7-25-2015
Polynomial Division
You do polynomial division the way you do long division of numbers. It’s difficult to describe the
general procedure in words, so I’ll work through some examples step-by-step.
Example. Find the quotient and remainder when x2 + 2x − 3 is divided by x − 2.
You may also see this kind of problem written like this: “Perform the division
x2 + 2x − 3
.”
x−2
Set up the division the way you’d set up division of numbers. To start, look at the first term (x) in
x − 2 and the first term (x2 ) in x2 + 2x − 3.
x2 + 2 x - 3
x-2
Ask yourself: “What times x gives x2 ?”
(?) · x = x2 .
You can see that x works, so put it on top:
x
x2 + 2 x - 3
x-2
Next, multiply the x − 2 by the x on top, and put the result x2 − 2x under x2 + 2x − 3. Line up terms
with the same power of x:
.
x
x2 + 2 x - 3
x2 - 2 x
x-2
=
2
2
Subtract: (x + 2x) − (x − 2x) = 4x.
x
x-2
x2 + 2 x - 3
2
- x -2x
4x
(When you subtract, be careful of the signs! In this case, the x2 terms cancel, but 2x−(−2x) = 2x+2x =
4x.)
Next, bring down the −3 and put it next to the 4x:
x
x-2
x2 + 2 x - 3
2
- x -2x
4x -3
Look at the first term (x) in x − 2 and the first term (4x) in 4x2 − 3.
x
x-2
x2 + 2 x - 3
2
- x -2x
4x -3
1
Ask yourself: “What times x gives 4x?”
(?) · x = 4x.
You can see that 4 works, so put it on top:
x+4
x-2
x2 + 2 x - 3
2
- x -2x
4x -3
Multiply the x − 2 by the 4 on top, and put the result 4x − 8 under 4x − 3. Line up terms with the
same power of x:
.
x+4
x-2
x2 + 2 x - 3
2
- x -2x
4x -3
4x -8
=
(You don’t multiply x − 2 by the x + 4 on top, just the 4; you already multiplied by the “x” in x + 4 in
an earlier step.)
Subtract: (4x − 3) − (4x − 8) = 5.
x+4
x-2
x2 + 2 x - 3
2
- x -2x
4x -3
- 4x -8
5
At this point, the “x” of “x − 2” doesn’t go into 5, so the division is finished.
The quotient is x + 4, the expression on the top. And the remainder is 5.
If you’re just asked for the quotient and remainder, you’re done.
x2 + 2x − 3
If the problem asked you to do the division
, then you’d write it this way:
x−2
5
x2 + 2x − 3
= (x + 4) +
.
x−2
x−2
The quotient x + 4 goes in front. The remainder 5 goes on top of the fraction. The expression x − 2,
which was on the bottom of the original fraction, goes on the bottom of the new fraction.
If you have trouble remembering where everything goes, think about how you convert improper frac38
tions to mixed numbers. Let’s say you want to convert
to a mixed number. To do this, you divide 38
7
by 7:
5
7 38
- 35
3
38
Then
= 5 73 .
7
2
However, if you think about it, 5 37 — which you read as “5 and
3
3
” — means “5 + ”, so
7
7
38
3
=5+ .
7
7
This is the same as what I did with the polynomials: The quotient 5 goes in front. The remainder 3
38
goes on top of the fraction. The expression 7, which was on the bottom of the original fraction
, goes on
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the bottom of the new fraction.
In other words, as often happens when you’re doing algebra, you can figure out “what to do with
variables” by thinking about “what you know to do with numbers”.
Example. Find the quotient and remainder when 2x3 − 2x + 1 is divided by x + 1.
Since 2x3 − 2x + 1 is missing an x2 term, I’ll write “0 x2 ” as a placeholder. It is optional, but this makes
it less likely that you’ll make a mistake when you do the subtraction.
Look at the first term (x) in x + 1 and the first term (2x3 ) in 2x3 − 2x + 1:
x+1
2x3 + 0 x2 - 2x + 1
Ask yourself: “What times x gives 2x3 ?”
(?) · x = 2x3 .
You can see that 2x2 works, so put it on top:
2x2
x+1
2x3 + 0 x2 - 2x + 1
Next, multiply x + 1 by the 2x2 on top, and put the result 2x3 + 2x2 under 2x3 − 2x + 1. Line up terms
with the same power of x:
.
2x2
x+1
=
2x3 + 0x2 - 2x + 1
2x3 + 2x2
2
Note that the 2x goes under the 0x2 I put in as a placeholder.
Subtract: (2x3 + 0x2 ) − (2x3 + 2x2 ) = −2x2 .
2x2
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2
Next, bring down the −2x and put it next to the −2x2 :
2x2
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2 - 2x
3
Look at the first term (x) in x + 1 and the first term (−2x2 ) in −2x2 − 2x:
2x2
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
x+1
-2x2 - 2x
Ask yourself: “What times x gives −2x2 ?”
(?) · x = −2x2 .
You can see that −2x works, so put it on top:
2x2 - 2x
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2 - 2x
Multiply x + 1 by the −2x on top (just the “−2x” — you multiplied by the “2x2 ”), and put the result
−2x − 2x under −2x2 − 2x:
.
2
2x2 - 2x
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2 - 2x
-2x2 - 2x
=
Subtract: (−2x2 − 2x) − (2x2 − 2x) = 0.
