2.5 Zeros of a Polynomial Functions


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1 .5 Zeros of a Polynomial Functions Section.5 Notes Page 1 The first rule we will talk about is Descartes Rule of Signs, which can be used to determine the possible times a graph crosses the xaxis and has a zero. A sign change is a place in f where you have + to or to +. Descartes Rule of Signs 1.) The number of positive real zeros of f equals the number of sign changes in f (x) or less an even integer..) The number of negative real zeros of f equals the number of sign changes in or less an even integer. Less an even integer means we keep subtracting from the number of sign changes until we get to zero or an negative number. Suppose that the number of sign changes was 6. Then we keep subtracting to get the other answers. We will get 6, 6, 6, 6 which is 6,,, 0. If we have 5, then our answers are 5,, or 1. EXAMPLE: Use Descartes Rule of Signs to find the number of possible positive and negative zeros of 6 x x 6x x. Positive Zeros: 6 We need to count the number of sign changes in x x 6x x. The brackets above indicate the place where we have + to or to +. In this case we have sign changes. To write our answer, we start with and then keep subtracting to get the other answers since it is less an even integer. So the number of positive real zeros is or or 0. You keep subtracting until you get a zero or a negative number. Negative Zeros: In order to find the negative zeros we must look at f ( x) : 6 x x x 6 x 5 Put in a x for x in the equation for f. 6 x x 6x x As we notice above there are two sign changes in f ( x). The number of sign changes will be or 0. Again we keep subtracting from our answer until we get to 0 or a negative number. EXAMPLE: Use Descartes Rule of Signs to find the number of possible positive and negative zeros of x x. Positive Zeros: We need to count the number of sign changes in x x. We see that there are sign changes, so as our answer we would write: or 1.
2 Negative Zeros: In order to find the negative zeros we must look at f ( x) : 5 5x x x x x Section.5 Notes Page We notice there are no sign changes in. This means there are no negative sign changes (0). Rational Zeros Theorem Let n a x a a n n 1 n 1 x... 0 be a polynomial. Then the number of possible real zeros of f is: factors of a0 factors of a n To review a factor is a number that evenly divides into something. For example, the factors of 6 are 1, and. EXAMPLE: List the possible real zeros of: x 7x x 8 Using the Rational Zeros Theorem we will write the factors of 8 over the factors of :,, 8 Now divide each number on top by each on the bottom. You will get the list of zeros: 1 8,, 8,,,,. These do not need to be in any special order. This list represents all the possible places the graph could cross the xaxis. EXAMPLE: Given x 8x 0, a.) Use Descartes Rule of Signs to find the number of c.) Find the zeros using synthetic division. a.) We need to find the number of sign changes of x 8x 0. There is only one sign change, so the number of positive real zeros is 1. For the negative zeros we need to look at : ( x) x 8 x 11 0 x 8x 0 f You will get: There are two sign changes. Therefore the number of negative real zeros is or 0.
3 b.) We need to use the Rational Zeros Theorem to find our list of possible zeros. Section.5 Notes Page,, 1 10, 0 Now divide each number on top by each on the bottom. You will get:,, 10, 0. This is our list of possible zeros. c.) Now we need to see which one is a zero. In order to do this, pick a zero to test and use synthetic division with the original equation. If we get a zero for the remainder we know that this is a zero. For example, let s first test the xvalue 1. We need to set up and do the synthetic division: We don t get a zero for the remainder, so we know 1 is not one of our answers Let s try a different number. We will now test x = We do get a zero, so x = 1 is one of our answers. We could go through this process for all the other possible zeros in our list, but since we found one we can find the other ones in an easier way. The last row of our synthetic division was This means our new equation is x 9x 0. We can set this equation equal to zero to get the other solutions. You will get x 9x 0 0. Factoring this we will get ( x )( x 5) 0, so our other answers are x =  and x = 5. So our answer for this one would be: x = 5, , and 1. This confirms Descartes Rule of Signs. We have one positive zero and two negative zeros. EXAMPLE: Given x x 19x 6, a.) Use Descartes Rule of Signs to find the number of c.) Find the zeros using synthetic division. a.) We need to find the number of sign changes of: x x 19x 6 There are two sign changes, so the number of positive real zeros is or 0. For the negative zeros we need to look at : ( x) x 11 x x 19 6 x x 19x 6 f You will get: There are two sign changes. Therefore the number of negative real zeros is or 0. b.) We need to use the Rational Zeros Theorem to find our list of possible zeros.,, 6 Now divide each number on top by each on the bottom. You will get: 1,, 6,,. I did not put 6, because this is the same as. We already have it.
