Download Division of polynomials

Division of polynomials
Division of polynomials closely resembles the division of integers. Examine the following side-by-side long
divisions. How are they alike? (Adapted from Demana and Waits, Precalculus, p. 207)
x 2 + 2x + 1
121
23 2789
23
Many polynomial division problems will not
have coefficients that match up to the
digits of numbers in the division of whole
numbers like in the example. However, this
example was cherry picked to illustrate a
key point: the process of performing long
division with whole numbers closely
resembles the process of performing long
division with polynomials. Every division,
whether integer or polynomial, involves a
dividend, divided by a divisor to obtain a
quotient and a remainder.
2 x + 3 2 x 3 + 7 x 2 + 8x + 9
2 x 3 + 3x 2
48
46
4 x 2 + 8x
4x 2 + 6x
29
23
2x + 9
2x + 3
6
6
Polynomial division problems usually come in one of two forms:
dividend polynomial
divisor polynomial
or
dividend polynomial ÷ divisor polynomial.
2 x 3 + 7 x 2 + 8x + 9
2x + 3
or
2 x 3 + 7 x 2 + 8 x + 9 ÷ (2 x + 3)
DIVISION ALGORITHM FOR POLYNOMIALS
Let p(x) and d(x) with the degree of p greater than or equal to the degree of d, and d ( x ) ≠ 0 . Then there are
unique polynomials q(x) and r(x), called the quotient and remainder, such that p ( x) = d ( x) • q ( x) + r ( x ) where
either r(x) = 0 or the degree of r is less than the degree of g. Say what?
WHAT THIS MEANS IN ENGLISH: Consider the integer division problem above:
2789
. Back in the day
23
(elementary school), when you finished a long division problem, you got to the “big R” which was the remainder,
and you would have written the answer 121 R 6 . In middle school, you would have written the answer as a mixed
number, so 121 R 6 becomes 121
6
. Changing the mixed number back into an improper fraction, you would
23
multiply 23 by 121 and then add 6, and put the result over 23.
In general,
2789
=
(23) (121)
+6
Dividend = (Divisor)(Quotient) + Remainder
So for polynomials, if p(x) is the dividend, d(x) is the divisor, q(x) is the quotient and r(x) is the remainder, then
the statement p ( x) = d ( x) • q ( x) + r ( x ) makes a little more sense. As illustrated above, this works for
integers as well as for polynomial division. When you divide the dividend by the divisor, you get a quotient and a
remainder. To check the problem, you multiply the divisor by the quotient and add the remainder to get the
dividend.
q ( x)
p( x)
So polynomial long division will look like this :
= d ( x) p ( x)
d ( x)
r ( x)
The solution to the polynomial division problem above now can be written
in the form
Dividend
remainder
2 x 3 + 7 x 2 + 8x + 9
6
= quotient +
where
, or in
= x 2 + 2x + 1 +
divisor
divisor
2x + 3
2x + 3
the form Dividend = (Divisor)(Quotient) + Remainder where 2 x + 7 x
3
2
(
)
+ 8 x + 9 = (2 x + 3) x 2 + 2 x + 1 + 6
LONG DIVISION
Follow these steps when performing long division of polynomials.
1. Write the dividend and divisor in descending powers of the variable.
2. Insert placeholders with zero coefficients for missing powers of the variable.
3. Perform the long division of the polynomial as you would with integers.
4. Continue the process until the degree of the remainder is less than the degree of the divisor.
2
3
EXAMPLE 1 Divide 13x − 10 x + 2 x + 17 . Write your solution in the form Dividend = quotient + remainder .
divisor
divisor
x−3
2 x 2 − 4 x +1
x − 3 2 x − 10 x + 13x + 17
3
2x − 6x
3
Divide 2x3 /x and put 2x2 in the quotient
2
2
Write the dividend in descending degree
Multiply 2x2(x – 3) and write product here
−4 x 2 + 13x
Subtract 2x3 – 6x2 from the first two terms of the dividend and write
difference here, bringing down the 13x. Now divide -4x2 /x, put -4x in
the quotient and repeat the whole process again.
