Polynomials. Dr. philippe B. laval Kennesaw State University. April 3, 2005


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1 Polynomials Dr. philippe B. laval Kennesaw State University April 3, 2005 Abstract Handout on polynomials. The following topics are covered: Polynomial Functions End behavior Extrema Polynomial Division Zeros of a polynomial Sign of a polynomial Factoring polynomial Sketch of the Graph of a polynomial 1 Definition and Examples Definition 1 1. A polynomial function is a function of the form f (x) =a n x n + a n 1 x n 1 + a n 2 x n a 2 x 2 + a 1 x + a 0 where a n 0, a 0,a 1,a n are real numbers, they are called the coefficients of the polynomial. n is any positive integer. 2. The above form is called the standard form of a polynomial. 3. The highest power of x is called the degree of the polynomial. 4. The term a n x n (corresponding to the highest power, not necessarily the first to appear) is called the leading term of the polynomial, a n is the leading coefficient. 5. The zeros (or the roots) of a polynomial f are the values of x for which f (x) =0. Example 2 f (x) =2 5x is a polynomial function. It is also a linear function. Linear functions are polynomials. The leading term is 5x, the leading coefficient is 5. 1
2 Example 3 f (x) =5 2x + x 2 is a polynomial function. It is also a quadratic function. Quadratic functions are polynomials. The leading term is x 2, the leading coefficient is 1. Example 4 f (x) = x 3 +7x +6 is a third degree polynomial. The leading term is x 3, the leading coefficient is 1. Example 5 f (x) =5is a constant polynomial, its degree is 0.The degree of any constant polynomial is 0. 2 Shape of the Graph 2.1 Practice In this section, we show a few examples. The graphs shown were obtained using computer software. You can also graph them on your calculator. You may have to specify a larger viewing window in the y direction as the y values tend to increase very quickly. As you look at these graphs, try to answer the following questions: 1. What seems to determine the end behavior of a polynomial? 2. What seems to determine how many xintercepts the polynomial has? 3. What seems to determine how many local maxima and minima the polynomial has? 4. How would you sketch (by hand) the graph of a polynomial? We will address these questions in the sections which follow. However, you should try to answer them without reading ahead. Then, you can read ahead to check your answers. Example 6 f (x) =x 3 7x 6 This is a third degree polynomial. Its graph is shown on Figure 1. Example 7 f (x) = x 3 +7x +6 This is also a third degree polynomial, very similar to the previous example. Its graph is shown on Figure 2. The main difference is the sign of the leading coefficient. It is negative, while in the previous example it was positive. Compare this graph with the previous one. Example 8 f (x) =x 5 +2x 4 23x 3 28x 2 +76x +80 This is a fifth degree polynomial. Its graph is shown on Figure 3 2
3 Figure 1: Third degree polynomial Figure 2: Third degree polynomial 3
4 Figure 3: Fifth degree polynomial Example 9 f (x) =x 4 x 3 13x 2 + x +12 This is a fourth degree polynomial. Its graph is shown on Figure 4. Example 10 f (x) = x 4 + x 3 +13x 2 + x +12 This is also a fourth degree polynomial. Its graph is shown on Figure 5. 4
5 Figure 4: Fourth degree polynomial Figure 5: Fourth degree polynomial 5
