#### Transcript 2.4 Real 0*s of Polynomial Functions

2.4 Real 0’s of Polynomial Functions Long Division q(x) called the quotient, r(x) called the remainder f(x) = dividend d(x) = divisor Remember when doing long division you must fill in holes with 0, including the constant example: if you are given x3 – 4x + 5 you will have to rewrite it as x3 + 0x2 – 4x + 5 Examples: f(x) = 4x3 – 8x2 + 2x - 1; d(x) = x – 1 f(x) = x3 – 1; d(x) = x + 1 Synthetic division Examples: f(x) = 2x3 – 3x2 – 5x – 12, d(x)=x -3 1) rewrite without the x’s, dividend: the signs stay the same, divisor: the sign changes 3 2 -3 -5 -12 multi. on the diagonal 6 9 12 add going down 2 3 4 0 last # is always the remainder Rewrite answer: 2x2 + 3x + 4 Examples: 2x4 – 5x3 + 7x2 - 3x -1 x -3 5x4 – 3x + 1 x-4 Remainder Theorem If a polynomial f(x) is divided by x – k, then the remainder is r = f(x) To use the remainder theorem: either solve the problem by long or synthetic division and find the remainder write your answer in the form f(x) = remainder Example: find the remainder when f(x) = 3x2 + 7x – 20 is divided by a) x – 2, b) x + 4 Answer: Factor Theorem A polynomial of function f(x) has a factor x – k if and only if f(k) = 0 To do: again either use long or synthetic division example: use the Factor theorem to determine whether the 1st polynomial is a factor of the second polynomial Example: x -3; x3 – x2 - x - 15 Finding rational 0’s To do: 1) find all factors of your constant (p) 2) find all factors of your leading coefficient (q) 3) find all values of p/q – list all of them 4) pick any p/q to determine if it’s a 0 by using synthetic division, need a remainder of 0 for it to work 5) rewrite answer & then factor it (if you get it down to x2) 6) find the 0’s by setting each factor = 0 & solving for x Example: f(x) = 3x3 + 4x2 – 5x - 2 Upper & Lower bounds k is an upper bound for the real 0’s of f if f(x) is never 0 when x>k k is a lower bound for the real 0’s of f if f(x) is never 0 when x<k Another way to say it: if k> 0 and every # in the last line is non-negative (positive or 0), then k is an upper bound for the real 0’s of f. if k <0 and the #’s in the last line are alternately nonnegative and non-positive, then k is a lower bound for all the real 0’s of f. Proving you have an upper or lower bound Prove that the number k is an upper bound k = 5, f(x) = 2x3 – 5x2 - 5x - 1 Prove that the number k is a lower bound k = -3, f(x) = x3 + 2x2 + 2x + 5 Finding real 0’s Use the Rational 0’s Theorem (which is what we use to find rational 0’s) Find p/q (p is the factors of your constant, q is the factors of the leading coefficient) Pick any one you want & use synthetic division to determine if it’s a 0 (your remainder will be 0) Rewrite your answer: if it’s x3 or higher, then you need to repeat what you did above. If it’s x2, you can factor it and then solve each factor Determine if the 0 is rational or irrational example Find all real 0’s of the function, finding exact values whenever possible. Identify each 0 as rational or irrational f(x) = x3 + 3x2 – 3x – 9 f(x) = 3x4 – 2x3 + 3x2 + x - 2