− 4 x 2 + 12 x
x + 17
x−3
20
The “big R” = remainder
2 x 3 − 10 x 2 + 13 x + 17
20
remainder
= 2x 2 − 4x +1 +
= quotient +
x−3
x−3
divisor
The following examples come from College Algebra Demystified:
EXAMPLE 2 Divide:
EXAMPLE 3: Divide:
Solution:
5x − 1
= x3 + x 2 − 2x + 4 + 2
x + 2x − 3
Solution:
= 3 x 3 + 12 x 2 + 48 x + 187 +
750
x−4
Both examples give a very clear illustration of how to subtract polynomials in the long division process. Dropping
signs is a very common source of errors. Moreover, Example 2 is a good example of how to divide a quadratic
divisor into a higher degree polynomial. Since 5x is not divisible by x2, you know that you are finished because
5x is one less degree than x2. Example 3 also demonstrates how to insert place markers for missing powers.
SYNTHETIC DIVISION
Synthetic division is division of polynomials using only the coefficients. It is a very useful shortcut when
dividing polynomials. It is mostly used when dividing by a linear divisor in the form (x- a), but it can be used to
divide a polynomial by any linear factor. The example below shows how synthetic division works in stages using
Example 1 above.
2x 2 − 4x +1
x − 3 2 x 3 − 10 x 2 + 13x + 17
2
−3 2
2x 3 − 6x 2
2
−4
− 10
− 4 x 2 + 12 x
− 3|
2
−6
−4
−4
−4 x 2 + 13x
1
+ 13 + 17
+ 13
+ 12
2
−10
13
17
−6
12
−3
−4
1
20
1 + 17
1 −3
1x + 17
1x − 3
20
20
Stage 1 Long Division
Stage 2 Variables suppressed
Stage 3 Collapsed vertically
Moving from stage to stage, focus on the coefficients and their relative positions. Moving from Stage 1 to
Stage 2, we suppress the powers of x, and then from Stage 2 to Stage 3, we eliminate unneeded duplications
and collapse vertically.
3|
2
17
−10 13
Finally, moving from Stage 3 to Stage 4 at right, we change the sign of the
coefficient representing the divisor and the sign of each of the numbers on the
6 − 12
3
second line. This means that the divisor 3 represents the factor (x – 3) like it
2
−4
1
20
would in the Remainder Theorem. Another advantage is that changing the signs
SYNTHETIC DIVISION!
of the second line allows us to add rather than subtract.
EXAMPLE 4 How to do SYNTHETIC DIVISION!! Divide:
2x 3 + 9x 2 + 7 x − 8
Follow these steps:
x+2
1) Write -2 (not 2) in the little box because -2 is what is being subtracted from x to get (x + 2) in the divisor.
2) Write the coefficients of the dividend to the right of the box. [Insert 0 for a place marker, if necessary.]
3) Skip a line and draw a line. Bring down the 2 (leading coefficient) below the line. See below left.
− 2|
2
9
7
−8
− 2|
2
−4
2
5
2
9
7
−8
−4
−10
6
5
−3
−2
4) Multiply -2 (the divisor) by 2 (the leading coefficient) and put the product (-4) in the blank below the 9.
5) ADD 9 and -4 to get 5.
6) Repeat Steps 4 and 5 until all the blanks are filled. See above right. The very last value is the remainder.
Interpreting the Results- If the remainder is zero, you have found a zero of the function p ( x ) , where
p ( x ) = dividend . The rest of the values are the coefficients of the quotient. Each term will be raised to the
one less power than the original dividend. In this case, you divided a cubic polynomial by a linear factor, so the
quotient will be quadratic.
Solution:
2x 3 + 9x 2 + 7 x − 8
−2
= 2 x 2 + 5x − 3 +
x+2
x +2
Remember, you should use synthetic division only when dividing by a linear factor in the form (x – a)!