1. I CAN USE LONG DIVISION AND SYNTHETIC DIVISION. I CAN APPLY THE FACTOR AND REMAINDER THEOREMS. Lesson 2-3 The Remainder and Factor Theorems

2. Long Division First divide 325 by 14 to review the process. Answer: 23 remainder 3 or 23 and 3/14 Now divide 2x 2 - 7x + 8 by x – 2. Answer: 2x – 3 remainder 2 or (2x – 3) + 2/(x-2) Finally divide 3x 4 – 2x 3 + 2 by x 2 – 1. (Note the missing terms.) Answer: 3x 2 – 2x + 3 remainder -2x + 5.

3. Review of Terms Dividend Divisor Remainder Quotient To check a problem, note that dividend = divisor x quotient + remainder If the remainder is zero, the divisor and quotient are factors of the dividend.

4. Synthetic Division This is a shortcut for dividing when the divisor is in the form x – a. Use synthetic division to divide 2x 2 – 7x + 8 by x – 2. Check to see that you get the same answer that you did using long division. Now divide x 3 + 5x 2 + 2x – 8 by x + 2. Answer: x 2 + 3x – 4 Can you use this answer to factor the function completely?

5. The Remainder Theorem First divide f(x) = 2x 2 – 3x + 4 by x -1 using synthetic division. Now find the f(1). What do you notice? Remainder Theorem When f(x) is divided by x-r, the remainder is f(r). To prove this for a quadratic, divide g(x) = ax 2 + bx + c by x – r using synthetic division. The remainder is ar 2 + br +c, which is f(r).

6. Factor Theorem Let f(x) = 2x 3 – 5x 2 + x – 6. a) Use the remainder theorem (also called synthetic substitution) to evaluate f(3). Can you see why this is easier than evaluating f(3) the “old” way? Answer f(3) = 6. b) Is x – 3 a factor of f(x)? No, because the remainder when diving by x – 3 is 6. Factor Theorem If f(x) is a polynomial function, f(r)=0 if and only if x – r is a factor of f(x).

7. Let f(x) = 2x 4 + x 3 – 11x 2 – 4x + 12 Is x + 2 a factor? Is x – 1 a factor? Factor f(x) as far as possible. Answer f(x) = (x+2)(x-1)(2x+3)(x-2)