Math 115 Spring 2014 Written Homework 6-SOLUTIONS Due Friday, March 14

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1 Math 115 Spring 2014 Written Homework 6-SOLUTIONS Due Friday, March 14 Instructions: Write complete solutions on separate paper (not spiral bound). If multiple pieces of paper are used, THEY MUST BE STAPLED with your name and lecture written on each page. Please review the Course Information document for more complete instructions. 1. For each function, determine whether the function is a polynomial function or not. If it is not a polynomial function, explain why not. If it is a polynomial function, determine the leading coefficient, the leading term, the degree, and the constant term. (a) f(x) := πx x The definition of a polynomial is that the coefficients are real numbers and the powers on the variable terms x n are always integers greater than or equal to 0. The function in (a) is a polynomial. The leading coefficient is π, the leading term is πx 5, the degree is 5, and the constant term is (b) f(x) := 7x 3 + x This is not a polynomial because of the x 3 term. The power is not a positive integer. (c) f(x) := 3x π +.7 This is not a polynomial because of the x π term. The power is not an integer. (d) f(x) := (x + 3)(2x 5)(2x 2 1)( 1 2 x + 5) This is a polynomial. The leading term is (x)(2x)(2x 2 )( 1 2 x) = 2x5. Thus the leading coefficient is 2 and the degree is 5. The constant term is (3)( 5)( 1)(5) = 75. (e) f(x) := (π) 4 This is a polynomial - it could be written as f(x) := (π) 4 x 0. The leading term, leading coefficient, and constant term are all (π) 4 and the degree is 0.

2 2. Determine the it as x and the it as x for each function. (a) f(x) := 2x 3 + 5x 2 3x + 11 x f(x) = x 2x3 [ ] = 2 x x3 = (b) h(x) := x 6 2x 3 + x 2 4x 12 f(x) = [ = 2 2x3 x3 ] = since (negative number) 3 < 0 h(x) = x x x6 = h(x) = x6 = since (negative number) 6 > 0

3 (c) g(x) := 3x 4 + 5x 3 + 6x 5 + x 2 + 3x 9 x g(x) = x 6x5 since this is the highest degree term [ ] = 6 x x5 = g(x) = [ = 6 6x5 x5 ] = since (negative number) 5 < 0 (d) p(x) := (πx 2 4)(3x + 2)(4x 2 5)(2 x) Here, we need the leading term for p(x). Note that the highest degree term for the polynomial is (πx 2 )(3x)(4x 2 )( x) = 12πx 6. ( p(x) = ) 12πx 6 x x [ ] = 12π x x6 = since the coefficient 12π is negative p(x) = [ = 12π ( 12πx 6 ) x6 ] = since 12π < 0 and (negative number) 6 > 0

4 3. For each function below (i) determine the long run behavior as x and x ; (ii) determine the intercepts of the graph of the function; (iii) draw a rough sketch of the graph of the function. (a) g(x) := (x 1)(2x 3)(3 x) 1. long-run behavior Need to examine the its to infinity. dictated by the leading term. Here, Recall that long-run behavior of a polynomial is g(x) = (x)(2x)( x) = x x x 2x3 =. Since g(x) is an odd degree polynomial (degree is 3), so g(x) = 2. intercepts g(x) is the opposite of g(x) x (a) y-intercept Since g(0) = ( 1)( 3)(3) = 9. The graph of g(x) intersects the y-axis at (0, 9). (b) x-intercept(s) Fortunately, g(x) is already factored. For the x-intercepts, we need to identify all of the roots of the polynomial. Here, we see that the roots are x = 1, x = 3 2, and x = 3 These correspond to the x-intercepts, ( ) 3 (1, 0), 2, 0, and (3, 0). 3. graph We use continuity to connect the end behavior and the intercepts. We need to test a point in the interval (1, 3) and a point in the interval ( 3, 3). First choose x = 5: g( 5) = ( 5 1)(2( 5) )(3 5 )) < 0. Now choose x = 2: g(2) = (2 1)(2(2) 3)(3 2)) > 0. Putting all of this 4 information together should give a rough sketch similar to this:

5 (b) f(x) := (x 5)(x 2 1)(11 3x) 1. long-run behavior Need to examine the its to infinity. dictated by the leading term. Here, Recall that long-run behavior of a polynomial is f(x) = x x (x)(x2 )( 3x) = 3x 4 =. x Since f(x) is an even degree polynomial (degree is 4), 2. intercepts f(x) = f(x) =. x (a) y-intercept Since f(0) = ( 5)( 1)(11) = 55. The graph of f(x) intersects the y-axis at (0, 55). (b) x-intercept(s) Fortunately, f(x) is already factored. For the x-intercepts, we need to identify all of the roots of the polynomial. Here, we see that the roots are These correspond to the x-intercepts, x = 5, x = ±1, and x = 11 3 (5, 0), (1, 0), ( 1, 0) and ( ) 11 3, 0.

6 3. graph We use continuity to connect the end behavior and the intercepts. We need to test points in the interval (1, 11 3 ). I choose x = 2: f(2) = (2 5)(22 1)(11 3(2)) < 0. We also need to test a point in the interval ( 11 3, 5). I choose x = 4: f(4) = (4 5)(42 1)(11 3(4)) > 0. Putting all of this information together should give a rough sketch similar to this: 4. Let f(x) := 3x 3 + x 2 38x (a) Determine all of the possible rational zeros of f(x) (according to the Rational Root Theorem). For a rational number p to be a zero, p must be a factor of c q 0 = 24 and q must be a factor of c 3 = 3. Thus, p can be ±1, ±2, ±3, ±4, ±6, ±8, ±12 or ±24, and q can be ±1 or ±3. The possible rational zeros, p, are q ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± 1 3, ±2 3, ±3 3, ±4 3, ±6 3, ±8 3, ±12 3, ±24 3. Removing the rational numbers that repeat when simplified, we get ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± 1 3, ±2 3, ±4 3, ±8 3.

