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Polynomial Functions Note: It is assumed that you are familiar with long division or synthetic division of polynomials. Examples of polynomial functions are f ( x ) 4x 3 3x 2 7 x 3 and and g( x ) 5x 4 2x 1 . The topic of this study guide is to find the x-intercepts also known as rational zeros of polynomial functions, and to be able to make a graph of these functions - finding the x- intercepts is essential to making the graph. Let us first look at some general aspects of polynomial functions. Typically they look as follows far right local max far left local min or maybe like 1 far left far right local max local min local min The local maxima and minima are sometimes collectively known as the turning points of the function. The following statements are considered basic knowledge concerning polynomial functions: I) A function of degree n can have at most n x-intercepts. Thus f ( x ) 2x 3 4x 5 , which is of degree 3, can have at most 3 x-intercepts. II) A function of degree n can have at most (n-1) turning points. This means that the function just mentioned can have at most 3-1=2 turning points. III) The far left and far right behavior of a graph is either up or down. This means that for "large" x values, say x = 20, the graph will continue to go steeply up or down. Likewise for x values on the far left like x = -20. The turning points will not be to the far left or right but around the origin. The far right and far left behavior of a polynomial can easily be determined by calculating f(20) and f(-20). If f(20) is positive the function goes up on the far right, if it is negative the function goes down on the far right. The same holds true for the far left. Thus the function mentioned earlier goes up to the far right since f(20)>0 and down to the far left since f(-20)<0. Let us turn our attention to the x-intercepts or real zeros of a polynomial function: if a function can be completely factored into binomials then it'll be easy to find the x- intercepts. Just set f(x)=0, factor the function, and use the zero product property to find the real zeros. If the function cannot be factored then we will have to look for other ways to find all the x intercepts. Let us first consider a few 'factorable' functions. 2 Example 1 : f ( x) 2 x 3 3 x 2 18 x 27 Set f(x) 0 to find the x intercepts : 0 x 2 (2 x 3) 9(2 x 3) 0 (2 x 3( x 2 9) 0 (2 x 3)( x 3)( x 3) The zeros are -3/2, -3 and 3. If you talk about zeros this notation is good enough, if you are calling them x-intercepts you should present the answer as (-3/2,0), (-3,0) and (3,0). Example 2 : g ( x ) x 4 25x 2 144 Set f x 0 to find the real zeros : 0 x 4 25x 2 144 0 ( x 2 16)(x 2 9) 0 x 4)(x 4 ( x 3)(x 3) so that the zeros are ±4 and ±3. However, the vast majority of polynomial functions will not easily be factorable if at all. There are a few techniques that are useful in such cases; amongst them are the Factor Theorem and the Rational Zero Theorem. 3 The Factor Theorem The factor theorem states that for any x=c, if f(c)=0 then (x-c) is a factor of f(x), and vice versa. As an example consider f ( x ) x 3 2x 2 5x 6 . Clearly, no factoring technique discussed thus far is able to bring relief here. The factor theorem, however, gets you started. Try an easy x-value, such as x=1 and you will find that f(1)=0 so that, according to the factor theorem (x-1) is a factor of f(x). Thus, the division of f(x) by (x-1) should have a remainder of zero. Performing the division ( x 3 2x 2 5x 6) ( x 1) , either by long division or by synthetic division, will give you the reduced polynomial x 2 x 6 so we can say that x 3 2x 2 5x 6 ( x 1)( x 2 x 6) and after factoring the trinomial we'll have x 3 2x 2 5x 6 ( x 1)( x 3)( x 2) . Thus the real zeros of the function are -2, 1 and 3. After plotting the x- and y-intercepts and looking at the far left and far right behavior we can now easily make the graph. All that we need to do still is to take a few simple x- values and insert them into f(x) in order to get a few extra points plotted and somewhat get a more accurate graph. It looks like this: As a second example try to