Download Divide Polynomials When dividing by a monomial, we divide

Divide Polynomials
When dividing by a monomial, we divide each term by the monomial.
By doing so, we basically create individual fractions which are then
reduced in the same way we reduced fractions when we discussed the
quotient rule for exponents.
Example 1: Divide: (6x2 – 8x + 12) ÷ (-2)
6x2 – 8x + 12
-2
6x2
8x
–
−2
−2
+
Divide each term in the numerator by -2
12
−2
Reduce individual fractions
-3x2 + 4x – 6
Example 2: Divide: 9x5 + 6x4 – 18x3 – 3x2
3x2
9x5
3x2
+
6x4
3x2
–
18x3
3x2
–
3x2
Divide each term in the numerator by 3x2
Reduce individual fractions
3x2
3x3 + 2x2 – 6x – 1
Example 3: Divide: (8x3 + 4x2 – 2x + 6) ÷ (4x2)
8x3 + 4x2 – 2x + 6
4x2
8x3
4x2
+
4x2
4x2
2x + 1 –
-
1
2x
2x
4x2
+
+
Divide each term in the numerator by 4x2
6
4x2
Reduce individual fractions
3
2x2
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Long division is required when dividing by a polynomial that is larger
than a monomial. Long division with polynomials works very similar to
long division with whole numbers. An example is given to review the
general steps that are used with whole numbers.
Example 4: 631 ÷ 4
1
4 631
-4
23
Divide the first numbers 6 ÷ 4 ≈ 1
Multiply 1· 4 and then subtract 6 - 4
Bring down the next number 3
15
4 631
-4
23
- 20
31
Divide 23 ÷ 4 ≈ 5
Multiply 5 · 4 and then subtract 23 - 20
Bring down the next number 1
157
4 631
-4
23
- 20
31
- 28
3
Divide 31 ÷ 4 ≈ 7
Multiply 7 · 4 and then subtract 31 – 28
Remainder of 3
The answer is 157
3
4
The same procedure is used to divide polynomials. The only difference
is the word “number” is replaced with the word “term”.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 5: Divide: (x2 + 2x – 24) ÷ (x – 4)
x
x + 6 x + 2x - 24 Divide the first terms x2 ÷ x = x
- x2 – 6x
Multiply x(x+6)=x2+6x and then subtract –(x2+6x) = -x2 – 6x
- 4x – 24
Bring down the next term -24
2
x-4
x + 6 x + 2x - 24
- x2 – 6x
- 4x – 24
Divide -4x ÷ x = -4
4x + 24
Multiply -4(x+6)=-4x-24 and then subtract –(-4x-24) = 4x + 24
0
No remainder
2
The answer is x – 4
To check the answer: (x – 4) (x + 6)
x(x + 6) – 4(x + 6)
x2 + 6x – 4x – 24
x2 + 2x – 24 Correct – same as the dividend
Example 6: Divide:
3x3 − 5x2 − 32x + 7
x−4
3x2
x – 4 3x3 – 5x2 – 32x + 7 Divide the first terms 3x3 ÷ x = 3x2
- 3x3 + 12x2
Multiply 3x2(x–4)=3x3-12x2 and subtract –(3x3-12x2)= -3x3+12x2
7x2 – 32x
Bring down the next term -32x
3x2 + 7x
x – 4 3x3 – 5x2 – 32x + 7
- 3x3 + 12x2
7x2 – 32x
Divide 7x2 ÷ x = 7x
- 7x2 + 28x
Multiply 7x(x–4)=7x2-28x and subtract –(7x2-28x) = -7x2+28x
- 4x + 7 Bring down the next term 7
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
3x2 + 7x – 4
x – 4 3x3 – 5x2 – 32x + 7
- 3x3 + 12x2
7x2 – 32x
- 7x2 + 28x