6 2.2 End Behavior of a Polynomial Here, we study the behavior of the polynomial as x gets very large (we write as x )andasx gets very small (we write as x ). Geometrically, we are studying what the graph of a polynomial looks like at both ends of the x axis. It turns out that this is entirely determined by the leading term. Remember, in the case of a quadratic function, the sign of the coefficient of x 2 determined whether the graph opened up or down. For polynomials in general, the sign of the leading coefficient as well as the degree of the polynomial determine the shape of their graph. For the discussion that follows, let us assume that the leading coefficient is a n. When n is even, both ends point in the same direction. The sign of a n determine whether they point up or down. When n is odd, the ends point in opposite directions. The sign of a n determine which end points down and which one points up. Thus, there are four possible cases case 1: n is even and a n > 0 The general shape of the graph at the ends will look like Figure 6. An example of such a graph is Figure 4. In this case, we say that y approaches as x approaches ±. In this case, we write y as x ±. Figure 6: n even, a n > 0 6
7 2.2.2 case 2: n is even and a n < 0 The general shape of the graph at the ends will look like Figure 7. An example of such a graph is Figure 5. Here, we have y as x ±. Figure 7: n even, a n < case 3: n is odd and a n > 0 The general shape of the graph at the ends will look like Figure 8. An example of such a graph is Figure 1. Here, y as x,andy as x case 4: n is odd and a n < 0 The general shape of the graph at the ends will look like Figure 9. An example of such a graph is Figure 2. Here, y as x,andy as x Conclusion It is not difficult to see why a polynomial behaves like its term of highest degree. We simply have to factor the leading term. Let f (x) be a polynomial of degree 7
8 Figure 8: n odd, a n > 0 Figure 9: n odd, a n < 0 8
9 n. Then, we have: f (x) = a n x n + a n 1 x n a 1 x + a 0 ( = a n x n 1+ a n 1x n 1 a n x n + a n 2x n 2 a n x n a 1x a n x n + a ) 0 a n x ( n = a n x n 1+ a n 1 a n x + a n 2 a n x a 1 a n x n 1 + a ) 0 a n x n As x ±, all the fractions containing x to some power in the denominator will get smaller and smaller. Therefore, as x ±, 1+ a n 1 a n x + a n 2 a n x a 1 a n x n 1 + a 0 a n x n 1. It follows that f (x) a nx n as x ±. 2.3 Local Extrema of a Polynomial A polynomial of degree n canhaveatmostn 1 local extrema, it may have fewer. Note that this implies that if a polynomial has n local extrema, then its degree must be at least n+1. Recall that in the case of a quadratic function, we found that it either had one maximum or one minimum. Knowing the number of maxima and minima is helpful is determining the shape of the graph of a polynomial. However, finding the maxima and minima of a polynomial given its formula is beyond the scope of this class. It is usually learned in a Calculus class. In this class, we will do it graphically. 9
10 Example 11 Select the graph from the four below that best represents the graph of f (x) =2x 4 9x 3 +11x 2 4 Thedegreeoff (x) is even (4) and the leading coefficient is positive, this reduces the choices to graphs 2 and 3. Since the degree of f (x) is 4, it has at most 3 local extrema, this rules out graph 2. Thus, graph 3 is the correct choice. 2.4 Xintercepts A polynomial of degree n can have at most n xintercepts, it may have fewer. Recall, a parabola (which is a polynomial of degree 2) can have 2, 1 or 0 x intercepts. Knowing the number of xintercepts is helpful is determining the shape of the graph of a polynomial. For polynomials of degree 2, one can use the quadratic formula to find the xintercepts. If the degree of the polynomial is higher, it is more difficult. If the degree is 3, there is a formula, but it is 10