7 (b) Show that x = 4 is a zero (or root) of the function f(x) := 3x 3 + x 2 38x + 24 (i.e. show that f( 4) = 0). What does this tell you about a factor of this polynomial function? To show x = 4 is a zero, we just need to show that f( 4) = 0. f( 4) = 3( 4) 3 + ( 4) 2 38( 4) + 24 = 3( 64) = = 0 The linear term x ( 4) = x + 4 is a factor of the polynomial function. (c) Determine the quotient and remainder when 3x 3 + x 2 38x + 24 is divided by the factor you found in part (b). (You may use either polynomial long division or synthetic division to do this.) 3x 2 11x + 6 x + 4 ) 3x 3 + x 2 38x x 3 12x 2 11x 2 38x 11x x 6x x 24 0 Hence, f(x) := 3x 3 + x 2 38x + 24 = (x + 4)(3x 2 11x + 6). (d) Use what you found in parts (a) - (c) to completely factor the polynomial function f(x) := 3x 3 + x 2 38x Continue the factoring by factoring the quadratic term; f(x) := (x + 4)(3x 2)(x 3)

8 5. Determine the zeros of the polynomial function g(x) := x 4 + 4x 3 10x 2 28x 15 and write it in fully factored form. (Hint: Use the Rational Root Theorem to determine all possible rational roots and then use the division technique of problem 4 to fully factor g(x).) For a rational number p to be a zero, p must be a factor of c q 0 = 15 and q must be a factor of c 4 = 1. Thus, p can be ±1, ±3, ±5 or ±15, and q can be ±1 only. The possible rational zeros, p, are q ±1, ±3, ±5, ±15. Plugging these potential roots into g(x), the first one we see is that x = 1 is a root. Hence, x ( 1) = x + 1 is a factor of g. x 3 + 3x 2 13x 15 x + 1 ) x 4 + 4x 3 10x 2 28x 15 x 4 x 3 3x 3 10x 2 3x 3 3x 2 13x 2 28x 13x x 15x 15 15x Hence, g(x) := x 4 + 4x 3 10x 2 28x 15 = (x + 1)(x 3 + 3x 2 13x 15). Continuing our use of the potential rational roots, we see that x = 3 is a root (since (3 2 ) 13(3) 15 = 0). Hence x 3 is a factor of g(x) (and of x 3 + 3x 2 13x 15). x 2 + 6x + 5 x 3 ) x 3 + 3x 2 13x 15 x 3 + 3x 2 6x 2 13x 6x x 5x 15 5x Hence, g(x) := x 4 + 4x 3 10x 2 28x 15 = (x + 1)(x 3)(x 2 + 6x + 5). We can continue using our list of potential rational roots, but the quadratic term is easy to factor. g(x) := x 4 + 4x 3 10x 2 28x 15 = (x + 1)(x 3)(x + 5)(x + 1) The roots/zeros of g(x) are x = 5, x = 1, and x = 3.

9 6. Use the long-run behavior, the intercepts, and continuity to draw a rough sketch of the graph of each polynomial function. (Note: you may use information you found in previous problems as part of these solutions.) (a) f(x) := 3x 3 + x 2 38x end behavior Need to examine the its to infinity. f(x) = x x 3x3 =. Since f(x) is an odd degree polynomial (degree is 3), 2. intercepts f(x) = f(x) =. x (a) y-intercept Since f(0) = 24. The graph of f(x) intersects the y-axis at (0, 24). (b) x-intercept(s) To determine the roots, we need to factor f(x). Fortunately, we factored f(x) in problem 4; f(x) := (x + 4)(3x 2)(x 3). Here, we see that the roots are x = 4, x = 3, and x = 2 3 These correspond to the x-intercepts, ( 4, 0), (3, 0), and ( ) 2 3, graph We use continuity to connect the end behavior and the intercepts. We need to test a point in the interval ( 7, 3). I choose x = 2: f(2) = (2 + 4)(3(2) 2)(2 3) < 0. Putting all of this 2 information together gives a rough sketch similar to this:

10 (b) g(x) := x 4 + 4x 3 10x 2 28x end behavior Need to examine the its to infinity. g(x) = x x x4 =. Since g(x) is an even degree polynomial (degree is 4), 2. intercepts g(x) = g(x) =. x (a) y-intercept Since g(0) = 15. The graph of g(x) intersects the y-axis at (0, 15). (b) x-intercept(s) To determine the roots, we need to factor g(x). Fortunately, we factored g(x) in problem 5; g(x) := (x + 1) 2 (x 3)(x + 5). Here, we see that the roots are These correspond to the x-intercepts, x = 1, x = 3, and x = 5 ( 1, 0), (3, 0), and ( 5, 0).

11 3. graph We use continuity to connect the end behavior and the intercepts. We need to test a point in the interval ( 5, 1). I choose x = 3: g( 3) = ( 3 + 1) 2 ( 3 3)( 3 + 5) < 0. Putting all of this information together gives a rough sketch similar to this: