The Fundamental theorem of algebra precalc precalculus larson 2 5 remainder roots descartes sign FTA

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• published: 23 Oct 2016
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Subscribe Here http://goo.gl/2XXaLS For more cool math videos visit our site at http://mathgotserved.com or http://youtube.com/mathsgotserved Good day students welcome to mathgotserved.com in this group were going to be going over how to find roots of polynomial functions by applying the fundamental theorem of algebra the instructions for the problem depths we're going to be sold in Omar us follows first we are to find a number of complex real imaginary positive and negative roots of the polynomial equation in number two we are to use irrational roots theorem to list the possible zeros of the polynomial and then will proceed to use synthetic division in any factoring technique to express the polynomial function as the product of lean your factors indole conclude by fine in all the zeros and coal from the signs using the carts prediction in parts one where we have to speak the number of positive and negative roots okay now let's take a look at the equation what if we have the polynomial equation 6X to the 4-20 9X to the third +18 X square plus minor X -4 equals zero okay let's start with the first part we are to find the number of complex roots for one number am possible this indicate number of complex roots number of complex roots now the number of complex roots on using the fundamental theorem of algebra is BZ Cleve the degree of the polynomial okay so what is the degree of the polynomial here the degree of the polynomial is the highest exponent for all the variable Om terms so if you look at the exponents we have for 321 and zero the highest degree exponent is for so the number of complex roots basically the degree in this case is equal to four okay now let's take a look at the next part the number of possible number of real roots okay we know for certain T that the will be for complex roots and beer could be a combination of real and imaginary roots so let's the The possible number of real roots out of the complex roots okay so puzzle number roots Om this is basically the degree or even number is less than degree then the degree okay so you just subtract 2468 from the degree on to the end up with Tom zero or greater so the degree or even numbers less than the degree okay so in this parts we know that the degree is what the degree from the previous part is for so the possible number of complex real roots is either for now what's the next the first even number two right so for subtract two from four we end up with to okay what's the next even numbers for so will subtract for from four we end up with zero so by looking at the degree or even numbers list we simply subtracting two from subsequent possible roots okay so have is a formula roots to roots or zero real roots now let's take a look at the possible number of imaginary okay so the roots for the polynomial am equation could either be real imagine that or imaginary want to combine them together we get the number of complex roots okay so possible number of imaginary roots can be done by carryout the simple arithmetic calculation so always simply do is the cat the car all complex roots minus the real okay the roots the complex roots that are not okay so let's carry out the calculations we have all total number of four complex roots okay now what is four of them are real is therefore complex roots of four of them are real in how many imaginary roots to we have absolutely non-because 4-4 is zero okay now what of all the four complex roots that you have two of them are even certain of them are real if 12 them are real then you going to have the other two will be imaginary because 4-2 is to now what if you have four complex roots and not of the complex roots are real what of that mean all four of them are imaginary right so you see the pattern here you have 4 to 0 I just reverse it in the case of a situation we haven't even degree okay so by turn out the subtraction you can easily see that the number of imaginary roots or 02 and four okay so remember the current the equation complex one is real is imaginary or imaginary plus real gives you complex okay now let's shift gears and take a look at the next part now it to find the possible number of positive roots okay so possible number of positive use roots okay so in this case what we looking at is the number of sine change use of the function okay limited at an additional veto here so we avoid confusion I we looking for the possible number of positive real roots okay now when I talk about Om sign change sign change for F of X we're looking at the Orient Om rule known as the to Descarte's rule of signs that's what we looking at here so the Council of sine basically involves look another sign change of the function and even numbers less than that so the number of sine changes we have of the original function and even numbers less almost the possible number of positive real roots that the related polynomial equation has okay so we're going to write down the equation of the function form basically