2.3: Polynomial Division Objectives: To divide polynomials using long and synthetic division To apply the Factor and Remainder Theorems to find real zeros of polynomial functions

Vocabulary As a class, use your vast mathematical knowledge to define each of these words without the aid of your textbook. Quotient Remainder Dividend Divisor Divides Evenly Factor

Exercise 1 Use long division to divide 5 into 3462. - - -

Exercise 1 Use long division to divide 5 into 3462. Quotient Divisor Dividend - - - Remainder

Exercise 1 Use long division to divide 5 into 3462. Dividend Remainder Divisor Divisor Quotient

Remainders If you are lucky enough to get a remainder of zero when dividing, then the divisor divides evenly into the dividend. This means that the divisor is a factor of the dividend. For example, when dividing 3 into 192, the remainder is 0. Therefore, 3 is a factor of 192.

Dividing Polynomials Dividing polynomials works just like long division. In fact, it is called long division! Before you start dividing: Make sure the divisor and dividend are in standard form (highest to lowest powers). If your polynomial is missing a term, add it in with a coefficient of 0 as a place holder.

How many times does x go into x2? Exercise 2 Divide x + 1 into x2 + 3x + 5 Line up the first term of the quotient with the term of the dividend with the same degree. How many times does x go into x2? Multiply x by x + 1 - - - - Multiply 2 by x + 1

Exercise 2 Divide x + 1 into x2 + 3x + 5 Quotient Dividend Divisor - - - - Divisor Remainder

Exercise 2 Divide x + 1 into x2 + 3x + 5 Dividend Remainder Divisor Quotient Divisor

Exercise 3 Divide 6x3 – 16x2 + 17x – 6 by 3x – 2

Exercise 4 Use long division to divide x4 – 10x2 + 2x + 3 by x – 3

Synthetic Division When your divisor is of the form x - k, where k is a constant, then you can perform the division quicker and easier using just the coefficients of the dividend. This is called fake division. I mean, synthetic division.

Synthetic Division Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. k a b c d = Add terms ka = Multiply by k a Remainder Coefficients of Quotient (in decreasing order)

Synthetic Division Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. Important Note: You are always adding columns using synthetic division, whereas you subtracted columns in long division. k a b c d = Add terms ka = Multiply by k a

Synthetic Division Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. Important Note: k can be positive or negative. If you divide by x + 2, then k = -2 because x + 2 = x – (-2). k a b c d = Add terms ka = Multiply by k a

Synthetic Division Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. Important Note: Add a coefficient of zero for any missing terms! k a b c d = Add terms ka = Multiply by k a

Exercise 5 Use synthetic division to divide x4 – 10x2 + 2x + 3 by x – 3

Exercise 6 Evaluate f (3) for f (x) = x4 – 10x2 + 2x + 3.

Remainder Theorem If a polynomial f (x) is divided by x – k, the remainder is r = f (k). This means that you could use synthetic division to evaluate f (5) or f (-2). Your answer will be the remainder.

Exercise 7 Divide 2x3 + 9x2 + 4x + 5 by x + 3 using synthetic division.

Exercise 8 Use synthetic division to divide f(x) = 2x3 – 11x2 + 3x + 36 by x – 3. Since the remainder is zero when dividing f(x) by x – 3, we can write: This means that x – 3 is a factor of f(x).

Factor Theorem A polynomial f(x) has a factor x – k if and only if f(k) = 0. This theorem can be used to help factor/solve a polynomial function if you already know one of the factors.

Exercise 9 Factor f(x) = 2x3 – 11x2 + 3x + 36 given that x – 3 is one factor of f(x). Then find the zeros of f(x).

Exercise 10 Given that x – 4 is a factor of x3 – 6x2 + 5x + 12, rewrite x3 – 6x2 + 5x + 12 as a product of two polynomials.

Exercise 11 Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42 given that f(7) = 0.

Rational Zero Test: we use this to find the rational zeros for a polynomial f(x). It says that if f(x) is a polynomial of the form: Then the rational zeros of f(x) will be of the form: Where p = factor of the constant & q = factor of leading coefficient Rational zero = Possible rational zeros = factors of the constant term___ factors of the leading coefficient Keep in mind that a polynomial can have rational zeros, irrational zeros and complex zeros.

Ex 1: Find all of the possible rational zeros of f(x)

Ex 2: Find the rational zeros of: Let’s start by listing all of the possible rational zeros, then we will use synthetic division to test out the zeros: 1. Start with a list of factors of -6 (the constant term): p = 2. Next create a list of factors of 1 (leading coefficient): q = 3. Now list your possible rational zeros: p/q = Testing all of those possibilities could take a while so let’s use the graph of f(x) to locate good possibilities for zeros. Use your trace button!

Ex 2 continued: Find all of the rational zeros of the function

Ex 3: Find all the real zeros of : p = Factors of 3: q = Factors of 2: Candidates for rational zeros: p/q = Let’s look at the graph: Which looks worth trying? Now use synthetic division to test them out.

Homework Dividing Polynomials Worksheet Page 127-128 36,38, 49-59 odd