11 complicated and few people remember it. If the degree is 4 or higher, there are no formulas like the quadratic formula. We will see in the next section how to find the xintercepts using other methods. There is one case for which it is easy to find the zeros of a polynomial: when it is factored. We illustrate this in the next section. 2.5 Zeros of a Factored Polynomial To find the zeros of a factored polynomial, we simply set each factor to 0 and solve for x. Remark 12 When a polynomial is factored, its degree is found by adding the exponent of each factor. Therefore, (x 1) (2x +4)(x 6) is of degree 3. (x 1) (x 3) 2 (x +3) 3 is of degree 6. Example 13 Find the zeros of f (x) =(x 1) (2x +4)(x 6) We set each factor equal to zero and solve for x. x 1=0 x =1. 2x +4=0 x = 2. x 6=0 x =6. So, the zeros of f (x) are x = 2, x =1and x =6. Definition 14 If (x c) k for k 1 is a factor of a polynomial P (x) then we say that c is a zero of multiplicity k. Whenk =1, c isalsocalledasinglezero. If k =2, c is called a double zero. If k =3, c is called a triple zero... Example 15 If p (x) =(x 1) (x 2) 2 then 1 is a single zero, 2 is a double zero. The multiplicity of a zero plays an important role in determining the shape of a graph. This is summarized in the proposition below. Proposition 16 Suppose that (x c) k for k 1 is a factor of a polynomial P (x), thatisc is a zero of multiplicity k for P (x), then the following is true: 1. If k is odd, then the graph of P (x) crosses the x axis at (c, 0). 2. If k is even, then the graph of P (x) is tangent (touches but does not cross) the x axis at (c, 0). Example 17 Consider the polynomial P (x) = (x 1) (x 3) (x +3). This polynomial has three zeros of odd multiplicity. They are x = 3, x =1and x =3. This means that the graph of P will cross the xaxis at ( 3, 0), (1, 0) and (3, 0). This can be verified on the graph of P which is shown below. 11
12 Example 18 Consider the polynomial P (x) =(x 1) (x 3) 2 (x +3) 3. This polynomial has three zeros. They are x = 3 (multiplicity 3), x = 1(multiplicity 1) and x =3(multiplicity 2). This means that the graph of P will cross the xaxis at ( 3, 0) and (1, 0) but will touch the xaxis at (3, 0). This can be verified on the graph of P which is shown below. 12
13 2.6 Yintercept When a polynomial is in standard form, its yintercept is the constant term. Otherwise, the technique to find the yintercept is always the same: we set x =0and solve for y. 2.7 Sketching the Graph of a Polynomial Using all the information we derived above, we can sketch by hand the graph of a polynomial with good accuracy. To graph a polynomial, use the follow the following outline: 1. Use the leading term to find the end behavior. 2. Find the zeros to find the xintercepts. 3. Use the multiplicity of the zeros to find if the polynomial crosses or touches the xaxis. 4. The zeros divide the xaxis into intervals. The sign of the polynomial is constant in each interval. Use test points to find this sign. 5. Find the yintercept. 13
14 6. Use the fact that a polynomial of degree n has at most n xintercepts and n 1 local maxima or minima. 7. If necessary, use additional polynomial values to get a better sketch. Example 19 Sketch the graph of f (x) =x 3 + x 2 6x We go through the above list and use the information gathered to sketch the graph. 1. The leading term is x 3. Since the degree of this polynomial is odd, both ends go in opposite directions. Since the leading coefficient is positive, the left end go to and the right end go to. In other words, the end behavior of this polynomial is like that of y = x To find the zeros, we solve x 3 + x 2 6x = 0 x ( x 2 + x 6 ) = 0 x (x 2) (x +3) = 0 So, the solutions are x = 0, x = 3 and x = 2. xintercepts are ( 3, 0), (0, 0) and (2, 0). It follows that the 3. Since the zeros all have multiplicity 1, it follows that the graph of the polynomial will cross the xaxis at its zeros. 