Question

5. You are given the function 8 (x)=2x°-11x +10x+8. Use this information to answer each of the following questions. (a) List all the possible rational solutions to 8(x). (Hint): Make sure that you use the Theorem for Bounds! [6 points] (b) Bob makes the following statement: “Based on all the knowledge we've learned in this class, I believe that that 8(x) could have a total of 2 real solutions and I negative solution." Do you agree with Bob? State Yes or No and explain your answer. (Note): You need to show mathematical work in your explanation. [12 points) (C) Can we use the Intermediate Value Theorem to show that there is a root somewhere between x = 0 and x = 3? If “Yes”, then use the IVT to show this is true. If “No”, then explain why the answer is “No”. [8 points)

Answer 1

The given function is 7(72) = 223 - 11** +102 +8. o Comparing 7131 ax²+ best cx+d, we get a:2, 6=-11, ca 10 and dc8, with Νσιο, Cay lel = 1-4) - la fel 1. = 5. 읽 1.1= 4. borgest of 4 / 5 and 4 is = 5.5 Cauchy's theorem for bounds the real Zeross of g(2) liels i The value Nola 3 says that all (-(5.5+), (5-5+1)) = (-6-5, 6-5) By the rationas nozess the overn, the on say possible rational Zeross are in the form b .. where p is the divissor of the Constant term d=8 and q is the divisor کمد

of the coefficient of the tom a=2 leading Then p should be 1, 22, 24, 28 and q should be + 1. +2 2 The only possible rational Zerols are +1, , 74, I 2, 18. ，

Since it does not lies on (-6.5,6-5) Henee the list of all possible rational Zeroy of f(x) are 41, 1, 24,+2.

(6) Yello with you on but there is a Imans mistake you say could have a total of 2 real solution agree that 7 (2) and 1 negative solution I think it will be a real positive portion and 1 solution. negative Explaination:

NEO, (1) = 2 - } + 10 +8 = 9+0 (-1) (4) 子 - 号+ y + 8 -2 - || -10 十 -15 全 0 ML 10 4 4 32 4 42 主) 3+新一告----(5-)2 2十 12 4 20 4 * 32 4 - - 彝 32 4 + 32 4 二)

8(1) = 128-176 + 40+8 = 0 84-4) =-128-176-40+8 = -336 #0 H2) 16-44 +20+8 91-2) = -16-44-20°8 72 70 = 0 This we have H-1)=0, 7(4) = 0 and H2) = 0 of g(2) Sinee degree So J(2) Ghould be three roots, real or comblexs. But here we see that H-1) = 0, 111) = 0 and 712) rie, all the moth of J (2) are real and there are a positive real solution and 2 and 4 one ne negative Bolution -

(C) yes. we have 7(29) = 2x3 = 1167 +10x +8. Then gees is a polynomial degree ees. 100) = ial of N600, = 870 73) = 54 - 99+30+85-7 <0, (3) are cargos. Now, the internal (0,3) lies on the interval (-6-5, 6-5) 810) and of opposite si

Using INT. we stay that there is a verdue e on (0, 3) He=0 ies there is a noot some where between Auch that x=0 and x=3. (082 3 + 76 C 0 (0,-1)] (3.-7)