Transcript Chapter Six 6.4

§ 6.4
Division of Polynomials
Division of Polynomials
1st case
Dividing a Polynomial by a
Monomial
To divide a polynomial by a monomial,
divide each term of the polynomial by
the monomial.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.4
Division of Polynomials
EXAMPLE
4 3
3 2
2
3 3
Divide: 40x y  20x y  50x y  10x y .
SOLUTION
40x 4 y 3  20x 3 y 2  50x 2 y
10x 3 y 3
Express the division in a vertical
format.
40x 4 y 3 20x3 y 2 50x 2 y



3 3
3 3
10x y 10x y 10x3 y 3
Divide each term of the polynomial
by the monomial. Note the 3
separate quotients.
2
5
 4x   2
y xy
Simplify each quotient.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.4
Division of Polynomials
EXAMPLE
2nd case
DIVISOR
QUOTIENT
4
x  2 4x  7
4x  8
1
REMAINDER
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.4
DIVIDEND
Division of Polynomials
EXAMPLE
3
2
Divide: x  5x  7 x  2 x  2 .
SOLUTION
x  2 x  5x  7 x  2
3
2
x2
x  2 x3  5x 2  7 x  2
Arrange the terms of the
dividend, x3  5x 2  7 x  2 , and
the divisor, (x + 2), in descending
powers of x.
Divide x3 (the first term in the
dividend) by x (the first term in
the divisor). Align like terms.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.4
Division of Polynomials
CONTINUED
x2
x  2 x  5x  7 x  2
3
2
x3  2 x 2
x2
x  2 x3  5x 2  7 x  2
x3  2 x 2
3x 2
Multiply each term in the divisor
(x + 2) by x 2 , aligning terms of
the product under like terms in
the dividend.
Subtract x 3  2x 2 from x3  5x 2
by changing the sign of each
term in the lower expression and
then adding.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.4
Division of Polynomials
CONTINUED
x2
x  2 x3  5x 2  7 x  2
x3  2 x 2
Bring down 7x from the original
dividend and add algebraically to
form a new dividend.
3x 2  7 x
x 2  3x
x  2 x3  5x 2  7 x  2
x3  2 x 2
3x 2  7 x
Find the second term of the
quotient. Divide the first term
of 3x 2  7 x by x, the first term
of the divisor.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.4
Division of Polynomials
CONTINUED
x 2  3x
x  2 x3  5x 2  7 x  2
x3  2 x 2
3x 2  7 x
Multiply the divisor (x + 2) by
3x, aligning under like terms in
the new dividend. Then subtract.
3x 2  6 x
x
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.4
Division of Polynomials
CONTINUED
x 2  3x  1
x  2 x3  5x 2  7 x  2
x3  2 x 2
3x  7 x
2
3x 2  6 x
x2
x2
0
Bring down 2 from the original
dividend and add algebraically to
form a new dividend.
Find the third term of the
quotient, 1. Divide the first term
of x + 2 by x, the first term of the
divisor.
Multiply the divisor by 1,
aligning under like terms in the
new dividend. Then subtract to
obtain the remainder of 0.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.4
Division of Polynomials
CONTINUED
The quotient is x 2  3x  1 and the remainder is 0. We will not list
a remainder of 0 in the answer. Thus,
x
3

 5x 2  7 x  2  x  2  x2  3x  1.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.4
Long Division of Polynomials
Long Division of Polynomials
1) Arrange the terms of both the dividend and the divisor in descending
powers of any variable.
2) Divide the first term in the dividend by the first term in the divisor. The
result is the first term of the quotient.
3) Multiply every term in the divisor by the first term in the quotient. Write
the resulting product beneath the dividend with like terms lined up.
4) Subtract the product from the dividend.
5) Bring down the next term in the original dividend and write it next to the
remainder to form a new dividend.
6) Use this new expression as the dividend and repeat this process until the
remainder can no longer be divided. This will occur when the degree of the
remainder (the highest exponent on a variable in the remainder) is less than the
degree of the divisor.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.4
Long Division of Polynomials
EXAMPLE
5
3
2
2
Divide: 3x  x  4x 12x  8 x  2.
SOLUTION
We write the dividend, 3x5  x3  4 x 2 12x  8 , as
3x5  0 x 4  x3  4 x 2  12x  8 to keep all like terms aligned. For the
same reason, we write the divisor, x 2  2 , as x 2  0 x  2.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.4
Long Division of Polynomials
CONTINUED
 5x  4
3x 3
x 2  0 x  2 3 x 5  0 x 4  x 3  4 x 2  12x  8
3x 5  0 x 4  6 x 3
5 x 3  4 x 2  12x
5 x 3  0 x 2  10x
4x2  2x  8
4x2  0x  8
 2x
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.4
Long Division of Polynomials
CONTINUED
The division process is finished because the degree of -2x, which
is 1, is less than the degree of the divisor x 2  2 , which is 2. The
answer is
3x 5  x 3  4 x 2  12x  8
 2x
3
 3x  5 x  4  2
.
2
x 2
x 2
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.4
Long Division of Polynomials
Important to Remember:
To divide by a polynomial containing more than one term, use long
division. If necessary, arrange the dividend in descending powers
of the variable. Do the same with the divisor. If a power of a
variable is missing in the dividend, add that term using a
coefficient of 0.
Repeat the four steps of the long-division process – divide,
multiply, subtract, bring down the next term – until the degree of
the remainder is less than the degree of the divisor.
When the degree of the remainder is less than the degree of the
divisor – you know you are done!
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.4
DONE