4. The zeros divide the xaxis into intervals. To find the sign of the polynomial in each interval, we pick a test point in each interval. This is a technique similar to the one used when we were finding the sign of a quadratic function. Interval <x< 3 3 <x<0 0 <x<2 2 <x< test point a f (a) f ( 4) = 24 f ( 1) = 6 f (1) = 4 f (3) = 18 sign of f To find the yintercept, we compute f (0). The yintercept is (0, 0). f (0) = 0 6. This polynomial is of degree 3, so it can have at most 3 xintercepts. We found all three of them. It will also have at most 2 local extreme values. 7. Using the information, the sketch of the graph is: 14
15 Example 20 Sketch the graph of g (x) =(x 1) (x 2) 2 (x +3) We go through the above list and use the information gathered to sketch the graph. 1. The leading term is x 4. Note that to find the leading term, we do not need to multiply everything. We only need to multiply the x terms. This means that both ends of the polynomial point in the same direction (even degree) and they point toward (positive leading term). 2. Since the polynomial is already factored, we can find its zeros easily. They are: x = 3 (multiplicity 1), x = 1(multiplicity 1) and x = 2(multiplicity 2). Therefore, the xintercepts are ( 3, 0), (1, 0) and (2, 0). 3. Since x = 3 and x =1have multiplicity 1, the graph of this polynomial will cross the xaxis at ( 3, 0) and (1, 0). x =2has multiplicity 2, sothe graph will touch the xaxis at (2, 0). 4. The three zeros of this polynomial determine four intervals. We use a table of values to find the sign of the polynomial in each interval, as in the previous example. Interval <x< 3 3 <x<1 1 <x<2 2 <x< test point a f (a) g ( 4) = 180 g (0) = 12 g (1.5) = g (3) = 12 sign of f
16 5. To find the yintercept, we compute g (0). We already did it in the table and found that g (0) = This polynomial will have at most 4 zeros and 3 local extrema. 7. Using the information derived, the shape of the graph is: 3 Real Zeros of a Polynomial If f (x) is a polynomial, then we said earlier that c was a root or a zero of f if f (c) =0. Geometrically, when we are finding the real zeros of a polynomial, we are finding its xintercepts. Finding the zeros of a polynomial is also closely related to factoring the polynomial. Thus, finding the zeros of a polynomial is very important in the study of this polynomial. We study how to find the zeros of a polynomial, and see how it relates to factoring it. We have already addressed the case when the polynomial was factored. In the following sections, we will look at polynomials which are not factored. First, we look at an important tool: polynomial division. 3.1 Dividing Polynomials Polynomials can be divided in a way similar to the way numbers are divided. The process is illustrated below as we show how to divide 6x 2 26x +12 by 16
17 x 4. 6x 2 x 4 )6x 2 26x +12 6x 2 24x 2x +12 2x +8 In this process, we use the following terminology: 6x 2 26x +12is called the dividend x 4 is called the divisor 6x 2 is called the quotient 4 is called the remainder andwewrite 6x 2 26x +12=(x 4) (6x 2) + 4 If the remainder had been 0, the polynomial would have been factored. It turns out that this process can be repeated for any polynomials. It is called the division algorithm. It says the following: Theorem 21 (Division Algorithm) If P (x) and D (x) are polynomials with D (x) 0, then there exist unique polynomials Q (x) and R (x) such that P (x) =D (x) Q (x)+r (x) where R (x) is either 0 or of degree less than the degree of D (x). P (x) is called the dividend, D (x) is called the divisor, Q (x) is called the quotient, and R (x) is called the remainder. Remark 22 If the divisor is a polynomial of degree 1, then the remainder will be a constant (since its degree has to be less than that of the divisor). Theorem 23 (Factor Theorem) Let P (x) be a polynomial. P (c) =0(i.e. c isazeroofp )ifandonlyifx c is a factor of P (x). Remark 24 If (x c) 2 is a factor of P (x), we say that c is a double zero. If (x c) 3 is a factor of P (x), we say that c is a triple zero. And so on. This theorem is very useful. Consider the examples below. 4 17
