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Algebra 2 - PreAP/GT Solving and Graphing Polynomial Equations #2 Name ____________________________________ Solve the following polynomial equations by factoring. 2 x3 5x2 4 x 10 0 x 4 3x 2 4 0 Notice that solutions can be rational, irrational, real, and/or complex/imaginary. IRRATIONAL ROOT THEOREM If a polynomial P x has rational coefficients and a b c is a root of the polynomial equation P x 0 , where a and b are rational and c is irrational, then a b c is also root of P x 0 . COMPLEX CONJUGATE ROOT THEOREM If a bi is a root of a polynomial equation with real number coefficients, then a bi is also a root of that polynomial equation. Write a polynomial function with zeros of 2i and 1. Write a polynomial function with zeros of 1 i , 2 , and -3. What happens when we try to solve x3 3x 2 10 x 24 0 by factoring. Not all polynomial equations will factor using the factoring rules we have been using. Could we solve the example above by using Square Root Property? Completing the Square? Quadratic Formula? Let’s look at a theorem that might help us solve the polynomial equations that we can not initially factor. RATIONAL ROOT THEOREM If the polynomial P x has integer coefficients, then every rational root of the polynomial equation p , where p is a factor of the constant term of P x and q is a q factor of the leading coefficient of P x . P x 0 can be written in the form x3 3x2 10 x 24 0 p (factors of constant): ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 q (factors of leading coefficient): ±1 POSSIBLE rational roots p : ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 q Find the possible rational roots of f x 8 x 4 2 x3 31x 2 8 x 4 . Once we have the possible rational roots, we can use the possible roots with some other theorems to help us solve and graph a polynomial function. Going back to the example from above, x3 3x2 10 x 24 0 POSSIBLE rational roots p : ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 q DESCARTES’ RULE OF SIGNS The number of positive real roots of a polynomial equation P x 0 with real number coefficients is equal to the number of sign changes between the coefficients of the terms of P x 0 or is less than this number by a multiple of 2. x3 3x2 10 x 24 0 + + – 1 sign change Therefore, the number of positive real roots is 1. – DESCARTES’ RULE OF SIGNS (Continued) The number of negative real roots of a polynomial equation P x 0 with real number coefficients is equal to the number of sign changes between the coefficients of the terms of P x 0 or is less than this number by a multiple of 2. x 3 3 x 10 x 24 0 2 x3 3x2 10 x 24 0 – + – + 1st sign change 2nd sign change Therefore, the number of negative real roots is either 2 or 0. Since we know that this polynomial has 3 solutions (including multiplicities and complex/imaginary solutions), we can make Descartes’ Chart based on Descartes’ Rule of Signs. + – C/i 1 1 2 0 0 2 each row must add to 3 since we know there has to be 3 solutions Using synthetic division, the Remainder Theorem, the Factor Theorem along with the fact that we have one real positive root, let’s try our possible rational roots to see if we can find the positive real root. (Note: there is no guarantee that the positive real root is a rational root) In case you don’t remember them – REMAINDER THEOREM If the polynomial function P x is divided by x a , then the remainder r = P a . FACTOR THEOREM For any polynomial P x , x a is a factor of P x if and only if P a 0 . Remember, we are guaranteed a positive root so we will start with the possible positive numbers – 1 1 3 1 -10 4 -24 -6 1 4 -6 -30 1 is not the positive real root since the remainder is not 0. 3 1 3 3 -10 18 -24 24 1 6 8 0 2 1 3 2 -10 10 -24 0 1 5 0 -24 2 is not the positive real root since the remainder is not 0. 3 is the positive real root since the remainder is 0. Since 3 is a root, then x3 3x 2 10 x 24 0 will factor to x 3 x2 6 x 8 which will then factor to x 3 x 2 x 4 . Therefore, the solutions to x3 3x 2 10 x 24 0 are x = 3, -2, -4. Now that we have solved the polynomial equation, we can also graph it using the x-intercepts, the y-intercept, and the end behavior as guidelines. x – intercept(s) of f x ___________________________ y – intercept of f x _______________ End behavior of f x _______________________________ Make Descartes’ Chart for the following polynomials: f x 8 x 4 2 x3 31x 2 8 x 4 f x x3 3x2 x 2 HOMEWORK For each problem below, list the possible rational roots (zeros) and make Descartes’ Chart. 1) x3 2 x 2 2 x 3 0 2) f ( x) x3 x2 10x 8 3) 3x3 7 x 2 15x 9 0 4) f x x 4 4 x3 3x 2 14 x 8 Bookwork: p 449 #20-22