Download presentation

Presentation is loading. Please wait.

Published byClaire Scott Modified over 9 years ago

1
Polynomial Division, Factors, and Remainders ©2001 by R. Villar All Rights Reserved

2
Polynomial Division, Factors, and Remainders In this section, we will look at two methods to divide polynomials: long division (similar to arithmetic long division) synthetic division (a quicker, short-hand method) Let’s take a look at long division of polynomials...

3
Example: Divide (2x 2 + 3x – 4) ÷ (x – 2) (x – 2) 2x 2 + 3x – 4 Rewrite in long division form... divisor dividend Think, how many times does x go into 2x 2 ? 2x Multiply by the divisor. 2x 2 – 4x Subtract. 7x – 4 Think, how many times does x go into 7x ? + 7 7x – 14 10 remainder 2x + 7 + 10 x – 2 divisor Write the result like this...

4
Example: Divide (p 3 – 6) ÷ (p – 1) (p – 1) p 3 + 0p 2 + 0p – 6 Be sure to add “place-holders” for missing terms... p2p2 p 3 – p 2 p 2 + 0p + p p 2 – p p – 6 p 2 + p + 1 – 5 p – 1 + 1 p – 1 –5 Let’s look at an abbreviated form of long division, called synthetic division...

5
Synthetic division can be used when the divisor is in the form (x – k). Example: Use synthetic division for the following (2x 3 – 7x 2 – 8x + 16) ÷ (x – 4) First, write down the coefficients in descending order, and k of the divisor in the form x – k : 4 2 –7 –8 16 k 2 Bring down the first coefficient. 8 Multiply this by k 1 Add the column. 4 –4 –16 0 These are the coefficients of the quotient (and the remainder) 2x 2 + x – 4 Repeat the process.

6
Example: Divide (5x 3 + x 2 – 7) ÷ (x + 1) –1 5 1 0 –7 Notice that k is –1 since synthetic division works for divisors in the form (x – k). place-holder 5x 2 – 4x + 4 – 11 x + 1 5 –5 –4 4 4 –11

7
2 2 1 –3 0 –5 f(2) = 23 2 4 5 10 7 14 Now, let f(x) = 2x 4 + x 3 – 3x 2 – 5 28 23 What is f(2)? f(2) = 2(2) 4 + (2) 3 – 3(2) 2 – 5 f(2) = 2(16) + 8 – 3(4) – 5 f(2) = 32 + 8 – 12 – 5 f(2) = 23 This is the same as the remainder when divided by (x – 2):

8
4 1 –6 8 5 13 f(4) = 33 1 4 –2 –8 0 0 5 Example: Use synthetic substitution to find f(4) if f(x) = x 4 – 6x 3 + 8x 2 + 5x + 13 20 33

9
–2 1 –13 24 108 This means that you can write x 3 – 13x 2 + 24x + 108 = (x + 2)(x 2 – 15x + 54) 1 –2 –15 30 54 –108 0 You can also use synthetic division to find factors of a polynomial... Example: Given that (x + 2) is a factor of f(x), factor the polynomial f(x) = x 3 – 13x 2 + 24x + 108 Since (x + 2) is a factor, –2 is a zero of the function… We can use synthetic division to find the other factors... This is called the depressed polynomial Factor this... = (x + 2)(x – 9)(x – 6) The complete factorization is (x + 2)(x – 9)(x – 6)

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google