18 Example 25 Factor f (x) =2x 2 +5x 3 By the theorem, we know that if c is a zero of f, x c will be a factor. Since this is a second degree polynomial, we know how to find all its zeros. They are given by the quadratic formula. The zeros are 3 and 1 2.Thusx+3 and x 1 are factors ( 2 of f. Be careful though, we cannot conclude that f (x) =(x +3) x 1 ). In 2 fact it is not. We can only conclude that x +3 and x 1 are factors of f. ( 2 However, since both f and (x +3) x 1 ) are of degree 2, the only thing we 2 are missing is a constant. In fact, ( (x +3) x 1 ) = x x 3 2 = 1 2 f (x) Thus, ( f (x) =2(x +3) x 1 ) 2 Example 26 Factor f (x) =x 3 7x +6 completely, given that 1 is a root of f (x). If 1 is a root, then x 1 is a factor of f (x). Thus, there exists a polynomial Q (x) such that f (x) = (x 1) Q (x) Q (x) = f (x) x 1 We divide f (x) by x 1, we obtain f (x) = (x 1) ( x 2 + x 6 ) = (x 1) (x +3)(x 2) Example 27 Find a polynomial F (x) of degree 4 that has zeros 3, 0, 1, and 5. By the factor theorem, x 0 (= x), x 1, x 5, andx +3 must be factors. Thus F (x) =x (x +3)(x 1) (x 5) is such a polynomial. It is not the only one though. The answer is no. However, sincewewantapolynomialofdegree4, any other solution will be a constant multiple of what we found. This brings us to the following remark. Remark 28 If we know all the zeros of a polynomial, we do not know the polynomial exactly, we know it up to a constant. To know the polynomial exactly, 18
19 we must also be given its value at a point. In the example above, we were asked to find a polynomial of degree 4, knowing 4 (thatisall)ofitszeros. Wefoundthat F (x) =x (x +3)(x 1) (x 5) was such a polynomial. All the polynomials which satisfy the given condition are of the form where C is a constant. F (x) =Cx(x +3)(x 1) (x 5) Example 29 Find all the polynomials of degree 3 having 1, 1, and2 as zeros. Since the polynomial is of degree 3, we are given all its zeros. So, we know all its linear factors. The polynomials we want are of the form where C is a constant. f (x) =C (x +1)(x 1) (x 2) Example 30 Find all the polynomial of degree 5 having 1, 1, and2 as zeros. Thelinearfactorofthepolynomialswewantare(x +1)(x 1) (x 2). However, since we want polynomials of degree 5, the polynomials we want are of the form f (x) =g (x)(x +1)(x 1) (x 2) where g (x) is a polynomial of degree Rational Zeros Theorem The factor theorem is very useful in factoring a polynomials, when we know some of its zeros. For example, if we have to factor a polynomial of degree 3, and we know one of its zeros, say c. Then, x c is a factor. The remaining factor must be a polynomial of degree 2. We know how to factor a polynomial of degree 2 since we know how to find its zeros. But, how did we happen to know one of the zeros of this polynomial to begin with? The next theorem gives us a method for finding all the rational zeros of a polynomial. Theorem 31 (Rational zeros theorem) Let f (x) =a n x n +a n 1 x n a 1 x + a 0 be a polynomial with integer coefficients. Then, every rational zero of f is of the form p q where p is a divisor of a 0 q is a divisor of a n Remark 32 In the above theorem, p q isassumedtobeinlowestterms. 19
