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Dividing Polynomials Depends on the situation. Situation I: Polynomial Monomial Solution is to divide each term in the numerator by the monomial. For example: xy xy y x y x 3 18 6 3 2 3 2 6 2 3 18 3 6 3 3 2 2 3 2 y x x xy xy xy y x xy y x

Dividing Polynomials Depends on the situation. Situation I: Polynomial Monomial Solution is to divide each term in the numerator by the monomial

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Page 1: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Depends on the situation.Situation I:

Polynomial

MonomialSolution is to divide each term in the numerator by the

monomial.For example:

xy

xyyxyx

3

1863 232

62

3

18

3

6

3

3

2

232

yxx

xy

xy

xy

yx

xy

yx

Page 2: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Situation II:

Polynomial

BinomialTwo methods of solution.

Method 1 Long DivisionRemember the process for long division of numbers.For example: 123 ÷ 3

3 123

41

– 12

3– 3

0

Page 3: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Basics of Division

Parts of Division

3 123

41

divisor dividend

quotient

Page 4: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Method 1 Long DivisionFor example:

4

281142811

212

x

xxxxx

x – 4 x2 – 11x + 28Focus on the first term of the divisor.

What does this term need to be multiplied by to equal the first term of the dividend?

x

Now multiply the entire binomial by this term.

x2 – 4x

Page 5: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Method 1 Long DivisionFor example:

4

281142811

212

x

xxxxx

x – 4 x2 – 11x + 28Now subtract the terms.

x

x2 – 4x

– 7x + 28

Now bring down the next term from the dividend.

Repeat the process. So once again, focus on the first term of the binomial.

Page 6: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Method 1 Long DivisionFor example:

4

281142811

212

x

xxxxx

x – 4 x2 – 11x + 28

x

– 7x + 28

What does x need to be multiplied by to get –7x?

– 7

Once again multiply the entire binomial by this term.– 7x + 28

Now subtract the terms.

x2 – 4x

0

So the answer or quotient is (x – 7) .

Page 7: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Practice

Divide the following using long division.

4263 2 nnn

1664263 2

nnnn

nn 186 2 416 n4816 n

52

3

52166

nnAnswer (Quotient):

remainder

Page 8: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

Method 2 Synthetic DivisionStep 1 Write the polynomial (dividend) in

descending order of exponents and be sure to account for all powers of the variable.

For example: (x4 – 2x3 + 2x – 1)(x + 1)-1

x4 – 2x3 + 0x2 + 2x – 1

Step 2 Write coefficients of the dividend in order.

1 –2 0 2 –1

Page 9: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

1 –2 0 2 –1

Step 3 Write constant value r of divisor (x – r) to the left of the coefficients. In this case with the divisor (x + 1) has constant –1.

Step 4 Bring down first coefficient. Multiply by constant and add result to second coefficient.

Repeat process successively until last coefficient.

–1

1

–1

–3

3

3

–3

–1

1

0

remainder

1

1202 234

x

xxxx

x – (– 1)

Page 10: Dividing Polynomials  Depends on the situation.  Situation I: Polynomial Monomial  Solution is to divide each term in the numerator by the monomial

Dividing Polynomials

1 –2 0 2 –1 –1

1

–1

–3

3

3

–3

–1

1

0

Step 5 Write the quotient with the numbers along the bottom row as the coefficients of the powers of the variable in descending order. Start with the power that is one less than that of the dividend. In this case it is 3.

Quotient: x3 – 3x2 + 3x – 1