find the x-intercepts of g ( x ) x 3 5x 2 . The usual factoring techniques yield nothing but the factoring theorem helps us out. Try a few easy x-values like x=1 or x=2 and you will find that g(1)≠0 so (x-1) is not a factor of g(x) but since g(2)=0 then (x-2) is a factor of g(x). Performing the division ( x 3 5x 2) ( x 2) , one way or the other, you will get the reduced polynomial x 2 2x 1 and so g( x ) ( x 2)( x 2 2x 1) . The trinomial can not be factored but you can nevertheless find the x-intercepts by setting g(x)=0. Using the zero product property means that x-2=0 or that x 2 2x 1 0 . This last equation can be solved with the quadratic formula and 4 gives you x 1 2 . The x-intercepts of g(x) are thus (2,0) (1 2 ,0) and (1 2 ,0) . As you see, the factor theorem is a very powerful and new factoring technique. The graph of this function can also be made after plotting all the intercepts and consider the far left and far right behavior. It looks like this: You may now think that it could take a long time before you finally find an x-value such that f(x)=0. This could have been true 20 years ago but ever since pc technology has been introduced to make our lives easier it is a different situation. Go to Microsoft Excel, open a new workbook and in column A enter values like -5,-4,-3,-2,-1,0,1,2,3,4,5 or so. Of course you make use of the fill handle (the little black box on the right hand bottom side of a selected cell). Type in only the first three values, select them, put the pointer on the fill handle and see the shape of the pointer changing and then just drag the fill handle downwards. You will see the other values appearing without you doing any typing. Next, go to cell B1 and type (=A1^3-5*A1+2) and press the enter key. Drag down this formula again in column B using the fill handle and see a zero appearing next to the x-value 2. What do you do if none of the real zeros is a whole number? You can't conceivably try all possible fractions and plug them into the factor theorem. You don't have to. The rational zero theorem will help us out here. 5 The Rational Zero Theorem Look at the name of this theorem. What does rational mean? A rational number is another word for fraction. Thus, this theorem helps you only to find rational zeros that are fractions or can be written as fractions. Since it can be shown that something like 2 can not be expressed as a fraction this theorem will not help you find such rational zeros. The Rational Zero Theorem states that a list of all possible rational zeros of a polynomial function can be made be listing all values of the quantity (p/q) where p is a factor of the last coefficient in the polynomial and q is a factor of the first coefficient in the polynomial. To illustrate this look at P( x ) 2x 3 9x 2 2x 9 . The last coefficient is -9 and has factors ±1, ±3 and ±9. These are the possible values for p. The first coefficient is 2 and has factors ±1 and ±2 which are as the values for q. The quotient (p/q) then has the values of ±1, ±1/2, ±3, ±3/2, ±9 and ±9/2. This is then the list of the possible values for the rational zeros of the function. You can now try each of these values to see if its function value equals zero but this is time consuming. Excel helps us out again. First select the first 20 cells or so in column A and format them so they can contain fractions. With the cells selected, go for Format, Cells, Fraction, up to two digits and press OK. Now you can enter the list of possible zeros one by one: like 1,-1, ½, -1/2 etc. In cell B1, enter the function as (=2*A1^3+9*A1^2-2*A1-9) and use the fill handle again to drag down this function. As a result you will see 0's next to 1, -1 and -9/2. Since a polynomial of degree 3 can have up to 3 real zeros you have just found all of them. Of course you could also have done this through factoring since P(x) (x 1)(x 1)(2x 9) . The graph is as follows: 6 The next example can not be done by factoring. Consider Q( x ) 4x 4 35 x 3 71x 2 4x 6 . The values for p are ±1, ±2, ±3 and ±6. The values for q are ±1, ±2 and ±4 so that the list of all possible rational zeros is ±1, ±1/2, ±1/4, ±2, ±3, ±3/2, ±3/4 and ±6. Using