Chapter 2 – Polynomial and Rational Functions 2.3 – Real Zeros of Polynomial Functions
Long Division Ex: What is / 19?
Long Division Ex: What is (6x 3 -19x 2 +16x – 4) / (x – 2)? x – 26x 3 – 19x x – 4 6x 2 – 7x+ 2 6x 3 – 12x 2 -7x 2 -7x x 2x 2x – x – 4 No remainder! Hooray!
Long Division Ex: What is (x 2 + 3x + 5) / (x + 1)? x + 1x 2 + 3x + 5 x+ 2 x 2 + x 2x 2x Remainder! Put this over the divisor! + 5
Long Division Ex: What is (4x 3 - 8x 2 +6) / (2x – 1)? 2x - 14x 3 – 8x 2 + 0x + 6 2x 2 – 3x- 3/2 4x 3 – 2x 2 -6x 2 -6x 2 + 3x -3x -3x +3/2 9/2 + 0x + 6
1. None Divide 4x 2 – 6x + 7 by x + 3. What is the remainder?
Divide x 3 – 1 by x – 1. ***Put a zero for the coefficient of missing terms!!!
Synthetic Division When dividing binomials of the form x – k, use a shortcut called synthetic division! To divide ax 3 + bx 2 + cx + d by x – k… k a b c d a ka r Coefficients of quotient (answer) Coefficients of dividend Remainder Vertically – add terms Diagonally – multiply terms
Synthetic Division When dividing binomials of the form x – k, use a shortcut called synthetic division! Ex: Divide x 3 – 2x 2 + 5x – 3 by x – 3. ◦Set up the coefficients first… ◦…then multiply diagonally and add vertically! ◦So the quotient is x 2 + x + 8, and the 21 becomes the remainder
Synthetic Division Ex: Divide x 5 – 2x x – 8 by x + 1. ◦Remember to put zeros for missing exponent terms!
Divide 2x 3 – 6x 2 + 7x – 4 by x – 2. What is the remainder? None
Rational Zero Test Every rational zero of a polynomial f will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. Ex: Find the possible rational zeros of f(x) = 2x 3 + 3x 2 – 8x + 3. ◦Possible rational zeros:
Factoring w/synthetic division If synthetic division gives a remainder of zero, then the divisor is a factor of the polynomial! Ex: Factor f(x) = 2x 3 + 3x 2 – 8x + 3. ◦On the previous slide, we determined the rational zeros, and of those rational zeros, only -1, 1, -3, and 3 were integers. ◦Guess and check using synthetic division! is not a zero! is a zero!
Factoring w/synthetic division Ex: Factor f(x) = 2x 3 + 3x 2 – 8x + 3. (cont.) ◦What’s left? ◦2x 2 + 5x – 3 ◦Because it’s a quadratic, you can factor this using either normal factoring or synthetic division. ◦2x 2 + 5x – 3 (2x – 1)(x + 3) ◦So the final factorization is (x – 1)(2x – 1)(x + 3).
Ex: Find all zeros of f(x) = 3x 3 + 4x 2 – 17x – 6. ◦Guess and check using synthetic division! ◦Our possible zeros are ±1, ±2, ±3, ±6, and ±1/3 ◦Answer: x = 2, -3, and -1/3 Ex: Find all zeros of f(x) = x 4 + x 3 - 4x 2 - 2x + 4. ◦Our possible rational zeros are ±1, ±2, ±4 ◦This factors to (x 2 – 2)(x + 2)(x – 1) ◦The (x 2 – 2) doesn’t factor, so use quadratic formula to find the REAL zeros of ◦Answer: Ex: Find all real zeros of f(x) = x 3 - x 2 - 4x - 2
Algebraically, find the zeros of f(x) = 3x 3 + x 2 – 8x , -1, 2/3 2. 1, -2, , 2/3, , 2, -1