Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 1 Chapter 6 Polynomial Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide Dividing Polynomials: Long Division and Synthetic Division

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 3 Dividing by a Monomial If A, B, and C are monomials and B is nonzero, then In words: To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 4 Example: Dividing by a Monomial Find the quotient.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 5 Solution Verify our work by finding the product:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 6 Using Long Division to Divide by a Binomial We can use long division to divide a polynomial by a binomial. The steps are similar to performing long division with numbers. Recall that we can verify our work by checking that Divisor ∙ Quotient + Remainder = Dividend

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 7 Example: Dividing by a Binomial Divide.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 8 Solution Divide 2x 2 (the first term of the dividend) by x (the first term of the divisor):

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 9 Solution To subtract 2x 2 + 6x, we change the signs of 2x 2 and 6x and add:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 10 Solution Next, bring down the 15:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 11 Solution The repeat the process.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 12 Solution We conclude: To check: which is true.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 13 Example: Dividing a Polynomial with Missing Terms Divide.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 14 Solution First, write the dividend in decreasing order: 5x 3 + 2x 2 – 4x + 1 Also, since the divisor doesn’t have an x term, we use 0x as a placeholder: x 2 + 0x – 3

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 15 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 16 Solution We conclude: To check: which is true.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 17 Synthetic Division When we divide a polynomial by a binomial of the form x – a, we can use a method called synthetic division, which uses the ideas of long division but is more efficient. To perform synthetic division, the divisor must be of the form x – a.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 18 Example: Performing Synthetic Division Use synthetic division to divide:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 19 Solution When performing synthetic division, there is no need to write the coefficient of x of the divisor x – 2, The method already takes the divisions by x into account. Also, instead of dividing by the constant term –2, we divide by its opposite: 2. That way, we can add without having to change signs first.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 20 Solution Start by writing 2 (of x – 2) and the coefficients of the dividend. Then bring down the 3.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 21 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 22 Solution So, the result is 3x 2 – x + 8 with a remainder 4.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 23 Solution We conclude:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 24 Example: Performing Synthetic Division when There Are Missing Terms Use synthetic division to divide:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.3, Slide 25 Solution Because the dividend does not have a p 2 term, we use 0 as a placeholder: We conclude: