Int Alg Lecture Notes, Section 5.3

Int. Alg. Notes Section 5.3 Page 1 of 5

Section 5.3: Dividing Polynomials; Synthetic Division Big Idea: Polynomials are the most important topic in algebra because any equation that can be written using addition, subtraction, multiplication, division, integer powers, or roots (which are rational powers) can be solved by converting the equation into a polynomial equation. The third step toward acquiring this awesome power is to be able to divide polynomials.

Big Skill: You should be able to divide polynomials using long division and, when appropriate, synthetic division.

Dividing a polynomial by a monomial: Divide the monomial into each term of the polynomial, and cancel when possible.

Practice: 24z5 1. = 18z2

9p4- 12 p 3 + 3 p 2 2. = 3p

x4 y 4+8 x 2 y 2 - 4 xy 3. = 4x3 y

Algebra is: the study of how to perform multi-step arithmetic calculations more efficiently, and the study of how to find the correct number to put into a multi-step calculation to get a desired answer. Int. Alg. Notes Section 5.3 Page 2 of 5

Dividing a polynomial by a polynomial using long division: Long division of polynomials is a lot like long division of numbers: a. Arrange divisor and dividend around the dividing symbol, and be sure to write them in descending order of powers with all terms explicitly stated (even the terms with zero coefficients). b. Divide leading terms, then multiply and subtract. c. Repeat until a remainder of order less than the divisor is obtained.

Comparison between dividing integers and dividing polynomials Dividend Remainder =Quotient + Divisor Divisor Compute 579 ÷ 16 Compute (5x2 + 7x +9) ÷ (x + 6)

Practice: 6x3- 7 x 2 + 6 x - 6 1. = 2x - 1

8- 9x + 2 x2 + 12 x 3 + 5 x 5 2. = x2 + 3

Dividing a polynomial by a binomial using synthetic division: THIS IS A SHORTCUT THAT ONLY WORKS WHEN THE DIVISOR IS A LINEAR BINOMIAL (I.E., THE DIVISOR IS x – c) !!!

Synthetic division is a shorthand way to divide a polynomial by the linear factor x – c: a. Write c outside the division bar and the coefficients of the dividend inside the bar. b. Bring the leading coefficient of the dividend straight down. c. Compute c times the number in the bottom row, and write the answer in the middle row to the right. d. Add and repeat until all coefficients are used up.

Comparison between long and synthetic division of polynomials Compute (2x3 – 3x2 – 4x + 11) ÷ (x – 2) using long Compute (2x3 – 3x2 – 4x + 11) ÷ (x – 2) using synthetic division. division.

Practice: 2x3+ x 2 - 7 x - 13 1. = x - 2

x4+8 x 3 + 15 x 2 - 2 x - 6 2. = x + 3

3x2 + 4 x - 7 3. = 2x + 5

Definition: Quotient of Functions f If f and g are two functions, then the new function that can be made by taking their quotient is called , and is g 骣f f( x) defined as: 琪 ( x) = , provided g(x)  0. 桫g g( x)

Practice:

3 2 骣f 骣f 1. If f( x) = x -2 x - 4 x - 5 and g( x) = x + 2 , find 琪 ( x) and 琪 (3) . 桫g 桫g

The Remainder Theorem If the polynomial P(x) is divided by x – c, then the remainder is the value P(c). This is because when we divide a polynomial by x – c, the remainder must be of degree less than x – c, which means the remainder has degree of zero, which is just a numeric constant R. So:

Dividend Remainder =Quotient + Divisor Divisor P( x) R =Q( x) + x- c x - c P( x) 骣 R �( x= c) + 琪 Q( x) � ( x c) x- c桫 x - c P( x) = Q( x)�( x + c) R P( c) = Q( c)�( c + c) R R= P( c)

Practice: 1. Use the remainder theorem to find the remainder when f( x) =3 x3 - 2 x + 6 is divided by x + 2 .

The Factor Theorem If P(x) is a polynomial function, then x – c is a factor of P(x) if and only if R = P(c) = 0. In other words, if P(c) = 0, then P(x) can be written as P(x) = (x – c)(Quotient(x)).

(This can be used to see if a divisor divides evenly into a dividend quickly.)

Practice: 1. Use the factor theorem to determine whether f( x) =2 x3 - x 2 - 16 x + 15 is divisible by x – 2 and x + 3.

Algebra is: the study of how to perform multi-step arithmetic calculations more efficiently, and the study of how to find the correct number to put into a multi-step calculation to get a desired answer.