20 Example 33 Use the rational zeros theorem to find the zeros of P (x) =2x 3 + x 2 13x +6, then factor f. In this case, a 0 =6, therefore p = ±1, ±2, ±3, ±6. Similarly, a n = a 3 =2, therefore q = ±1, ±2. The possible values for p q are ±1 1, ±2 1, ±3 1, ±6 1, ±1 2, ± 2 2, ±3 2, ±6. If we simplify, and eliminate the duplicates, we see that the 2 possible values for p q are: ±1, ±2, ±3, ±6, ±1 2, ±3. We check each value. 2 With Scientific Notebook, this amounts to doing P 2 = Thus, the zeros are 3, 2 and 1 2.Hence, ( P (x) =C x 1 ) (x 2) (x +3) 2 Since P (0) = 6, it follows that 6 = C 6 = 3C ( 1 ) ( 2) (3) 2 Or, C =2.Thus, P (x) = ( 2 x 1 ) (x 2) (x +3) 2 = (2x 1) (x 2) (x +3) 4 Intermediate Value Theorem This is a very important theorem in mathematics. Here, we give a simplified version of it. You will learn the complete version in Calculus. This theorem 20
21 is based on the fact that the graph of a polynomial is unbroken. Therefore, if at x = a the polynomial is say positive (above the xaxis) and at x = b it is negative (below the xaxis), then it must have crossed the xaxis somewhere between a and b. Thus, we know that there must be a zero between x = a and x = b. Here is the statement of the theorem. Theorem 34 Let P (x) be a polynomial and suppose that a and b are two numbers such that P (a) and P (b) have opposite signs. Then, P (x) has a real zero between a and b. Remark 35 The converse is not true. That is, the fact that P (a) and P (b) have the same sign does not mean the polynomial has no zero between a and b. Example 36 Prove that P (x) =x 2 7x +2x 3 6 has a zero between 1 and 3. We see that P (1) = 10, P (3) = 36. P (1) and P (3) have opposite signs, therefore P has a zero between 1 and 3. Not only we can use this theorem to help us determine if a polynomial P (x) has a zero between two numbers, we can also use it to approximate the zero. Thekeyistorealizethatifwefindtwoconsecutivevaluesa and b for which P (a) and P (b) have opposite signs, then P has a zero between a and b. Example 37 FindazeroforP (x) =x 2 2, correct to 2 decimal places. It is easy to see that P (1) = 1 and P (2) = 2. Thus, P has a zero between 1 and 2. Now, starting at 1, welookatp (1.1), P (1.2),..., P (1.9) until we find two consecutive values for which the sign of P is opposite. P (1.1) = 0.79, P (1.2) = 0.56, P (1.3) = 0.31, P (1.4) = 0.04, P (1.5) = Thus, we now know that the zero is between 1.4 and 1.5. We repeat the procedure with 1.41, 1.42, 1.43,... P (1.41) = , P (1.42) = Thus the zero is between 1.41 and So, we already know the zero correct to 1 decimal place. We do this one more time with 1.411, 1.412, 1.413,... P (1.411) = , P (1.412) = , P (1.413) = , P (1.414) = , P (1.415) = So, the zero is between and So, the zero correct to two decimal places is Can you guess which number we are approximating? Remark 38 The above procedure may seem tedious by hand. It is. However, with modern computers, it can be implemented easily and performed very quickly. 5 Sign of a Polynomial The method of finding the sign of a polynomial function is similar to the method used for a quadratic function. First, we find the zeros of the polynomial. The zeros of the polynomial divide the real line into intervals. The sign of the polynomial is constant in each interval. To find it, we pick a number in each interval, and find the sign of the polynomial at that number. 21
