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Finding Roots of Polynomials SHOW ALL WORK and JUSTIFY ALL ANSWERS

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Finding Roots of Polynomials SHOW ALL WORK and JUSTIFY ALL ANSWERS Name: Math 1314 College Algebra Finding Roots of Polynomials SHOW ALL WORK AND JUSTIFY ALL ANSWERS. Find all zeroes of P(x) = 2x6 − 3x5 − 13x4 + 29x3 − 27x2 + 32x − 12. 1. Make sure all terms of P(x) are written in descending order. 2. Can we factor out a greatest common factor from each term? 3. Is 0 a zero? 0 is a zero if P(0) = 0: 4. Look for sign changes in P(x): P(x) = 2x6 − 3x5 − 13x4 + 29x3 − 27x2 + 32x − 12 There are sign changes in P(x). Summary of Descartes’ Rule of Signs: For potential positive real zeroes: Count sign changes in P(x). Take that number and continue to subtract 2 until you get 0 or 1: The original value and each result of the subtraction could be the number of positive real zeroes. Using Descartes’ Rule of Signs, we could have or or positive real zeroes. 5. Find P(−x) To find P(−x), change signs of all coefficients of xodd power P(−x) = There are sign changes in P(−x). Summary of Descartes’ Rule of Signs: For potential negative real zeroes: Count sign changes in P(−x). Take that number and continue to subtract 2 until you get 0 or 1: The original value and each result of the subtraction could be the number of negative real zeroes. Using Descartes’ Rule of Signs, we will have negative real zero. 6. Fill in the values for possible zeroes. I have started for you. Remember that every row must add up to the degree of the polynomial, which is 6 in this example. 0 is not a zero positive real zeroes negative real zeroes nonreal zeroes 0 5 NOTE: If 0 was a zero, we’d 0 3 put 1’s in the first column. 0 1 7. Factor the leading coefficient. Factor the constant term. Use the Rational Root Theorem to list all possible rational roots. Factors of leading coefficient: Factors of constant term: Summary of Rational Root Theorem: Potential rational roots look like: factor of constant term factor of leading coefficient 1 1 2 3 4 5 12 1 2 3 4 5 12 Possible rational zeroes: ± ;± ;± ;± ;± ;± ;± ;± ;± ;± ;± ;± 1 1 1 1 1 1 2 2 2 2 2 2 (I have filled in the options for factors of the leading coefficient. You need to fill in all factors of the constant term for each blank). 1 3 We can reduce the list above to these possible rational zeroes: ±1;±2;±3;±4;±6;±12;± ;± . 2 2 8. We are now going to use synthetic division to find a root. (Remember to write 0 in place of any missing terms). Which possible rational zero should we try first? We have 16 choices. I hope we choose wisely! Since we have one negative real root, let’s try a negative possibility first. How about x = −3? NOTE: We will have exactly one negative real zero, but this zero does not have to be a rational number. But it could be! 9. Use synthetic division to see if x = −3 is a zero of P(x). If the last entry in our synthetic division is 0, we found a zero. -3 2 -3 -13 29 -27 32 -12 -32 -3 -13 29 -27 32 -12 NOTE: x = −3 is a zero, but we can also conclude some- Recall: Lower bounds look like: thing else that is very useful! What do you notice about -# coefficients of the polynomial P(x) the signs of the numbers in the last row? They alternate! -32 -3 -13 29 -27 32 -12 Thus, x = −3 is a lower bound. We do not need to check -3+-+-+-+ x = −4;x = −6 or x = −12. OR 0 can take the place of any + or -: We have reduced our list of possible rational zeroes to -# coefficients of the polynomial P(x) 1 3 ±1;±2;±3;4;6;12;± ;± . -32 -3 -13 29 -27 32 -12 2 2 -300+-+-0 10. The last row from our synthetic division gives us a depressed equation. Thus, we can rewrite P(x) in terms of the new factor we found and the depressed equation. P(x) = (x + 3)(2x5 − 9x4 + 14x3 − 13x2 + 12x − 4) Recall: Factor Theorem: If x = a is a zero of P(x), then (x − a) is a factor of P(x) and vice versa. 11. Check to see if we can factor out a greatest common factor from the depressed equation. We cannot, so we continue in our process. 