- 4x + 7
4x – 16
-9
Divide -4x ÷ x = -4
Multiply -4(x–4)=-4x+16 and subtract –(-4x+16)= 4x - 16
Remainder of -9
The answer is 3x2 + 7x – 4 –
9
x−4
To check the answer: (x – 4)(3x2 + 7x – 4)
x(3x2 + 7x – 4) – 4(3x2 + 7x – 4)
3x3 + 7x2 – 4x – 12x2 – 28x + 16
3x3 + 7x2 – 12x2 – 4x – 28x + 16
3x3 – 5x2 – 32x + 16
3x3 – 5x2 – 32x + 16 – 9 Add the remainder
3x3 – 5x2 – 32x + 7
Correct – same as the dividend
Example 7: Divide: (6x3 – 8x2 + 10x + 103) ÷ (2x + 4)
3x2
2x + 4 6x3 – 8x2 + 10x + 103 Divide the first terms 6x3 ÷ 2x = 3x2
- 6x3–12x2
Multiply 3x2(2x+4)=6x3+12x2 and subtract –(6x3+12x2)= -6x3-12x2
- 20x2 + 10x
Bring down the next term 10x
3x2 – 10x
2x + 4 6x3 – 8x2 + 10x + 103
- 6x3–12x2
- 20x2 + 10x
Divide -20x2 ÷ 2x = -10x
20x2 + 40x Multiply -10x(2x+4)=-20x2-40x and subtract –(-20x2-40x)= 20x2+40x
50x + 103
Bring down the next term 103
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
3x2 – 10x + 25
2x + 4 6x3 – 8x2 + 10x + 103
- 6x3–12x2
- 20x2 + 10x
20x2 + 40x
50x + 103
- 50x – 100
3
Divide 50x ÷ 2x = 25
Multiply 25(2x+4)=50x+100 and subtract –(50x+100)= -50x-100
Remainder of 3
The answer is 3x2 – 10x + 25 +
3
2x+4
To check the answer: (2x + 4)(3x2 – 10x + 25)
2x(3x2 – 10x + 25) + 4(3x2 – 10x + 25)
6x3 – 10x2 + 50x + 12x2 – 40x + 100
6x3 – 10x2 + 12x2 + 50x – 40x + 100
6x3 + 2x2 + 10x + 100
6x3 + 2x2 + 10x + 100 + 3 Add the remainder
6x3 + 2x2 + 10x + 103
Correct – same as the dividend
In long division, it is very important that the terms are written in
descending powers and no exponent skipped. If the polynomial is not
written in descending powers, it must be put in order. If an exponent is
skipped, a “filler” term, with zero as its coefficient, must be added.
Example 8: Divide:
2x3 +42−4x
x+3
2x3 – 4x + 42
Reorder the dividend so exponents are descending
2x3 + 0x2 – 4x + 42
Add a “filler” term 0x2
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
2x2
x + 3 2x3 + 0x2 – 4x + 42 Divide the first terms 2x3 ÷ x = 2x2
- 2x3 – 6x2
Multiply 2x2(x+3)=2x3+6x2 and subtract –(2x3+6x2) = -2x3+6x2
- 6x2 – 4x
Bring down the next term -4x
2x2 – 6x
x + 3 2x3 + 0x2 – 4x + 42
- 2x3 – 6x2
- 6x2 – 4x
Divide -6x2 ÷ x = -6x
6x2 + 18x
Multiply -6x(x+3)=-6x2-18x and subtract –(-6x2-18x) = 6x2+18x
14x + 42 Bring down the next term 42
2x2 – 6x + 14
x + 3 2x3 + 0x2 – 4x + 42
- 2x3 – 6x2
- 6x2 – 4x
6x2 + 18x
14x + 42
- 14x – 42
0
Divide 14x ÷ x = 14
Multiply 14(x+3)=14x+42 and subtract –(14x+42) = -14x-42
No remainder
The answer is 2x2 – 6x + 14
To check the answer: (x + 3)(2x2 – 6x + 14)
x(2x2 – 6x + 14) + 3(2x2 – 6x + 14)
2x3 – 6x2 + 14x + 6x2 – 18x + 42
2x3 – 6x2 + 6x2 + 14x – 18x + 42
2x3 + 0x2 – 4x + 42
2x3 – 4x + 42 Correct – same as the dividend
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)