, has "meaning") and produces unique results. You can find out more using Google to search the internet for material on 'division of polynomials'.
 
  • #8
808
37
It involves a bit of work to show that the operation of division actually exists (i.e, has "meaning") and produces unique results. You can find out more using Google to search the internet for material on 'division of polynomials'.
I might research the former if I have spare time. Thanks for all of your help, guys.
 

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