12. Let’s use the Rational Root Theorem on the depressed equation (2x5 − 9x4 + 14x3 − 13x2 + 12x − 4) = 0 to see if we can eliminate any more rational zeroes. Factors of leading coefficient: Factors of constant term: Summary of Rational Root Theorem: Potential rational roots look like: factor of constant term factor of leading coefficient 2 1 2 4 1 2 4 Possible rational zeroes: ± ;± ;± ;± ;± ;± 1 1 1 2 2 2 (I have filled in the options for factors of the leading coefficient. You need to fill in all factors of the constant term for each blank). 1 We can reduce the list above to these possible rational zeroes: ±1;±2;±4;± . 2 Now, go back to our chart from Descartes’ Rule of Signs. We had only one negative real zero and we found it, so we do not have to check any more negatives! 1 Thus, our list is reduced further to 1;2;4; . 2 What should we try next? How about x = 1? NOTE: Once you find a zero, it is a good idea Use synthetic division to see if x = 1 is a zero of P(x). to test it again, in case it is a zero of 13. 1 2 -9 14 -13 12 -4 multiplicity > 1. However, we do not need to -32 -3 -13 29 -27 32 test x = −3 again since P(x) has only one negative real zero. Drat! x = 1 is not a zero. Recall: Upper bounds look like: Also, we cannot conclude that it is an upper bound. +# coefficients of the polynomial P(x) 1 14. 1 is not a zero, so our possible rational roots are 2;4; -32 -3 -13 29 -27 32 -12 2 -3+++ ++ ++ Let’s try another possible rational root on our list: OR 0 can take the place of any + (the entries What about x = 2? must be nonnegative): 15. Use synthetic division to see if x = 2 is a zero of P(x). +# coefficients of the polynomial P(x) 2 2 -9 14 -13 12 -4 -32 -3 -13 29 -27 32 -12 -32 -3 -13 29 -27 32 -300+ ++ +0 16. x = 2 is a zero. Is x = 2 an upper bound? No. 17. Rewrite P(x): P(x) = (x + 3)(x − 2)(2x4 − 5x3 + 4x2 − 5x + 2). Check to see if we can factor the depressed equation. We cannot, so we continue in our process. 18. Using the Rational Root Theorem on our depressed equation does not eliminate any possible rational zeroes. 19. Let’s use synthetic division to see if x = 2 is a zero of multiplicity > 1. 2 2 -5 4 -5 2 32 -3 -397 3 20. x = 2 is a zero of multiplicity because synthetic division shows us it is a zero again. Rewrite P(x): P(x) = (x + 3)(x − 2)(x − 2)(2x3 − x2 + 2x − 1) Check to see if we can factor the depressed equation. If we want to continue testing zeroes with syn- We can factor by grouping. 1 thetic division, we can. We would test x = , see 2 that it is a zero and continue from there. 21. Factor 2x3 − x2 + 2x − 1: Factor the greatest common factor from the first and second terms. Then factor the greatest common factor from the third and fourth terms. 2x3 − x2 + 2x − 1 = x2( ) + 1() Thus, 2x3 − x2 + 2x − 1 = (x2 + 1)(2x − 1) 22. Rewrite P(x): P(x) = (x + 3)(x − 2)(x − 2)(2x − 1)(x2 + 1) Now, we have written P(x) as a product of linear factors and one quadratic factor. We can solve x2 + 1 = 0 by factoring (if possible), completing the square, or quadratic formula. We cannot factor (x2 + 1) over the real numbers. Solve x2 + 1 = 0. Did you get a complex number? If so, did you get its conjugate? Recall: Conjugate Pairs Theorem: If a + bi is a root of P(x), a − bi is also a root of P(x). List all zeroes of P(x): 23. Strategy to find all zeroes of a polynomial P(x): (a) Make sure all terms of P(x) are written in descending order. (b) Factor P(x) if possible. (c) Check to see if x = 0 is a zero of P(x). (d) Use Descartes’ Rule of signs to determine the number of positive, negative, and nonreal roots. (e) Use the Rational Root Theorem to list all possible rational roots. (f) Use synthetic division to find a root. (Remember to write 0 in place of any missing terms). (g) Rewrite P(x), using the root to write as a product of factors. (h) Look at the depressed equation to see if some rational roots can be eliminated. Use this and results from Descartes’ Rule to choose the next potential zero to try. (i) Continue until P(x) is written as a product of linear factors and one quadratic factor (i.e. depressed equation is quadratic). Solve the quadratic equation by factoring, completing the square, or quadratic formula. 4.
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