4 Section.5 Notes Page c.) Now we need to see which one is a zero. We need to start testing zeros. We will start with x = 1 again We do get a zero here. This will leave us with an x cubed equation which we will not be able to factor very easily. We need to find another zero so that we can reduce this cube to a square After more trail and error we get x = is another zero Let s look at the first row of numbers we got when we used x = 1. We got Since we know x =  is also a zero, let s use synthetic division with this row and the second zero we found, x = : This will leave us with x 8x which we can set equal to zero to find the other answers. (We also could have done synthetic division with the other row 80 above: and used x = 1. Does not matter which one you do). Now we can solve x 8x 0. Factoring we will get: ( x )(x 1) 0. Solving this we will get x =, x = 1/. Putting this all together our final answer for part c will be x = 1, 1/,, and. This confirms our Descartes Rule of Signs. We got two positive zeros and two negative zeros. EXAMPLE: Given x x x x, a.) Use Descartes Rule of Signs to find the number of c.) Find the zeros using synthetic division. a.) We need to find the number of sign changes of: x x x x There is only one sign change, so the number of positive real zeros is 1. For the negative zeros we need to look at : ( x) x x x x x x x f You will get: There are three sign changes. Therefore the number of negative real zeros is or 1. b.) We need to use the Rational Zeros Theorem to find our list of possible zeros., 1 Now divide each number on top by each on the bottom. You will get:,. c.) Now we need to see which one is a zero. We need to start testing zeros. We will start with x = 1 again We do get a zero here. This will leave us with an x cubed equation which we 10 will not be able to factor very easily. We need to find another zero so that we can reduce this cube to a square.
5 We find that x = 1 is another zero Section.5 Notes Page 5 Let s look at the first row of numbers we got when we used x = 1. We got Since we know x = 1 is also a zero, let s use synthetic division with this row and the second zero we found, x = 1: This will leave us with x x which we can set equal to zero to find the 1 other answers. (We also could have done synthetic division with the other row 1 0 above: and used x = 1. Does not matter which one you do). Now we can solve x x 0. Factoring we will get: ( x )( x ) 0. Solving this we will get x = . Putting this all together our final answer for part c will be x. But what about Descartes Rule of Signs? We got only two negative zeros and one positive zero. Descartes told us that we should have three negative zeros. When we solved ( x )( x ) 0 this actually gave us a double root,  and . So technically the answers are x, which would satisfy Descartes. We don t need to write a repeated root more than once. EXAMPLE: Given x x 65, a.) Use Descartes Rule of Signs to find the number of 1 c.) Find the zeros using synthetic division given f ( 5) 0 and f 0. a.) We need to find the number of sign changes of: x x 65 There are two sign changes, so the number of positive real zeros is or 0. For the negative zeros we need to look at : x x 65 ( x) ( x) 5( x) 11( x) 65 You will get: There are two sign changes. Therefore the number of negative real zeros is or 0. b.) We need to use the Rational Zeros Theorem to find our list of possible zeros. 65 Dividing gives us ,,,,. c.) We were given the first two zeros so we don t need to do trial and error on this one. We will do synthetic division twice to get a quadratic. It doesn t matter which zero you start with. I will start with x =
6 Section.5 Notes Page 6 ½ Again it does not matter what zero you start with. Since we started out with a fourth power, doing synthetic division twice brought this down to a quadratic. We now have to solve x 1x 6 0. We notice there is a common factor of. We can divide both sides of the equation by to make our calculations easier. You will get x 6x 1 0. This cannot be factored, so we must use the quadratic formula: 6 6 (1)(1) i x i. (1) We do not get a real number as a result, but we still need to include this in our answer. We write our answer as x = 1, 5, i. Does Descartes Rule take into consideration the complex numbers? NO! Descartes only gives you the possible REAL negative and positive zeros. In part a it said we would either get positive real zeros or no positive real zeros. This was the same for the negative zeros. We should get either two or none of them. In our case we have two positive zeros and no negative zeros, so this does fit Descartes Rule of Signs.
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