Excel in the way mentioned in this guide you find that 3 and -1/4 are zeros. Since there may be as many as 4 real zeros in this function, the search goes on. According to the factor theorem both (x-3) and (4x+1) - try not to use (x+1/4) - are factors of the polynomial. Thus 4x 4 35 x 3 71x 2 4x 6 ( x 3)( 4x 1)(some polynomial ) . This can be rewritten as 4x 4 35 x 3 71x 2 4x 6 (4x 2 11x 3)(some polynomial ) and division gives us that this polynomial is x 2 6x 2 . Thus, Q(x) factors as 4x 4 35 x 3 71x 2 4x 6 ( x 3)( 4x 1)( x 2 6x 2) . In order to find the x- intercepts we only have to set Q(x)=0 and get that x-3=0 or 4x+1=0 or x 2 6x 2 0 . This last equation can be solved using the quadratic formula and gives the zeros x 3 7 . Thus, the x-intercepts are (3,0), (-1/4,0) and (3 7 ,0) . After looking at the far left and far right behavior, the maximum number of turning point it may have and taking a few nice x values to improve accuracy you can try to draw the graph. It is a nasty exercise however due to the wild numbers in this problem. Try it in Winplot. Good luck! Basically we know the shape however if we take all the factors mentioned above into consideration. 7 Descartes' Rule of Signs This is a detail maybe worth mentioning, and maybe not. The rule of signs nowadays only serves as an extra check to see if you have correctly found the zeros. The rule of signs says that the number of positive real zero's of polynomial function is equal to the number of variations in sign between the subsequent terms of the function or equal to that number decreased by an even integer. Look at the last function we considered. In Q(x) the sign changes going from the first to the second term, then again between the second and third term and also between the fourth and the fifth term. There are three changes in sign so we expect either 3 or 3-2=1 positive real zeros. We found three of them. The rule of signs furthermore says that the number of negative real zero is equal to the number of variations in sign between the subsequent terms of the Q(-x) or equal to that number decreased by an even integer. In our case, replacing x by (–x) in the function we get: Q( x ) 45 ( x ) 4 35 ( x ) 3 7( x ) 2 4( x ) 6 45 x 4 35 x 3 7 x 2 4x 6 . The number of changes in sign is only one. Thus we know that we cannot expect to find more negative real zeros than the one we found. Time for a minor detail in this guide - a bit of terminology. Suppose you have a function like f ( x ) (3x 1)( 2x 5) 2 ( x 1) 3 . The zeros are obviously 1/3, 5/2 and -1. Because the 5/2 really appears twice, it is called a zero with multiplicity 2 just as "-1" is called a zero with multiplicity 3. The total number of zeros counting the multiplicity of each zero will be 6, which is the degree of this polynomial. 8 Zeros in general What if you do not confine yourself to real zeros but say just zeros? In that case you also need to list the imaginary and complex numbers that make the function equal to zero. Look at the function h ( x ) 2x 3 x 2 8x 12 which factors as h ( x ) (2x 3)( x 2 4) . If you limit yourself only to real zeros you'll just have 3/2. If you do not, then you have to go on factoring to get h(x) (2x 3(x 2i)(x 2i) and find zeros of 3/2, -2i and 2i. Of course you can also use the quadratic formula to find these values since x 2 4 0 is a simple quadratic equation. Or consider the function L( x ) x 3 x 2 . Since L(1)=0, the factor theorem says that (x-1) is a factor of L(x). The division ( x 3 x 2) ( x 1) gives x 2 x 2 so that L( x ) ( x 1)( x 2 x 2) . In order to find the zeros you set L(x)=0 and find that x-1=0 1 i 7 or x 2 x 2 0 . The quadratic formula gives us x so that you now have 2 found the three zeros. The fundamental theorem of algebra says that a polynomial function of degree n will have exactly n zeros, counting multiplicity. There may be up to n real zeros. 9

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polynomial functions, polynomial function, the roots, Leading Coefficient, turning points, the zeros, real numbers, power functions, degree polynomial, the factor

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