2x2 - 2x
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2 - 2x
2
- -2x - 2x
0
Bring down the 1 — since 0 + 1 = 1, I just write the 1:
2x2 - 2x
x+1
2x3 + 0x2 - 2x + 1
3 + 2x2
2x
-2x2 - 2x
2
- -2x - 2x
0+1
Since x + 1 does not go into 1, the division is finished:
2x2 - 2x
x+1
2x3 + 0x2 - 2x + 1
- 2x3 + 2x2
-2x2 - 2x
2
- -2x - 2x
1
4
The quotient is 2x2 − 2x and the remainder is 1.
In fraction form, this would be written as
2x3 − 2x + 1
1
= (2x2 − 2x) +
.
x+1
x+1
Example. Find the quotient and remainder when 3x3 − x2 − 7x + 12 is divided by x2 − 2x + 1.
Look at the first term (x2 ) in x2 − 2x + 1 and the first term (3x3 ) in 3x3 − x2 − 7x + 12:
x2 - 2 x + 1
3 x 3 - x2 - 7 x + 12
Ask yourself: “What times x2 gives 3x3 ?”
(?) · x2 = 3x3 .
You can see that 3x works, so put it on top:
3x
2
x -2x+1
3 x 3 - x2 - 7 x + 12
Next, multiply x2 − 2x + 1 by the 3x on top, and put the result 3x2 − 6x2 + 3x under 3x3 − x2 − 7x + 12.
Line up terms with the same power of x:
.
3x
x2 - 2 x + 1
3 x 3 - x 2 - 7 x + 12
3 x 3- 6 x 2+ 3 x
=
Subtract: (3x3 − x2 − 7x + 12) − (3x2 − 6x2 + 3x) = 5x2 − 10x.
3x
2
x -2x+1
3 x 3 - x 2 - 7 x + 12
3
2
- 3x -6x +3x
5 x2 - 10 x
Next, bring down the 12 and put it next to the 5x2 − 10x:
3x
2
x -2x+1
3 x 3 - x 2 - 7 x + 12
3
2
- 3x -6x +3x
5 x2 - 10 x + 12
Look at the first term (x2 ) in x2 − 2x + 1 and the first term (5x2 ) in 5x2 − 10x:
3x
2
x -2x+1
3 x 3 - x 2 - 7 x + 12
3
2
3
- x -6x +3x
5 x2 - 10 x + 12
5
Ask yourself: “What times x2 gives 5x2 ?”
(?) · x2 = 5x2 .
You can see that 5 works, so put it on top:
3x+5
x2 - 2 x + 1
3 x 3 - x 2 - 7 x + 12
3
2
- 3x -6x +3x
5 x2 - 10 x + 12
Multiply x2 − 2x + 1 by the 5 on top (just the “5” — you already multiplied by the “3x”), and put the
result 5x2 − 10x + 5 under 5x2 − 10x + 12:
3x+5
2
x -2x+1
3 x 3 - x 2 - 7 x + 12
3
2
- 3x -6x +3x
5 x2 - 10 x + 12
5 x2 - 10 x + 5
Subtract: (5x2 − 10x + 12) − (5x2 − 10x + 5) = 7.
3x+5
x2 - 2 x + 1
3 x 3 - x 2 - 7 x + 12
3
2
- 3x -6x +3x
5 x2 - 10 x + 12
2
- 5 x - 10 x + 5
7
Since x2 − 2x + 1 does not go into 7, the division is finished.
The quotient is 3x + 5 and the remainder is 7. In fraction form, this would be written as
7
3x3 − x2 − 7x + 12
= 3x + 5 + 2
.
x2 − 2x + 1
x − 2x + 1
Example.
x4 − 4x2 − 3
6
= x2 − 3 − 2
,
x2 − 1
x −1
or
x4 − 4x2 − 3 = (x2 − 3)(x2 − 1) − 6.
Example.
x3 + 4x2 − 5x − 20
= x2 − 5,
x+4
or
x3 + 4x2 − 5x − 20 = (x2 − 5)(x + 4).
Sine the remainder is 0, this shows that x3 + 4x2 − 5x − 20 factors as (x + 4)(x2 − 5).
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The Remainder Theorem. If p(x) is a polynomial, the remainder when p(x) is divided by x − a is p(a).
Example. Divide p(x) = 3x4 − 2x + 3 by x − 2:
3x4 − 2x + 3 = (x − 2)(3x3 + 6x2 + 12x + 22) + 47.
The remainder is 47. On the other hand,
p(2) = 3 · 16 − 4 + 3 = 47.
The remainder is the same as p(2).
The Root Theorem. x − a divides p(x) evenly if and only if p(a) = 0 (i.e. x = a is a root of p(x)).
Example. Let p(x) = 2x4 + 3x3 + 1.
p(−1) = 2 − 3 + 1 = 0.
Therefore, x − 1 should divide 2x4 + 3x3 + 1. In fact,
2x4 + 3x3 + 1 = (x − 1)(2x3 + x2 − x + 1).
Example. Let p(x) = x2 − 6x + 8. Then
p(x) = (x − 2)(x − 4).
Since x − 2 and x − 4 are factors, x = 2 and x = 4 should be roots — and they are:
p(2) = 4 − 12 + 8 = 0,
p(4) = 16 − 24 + 8 = 0.
Example. If you know or can guess a factor, you can sometimes complete a factorization by long division.
For example x3 − 27 has x = 3 as a root. By the fact I stated earlier, this means that x − 3 is a factor.
Divide x − 3 into x3 − 27 to get x2 + 3x + 9.
Therefore,
x3 − 27 = (x − 3)(x2 + 3x + 9).
c 2012 by Bruce Ikenaga
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