22 Example 39 Find the sign of f (x) =x 3 7x +6. We found above that f (x) =(x 1) (x +3)(x 2). Thus its zeros are 3, 1 and 2. Three numbers determine four intervals. We find the sign of f in each interval. Interval (, 3) ( 3, 1) (1, 2) (2, ) Test point a f (a) f ( 4) = 30 f (0) = 6 f (1.5) = f (3) = 12 Sign of f + + Therefore, f is positive in ( 3, 1) (2, ). It is negative everywhere else, that is in (, 3) (1, 2). 6 Problems 1. Problems assigned in the book (a) # 1, 2, 3, 4, 1125, 2931, 45, 46, 48, 49, 74, 75, 80 on pages (b) # 1, 2, 3, 5, 6, 7, 39, 40, 41 on page 273, 274. (c) # 1, 2, 13, 14, 43, 49, 50, 51, 71, 73, 74 on page (d) # 1, 3, 5, 6, 8, 15, 16, 20, 29 on pages 310, Find the degree, leading term and leading coefficient of each polynomial below. (a) f (x) =3x 2 +9x 10x 3 4 (b) f (x) = x 3 +7x 2 +36x 240 (c) f (x) = 2x 4 + x 3 +13x 2 +8x 1 (d) f (x) =(x 1) ( x 3 x 2 x 1 ) 3. Using your calculator, or the graphing applets PlotSolve, sketchthe graph of the polynomials below. For each graph you obtained, try to find the coordinates of the x and yintercepts as well as the coordinates of the max and min. Graphing polynomials with a calculator usually involves playing with the viewing window. (a) f (x) =3x 2 +9x 10x 3 4 (b) f (x) = x 3 +7x 2 +36x 240 (c) f (x) = 2x 4 + x 3 +13x 2 +8x 1 (d) f (x) =x 4 2x Select the graph that best represents the given polynomial. Explain your choice. In making your choice, you will have to look at the degree of the polynomial, its maximum number of xintercepts, its yintercept and the number of max and min it can have. 22
23 (a) f (x) = x 3 + x 2 + x 1 (b) g (x) =x 3 6x 2 +9x +5 (c) h (x) =x 4 3x 3 +3x 2 x 23
24 5. Select the graph from the graphs below that best represent the given polynomial. (a) f (x) =x 3 2x 2 (b) f (x) =x 4 x 3 x 6 + x (c) f (x) =2x 4 5x 3 +4x 2 x (d) f (x) =2x 4 + x Find the quotient and the remainder in each case below: (a) x3 +2x 2 +2x +1 x +2 (b) x3 8x +2 x +3 (c) 6x3 +2x 2 +22x 2x 2 +5 ( x 2 +3x +1 ) (x +1) (d) (x +1) 7. Find a polynomial of the specified degree that has the given zeros. 24
25 (a) Degree 3, zeros: 1, 1, 3. (b) Degree 4, zeros: 2, 0, 2, 4 (c) Degree 4, zeros: 1, 1, 3, 5 8. Find all the zeros of the given polynomials, then factor them and study their sign. (a) (x 1) 2 ( x 2 x 6 ) (b) x 3 x 2 14x +24given that 2 is a zero. (c) x 4 x 3 7x 2 +13x 6 given that 1 is a double zero. (d) x 4 5x 2 +4given that 1 and 1 are zeros. 25
26 7 Problems (solutions) 2. (a) degree: 3, leading term: 10x 3,leading coefficient: 10 (b) degree: 3, leading term: x 3,leading coefficient: 1 (c) degree: 4, leading term: 2x 4,leading coefficient: 2 (d) degree: 4, leading term: x 4,leading coefficient: 1 3. (a) f (x) =3x 2 +9x 10x 3 4 yintercept: (0, 4) xintercepts:( 1, 0), (.5, 0) and (.8, 0) local max: (.656,.372) local min: (.456, 6.532) (b) f (x) = x 3 +7x 2 +36x
27 yintercept: (0, 240) xintercepts: ( 5.922, 0), (5.357, 0), and(7.565, 0) local max: (6.51, ) local min: ( 1.843, ) (c) f (x) = 2x 4 + x 3 +13x 2 +8x 1 27
28 yintercept: (0, 1) xintercepts: ( 1.794, 0), (.859, 0), (.106, 0), and(3.047, 0) local max:( 1.418, 2.858), and(2.125, 43, 517) local min: (.332, 2.284) (d) f (x) =x 4 2x (a) #2 (b) #3 (c) #1 5. (a) #6 (b) #2 (c) #3 (d) #5 yintercept: (0, 1) xintercepts: (1, 0), and(1.839, 0) local max: none local min:(1.5,.686) 6. (a) Quotient: x 2 +2 Remainder: 3 28
29 (b) Quotient: x 2 3x +1 Remainder: 1 (c) Quotient: 3x +1 Remainder: 7x 5 (d) Quotient: x 2 +3x +1 Remainder: 0 7. Let C denote a constant (a) C (x +1)(x 1) (x 3) (b) Cx(x +2)(x 2) (x 4) (c) C (x +1)(x 1) (x 3) (x 5) 8. Find all the zeros of the given polynomials, then factor them and study their sign. (a) Zeros: 2, 1 (double zero), 3 Factored form: (x 1) 2 (x +2)(x 3) Sign: negative in ( 2, 3), positive in (, 2) (3, ) (b) Zeros: 4, 2, 3 Factored form: (x 2) (x 3) (x +4) Sign: negative in (, 4) (2, 3), positive in ( 4, 2) (3, ) (c) Zeros: 3, 1 (double), 2 Factored form: (x 1) 2 (x +3)(x 2) Sign: negative in ( 3, 2), positive in (, 3) (2, ) (d) Zeros: 2, 1, 1, 2 Factored form: (x 1) (x +1)(x 2) (x +2) Sign: Negative in ( 2, 1) (1, 2), positive in (, 2) ( 1, 1) (2, ) 29
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