Video: Describing the Possible Real Zeros Using Descartes’s Rule of Signs

Using Descartes’s rule of signs, find which of the following describes the possible real zeros of ℎ(𝑥) = 3𝑥⁵ + 2𝑥⁴ − 3𝑥³ + 𝑥² + 6𝑥 + 2.

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Video Transcript

Using Descartes’s rule of signs, find which of the following describes the possible real zeros of ℎ 𝑥 is equal to three 𝑥 to the power of five plus two 𝑥 to the power of four minus three 𝑥 cubed plus 𝑥 squared plus six 𝑥 plus two.

So we’re gonna use Descartes’s rule. And what we’re actually gonna try and do is actually find the possible real zeros. Zeros can also be known as roots. This is where we’re actually gonna find our solutions to our function. Well, Descartes’s rule of signs, first of all, tells us that the number of positive real zeros or roots in 𝑦 equals 𝑃 𝑥, so our polynomial function is equal to the number of changes of sign in front of each term or less than this by an even number.

Okay, so that’s our positive real zeros. What about our negative real zeros? Well, the number of negative real zeros in our polynomial function is equal to the number of changes in sign in front of the terms in our polynomial function, but with negative 𝑥, or is less than this by an even number.

Okay, great. So we’ve now got Descartes’s rules of signs, what do we do? Well, in order for this to work, our terms must be in descending powers. So the terms of our polynomial must be in descending powers. Well, if we look at our polynomial, well this is already the case cause we’ve got our powers. So we’ve power of five, power of four, power of three, power of two, one, and zero. So yes, we’re in descending powers. So what now? What do we do?

Well, step two is to actually count the number of changes in sign for each of our coefficients. So in order to help us do this, what I’ve done is I’ve actually written our coefficients out. And I’ve actually included signs. So we’ve got positive three, positive two, negative three, positive one, positive six, and positive two. So now what I’ve done is I’ve actually put on where it changes. So from the first coefficient to the second coefficient no change cause they’re both positive. Then we go from positive two to negative three. So yes, there’s a change. Then from negative three to positive one, yes, another change. Then positive one to positive six, no change. And then from positive six to positive two, no change.

So therefore, we can say that the possible number of zeros are two because when we look back at Descartes’s rule of signs, it says that it’s equal to the number of changes in sign in front of each term. Or, we can actually have no zeros. And this is because it actually states that it’s “or less than this by an even number”. So therefore, if we subtract two from two, we get to zero. Which means that the number of positive real zeros is either two or zero. Okay, we’ve dealt with the positive real zeros. Now let’s look at the negative real zeros.

Now in order to find the number of negative real zeros, what we need to actually do is substitute negative 𝑥 for 𝑥 in our function. So when we do that, we actually get three multiplied by negative 𝑥 to the power of five plus two multiplied by negative 𝑥 to the power of four minus three multiplied by negative 𝑥 to the power of three plus negative 𝑥 squared plus six multiplied by negative 𝑥 plus two. Okay, so now what we need to do is actually work out which of these terms is gonna be positive and which of them is gonna be negative.

Well, the first term is gonna be negative three 𝑥 to the power of five. And that’s because if we have negative 𝑥 to the power of an odd number, we always get a negative result. For example, if we have negative two cubed, that’s like negative two multiplied by negative two which is positive. So it gives us positive four. Then multiply again by negative two which gives us a negative, negative eight. And if we have negative 𝑥 to the power of an even number, that’s gonna give us a positive result. For example, if we had negative three squared, we get nine. Because again, negative three multiplied by a negative gives us a positive because a negative multiplied by a negative gives a positive. Okay, great. So we know the reasoning why. Let’s move on and find out what the next terms will be.

So then we’ve got plus two 𝑥 to the power of four plus three 𝑥 cubed, and that’s because we’ve actually got a negative multiplied by a negative which gives us a positive, plus 𝑥 squared minus six 𝑥 plus two. Okay, great. So now we actually have all of our terms. What we need to do is see are there any changes of sign. Again, I’m just writing out our coefficients to make it easier to see. Well, we can see that the first term to the second term, we’ve got negative to a positive. So yes, this is a change. Then we’ve got a positive to a positive, no change. Positive to positive, no change. Then another positive to a negative, so yes, this is a change. And then a negative back to a positive, so there’s another change. So therefore, we can say that there’re gonna be three zeros. But again, we have to include “or” another value. And in this case, it’s going to be one. And that’s because it can be less by an even number. Well, if you’ve got three and we subtract an even number, so subtract two. We get down to one.

So therefore, we can say that function ℎ has either two or zero positive zeros and either three or one negative zeros. And just to remind us how we actually did that using Descartes’s rule, we first of all make sure that it’s actually in descending powers. Our terms are in our function. Then to find the number of positive real zeros, what we do is we look at the number of changes in sign. Once we’ve actually found out the number of changes of sign, this will be our number of zeros. And we can see if there are any other number of zeros by subtracting even numbers. Then to find the negative real zeros, we substitute negative 𝑥 in for 𝑥 into our function. Again, we count out the number of changes of sign of our coefficients. And then finally, we count up these changes. And again, that gives us the number of zeros. But to find out if there are any other possible number of zeros, what we do is subtract even numbers, yet again!

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