Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Rational 0 theorem. Given some polynomial with all of its coefficients on each term being integers, and the fact that we may have a rational zero of the form P / Q means that P would have to be a factor of the constant of the polynomial, and Q has to be a factor of the leading coefficient. A polynomial of degree N has N roots. We need to count multiple roots individually, so if it's a third degree, we're going to have three roots, although it may be the same root occurring three times. Imaginary roots and irrational roots always occur in conjugate pairs A+ bi, A minus BI, or a plus the square root of BA minus the square root of B. Discarte's rule of sines. Let F of X be a polynomial with real coefficients. Then the number of positive real zeros is the number of sign changes in F of X, or possibly that number of sign changes minus an even number until zero or one is reached. It needs to be minus an even number because complex numbers and irrational numbers are always coming in pairs. The number of negative real zeros may be found by replacing X with negative X and counting the sign changes. The possible negative real zeros will be found by counting the sign changes or possibly that number minus an even number until zero or one is reached. Let's look at an example. If we have F of X equal X ^3 -, 4 X squared plus eight X -, 5, we're going to talk about the possible rational roots. All of the possible rational roots are the factors of the constant, in this case 1 and 5, divided by the factors of the leading coefficient, in this case one. So we'd have positive and -1 divided by positive -1, positive -5 divided by positive -1. The numbers we would need to test are one negative 1/5, negative 5. Now Discarte's rule of sign helps us because the rule of sign says we're going to have positives, negatives or imaginaries. If we count how many sign changes this was a positive term going to a negative term, hence one sign change a negative term going to a positive term, hence 2 sign changes a positive term going to a negative, hence 3 possible positive real roots. Now if I stick a negative X, that's going to help me figure out how many possible negative real roots. So when I look at this and simplify it, F of negative X would give me negative X ^3 -, 4 X squared -8 X -5. There were no sign changes. Negative to negative, negative to negative, negative to negative, no sign changes. So there's no possible negative real roots. If we look at the degree, we know that there should be 3 roots total. So if there were 3 positive and zero negative, that would make 0 imaginary. Now the positives could be any multiple less than three, taking away a pair. So any multiple of 2 less than 3. So 3 -, 2, we could have one positive, we'd still have no negative. But imaginaries always occur in pairs. When we look at this, it wouldn't make any sense at all to try to use the negative roots up here, because there aren't going to be any negative zeros. So now I've just taken the quantity that I need to test by synthetic division down to two. If I put one in for my synthetic division, one negative 4/8, negative 5. If it's a root, we're going to have a remainder of 0, so we bring down the one we multiply. If we're above, we add. If we're below, we multiply. If we're above, we add. If we're below, we multiply and 1 is going to be a root. So we know that X -, 1 times something would give us back this X ^3 -, 4 X squared plus eight X -, 5. Well, that something can be found right here. This is going to be X ^2 -, 3 X +5. So that equals the original that we started with. We found 10. So far the 10 we have found is 1. Now we can do possible roots again, but it's a square, so it's going to be easier if we either factor it or use quadratic formula. If we look at this, it's not going to factor. There aren't 2 numbers that multiply to give me 5 and add to give me -3. So we're going to use quadratic formula. We're going to get X equal 1 from the first parenthesis. X equal the opposite of B plus or minus the square root b ^2 -4 * a * C all over 2A so X is going to be +3 plus or minus the square root 9 - 29 minus 20s negative 11 / 2, so X equal 3 ± I sqrt 11 / 2. We really end up with three roots because we have one back here, so one, zero. We have 3 + I square roots of 11 / 2, zero and we have 3 -, I square roots of 11 / 2, 0. Thank you and have a wonderful day.