1. Chapter 5-4: Dividing Polynomials St. Augustine Preparatory School October 5, 2015

2. Last Class (Ch. 5-1, pg 288) -Finding zeros of a polynomial function y = (x+2)(x-1)(x-3) has zeros: -2, 1, 3 -Writing a polynomial from its zeros if a polynomial has zeros: -2, 2, 3, then the polynomial is y = (x+2)(x-2)(x-3), which then multiplies out to be y = x 3 – 3x 2 – 4x +12 -Relative maximum or relative minimum -R. Max: value of the function at an up-to-down turning point -R. Min: Value of the function at a down-to-up turning point

3. Relative maximum/minimum

4. Factoring When the variable with the highest exponent down not have a constant of 1 (ie. 9x 3 + 6x 2 - 3x) Step 1: Write down the function Step 2: Factor out 3x from each term Step 3: Find two numbers that multiply to be -3x 2 and add together to be 2x. Insert them into the equation. Step 4: Factor again if necessary Step 5: Use the term before each of the identical binomials to make anew binomial and multiply it by the binomial that both terms had in common.

5. Chapter 5-4: Dividing Polynomials Similar to long division

6. Polynomial Long Division Example: 4x 2 + 23x – 16 divided by x+5

7. Checking factors using long division You can use long division to figure out if one function is a factor of another. If the two functions are divided and the remainder = 0, then the divisor is a factor of the function. If it does not = 0, it is not a factor. Continue using the long division process until the degree of the remainder is less than the degree of the divisor.

8. Checking factors using long division

9. Example: Is x 2 + 1 a factor of 3x 4 – 4x 3 + 12x 2 + 5? ***Do not forget to add in the terms that are missing*** This means: x 2 + 1 should be wrote as x 2 + 0x + 1 and 3x 4 – 4x 3 + 12x 2 + 5 must be rewrote to include a 0x term as well. Answer: no it is not a factor

10. Assignment Questions Page 308: 9, 13, 14, 15, 18, 20 If finished, read through problem 3 on how to do synthetic division

11. Synthetic Division Step 1: Reverse the sign of the constant in the expression that is being divided by. Write the coefficients of the polynomial to the right of it. Step 2: Bring down the first coefficient Step 3: Multiply the coefficient by the the divisor. Add it to the next coefficient in line Step 4: Continue until all terms have been added by. The last term is the remainder.

12. Example Divide x 3 – 14x 2 + 51x -54 by x+2

13. Another example Divide x 3 – 57x + 56 by x- 7. What is the quotient and the remainder? Quotient: x 2 + 7x – 8 Remainder: 0 Answer: x 2 + 7x – 8, R 0

14. Remainder Theorem If you divide a polynomial P(x) of degree n ≥ 1 by x – a, then the remainder is P(a) Example: If you divide P(x) = x 5 – 2x 3 – x 2 + 2, by x – 3, then the remainder is equal to P(3) P(3) = 3 5 – 2(3) 3 – (3) 2 + 2 P(3) = 182 The remainder is 182

15. Remainder Theorem Note: this only works when dividing by x – a. If you have a polynomial that is x + a, remember that it could be wrote as x – (-a). Ex. If you were to divide P(x) by: 1. x – 4, you would find P(4) 2. x + 3, you would find P(-3) 3. x + 9, you would find P(-9) 4. x – 1, you would find P(1)

16. Assignment Questions Page 308, 21, 24, 25, 28, 34, 37, 38