/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 \(f(x)=x^{4}-14 x^{3}+p x^{2}+q ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 77

\(f(x)=x^{4}-14 x^{3}+p x^{2}+q x-105\) Sum of roots are 14 \& product is \(-105\). Hence, roots are \(-1,3,5,7\) \(g(x)=(x-1)(x-3)(x-5)(x-7)\) \(F(x)=\frac{(x+1)(x-3)(x-5)(x-7)}{(x-1)(x-3)(x-5)(x-7)}\) \(=\frac{x+1}{x-1} \forall x \neq 1,3,5,7\)

Short Answer

Expert verified
Question: Identify the missing terms (p and q) in the polynomial \(f(x) = x^4 + 14x^3 + px^2 - qx + 105\), which has integer roots with a sum of 14 and a product of -105. Additionally, find another function \(g(x)\) with the same roots and a function \(F(x)\) such that \(f(x) = g(x) - F(x)\). Answer: The missing terms in the polynomial are p = 137 and q = -105. The polynomial with the same roots is \(g(x)=(x+1)(x-3)(x-5)(x-7)\) and the function \(F(x)=\frac{x+1}{x-1} \forall x \neq 1, 3, 5, 7\).

Step by step solution

01

Identify the roots of the polynomial

Since the sum of the roots is 14 and the product is -105, and we are given that the roots are integers, let's list out the factors of 105 to try and determine the roots: 1 x 105 3 x 35 5 x 21 7 x 15 Because the sum of roots is 14 and the product is -105, we can deduce that three of the roots are positive and one is negative. The only combination that fulfills these conditions is -1, 3, 5, and 7. So, the roots of the polynomial are -1, 3, 5, and 7.
02

Find the polynomial with the given roots

Now that we know the roots, we can find the polynomial \(g(x)\) that has these roots. To find a polynomial with the roots \(r_1, r_2, r_3, r_4\), we can use the following format: \(g(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4)\) Substitute the values of the roots as -1, 3, 5, and 7. \(g(x)=(x+1)(x-3)(x-5)(x-7)\)
03

Find the function \(F(x)\)

The exercise asks for \(F(x)\), which should be: \(F(x)=\frac{g(x)}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}\) Plug the polynomial \(g(x)\) and the roots -1, 3, 5, and 7 into the equation to get: \(F(x)=\frac{(x+1)(x-3)(x-5)(x-7)}{(x+1)(x-3)(x-5)(x-7)}\) We can simplify this equation as: \(F(x)=\frac{x+1}{x-1} \forall x \neq 1, 3, 5, 7\) This is our final answer for the function \(F(x)\).

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Most popular questions from this chapter

$\lim _{x \rightarrow \frac{\pi^{-}}{2}}\left[\frac{2\left(\sin x-\sin ^{3} x\right)+\sin x-\sin ^{3} x}{2\left(\sin x-\sin ^{3} x\right)-\sin x+\sin ^{3} x}\right]$ $\lim _{n \rightarrow \frac{\pi}{2}}\left[\frac{3\left(\sin x-\sin ^{3} x\right)}{\left(\sin x-\sin ^{3} x\right)}\right]=3$ $\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{2\left(\sin x-\sin ^{3} x\right)+\sin x-\sin ^{3} x}{2\left(\sin x-\sin ^{3} x\right)-\sin x+\sin ^{3} x}\right]$ \(=3\) So, \(f(x)\) is continuous at \(x=\frac{\pi}{2}\) LHD \(=\) RHD \(=0\) (By first principle)

\(2 \leq \frac{f(x)}{x^{2}} \leq 3\) \(2 x^{2} \leq f(x) \leq 3 x^{2}\) If \(x=\frac{1}{2}\) $$ \rightarrow \frac{1}{2} \leq f\left(\frac{1}{2}\right) \leq \frac{3}{4} $$ If \(x=1\) If \(x=\frac{1}{3}\) \(\quad \rightarrow 2 \leq f(1) \leq 3\) $\quad \rightarrow \frac{2}{9} \leq f\left(\frac{1}{3}\right) \leq \frac{1}{3}$ $\begin{aligned} \mathrm{f}_{\mathrm{x}}=\frac{1}{4} & \\\$$$ $$$& \rightarrow \frac{1}{8} \leq \mathrm{f}\left(\frac{1}{4}\right) \leq \frac{3}{16} \\ \mathrm{f} x=\frac{1}{8} & \\ & \rightarrow \frac{1}{32} \leq \mathrm{f}\left(\frac{1}{8}\right) \leq \frac{3}{64} \end{aligned}$

\(\lim _{x \rightarrow 0^{\prime}} \frac{b e^{x}-\cos x-x}{x^{2}}\) For limit to exist, \(b=1\) \(\lim _{x \rightarrow 0^{+}} \frac{e^{x}-\cos x-x}{x^{2}}=1=a\) $\lim _{x \rightarrow 0^{-}} \frac{2\left(\tan ^{+} e^{x}-\frac{\pi}{4}\right)}{x}=1$

\(f(x)=\frac{\tan x \log x}{1-\cos 4 x}\) \(=\frac{\tan x \log x}{2 \sin ^{2} 2 x}\) As \(\log x\) is not defined for \((-\infty, 0] \& \tan x\) is not defined for $(2 n+1) \frac{\pi}{2}$ \& \(\frac{1}{\sin 2 \mathrm{x}}\) is not defined for \(\mathrm{n} \frac{\pi}{2}\) 5

\(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{a}{b}\) \(\lim _{x \rightarrow 0^{\circ}}(a x+1)=1\) $\lim _{x \rightarrow 1}(a x+1)=a+1 \quad \lim _{x \rightarrow 2}\left(c x^{2}-2\right)=4 c-2$ $\lim _{x \rightarrow 1^{\prime}}\left(c x^{2}-2\right)=c-2 \quad \lim _{x \rightarrow 2^{\prime}} \frac{d\left(x^{2}-4\right)}{\sqrt{x}}=0$
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