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Chapter 2: Problem 96

Find the constant \(c\) such that the denominator will divide evenly into the numerator. $$\frac{x^{5}-2 x^{2}+x+c}{x+2}$$

Short Answer

Expert verified
The constant \(c\) that makes in the numerator such that denominator divides evenly into it is 3.

Step by step solution

01

- Identify coefficients and constant

First, the polynomial is divided using the synthetic division method. Identify the coefficients of the terms and list them. The coefficients on the polynomial \(x^{5}-2 x^{2}+x+c\) are 1, 0, -2, 1 and \(c\). Notice that if a term in the polynomial is missing its coefficient is zero.
02

- Synthetic Division

Perform synthetic division using the number -2 instead of \(x+2\). Place the coefficients and \(c\) in a row and bring down the first number (which is 1). Multiply -2 by the first number and add the result to the second number. Repeat this process until performing the last row addition. Note that the final number will be our remainder.
03

- Setting up and Solving the Equation

For the denominator to divide evenly into the numerator, the remainder must be zero. According to the synthetic division, this remainder is \(-3+c\). So we can set up an equation \(-3 + c = 0\). Solving this equation will give us the value of \(c\) that makes the denominator divide evenly into the numerator.

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Most popular questions from this chapter

Cube each complex number. (a) \(-1+\sqrt{3} i\) (b) \(-1-\sqrt{3} i\)

Write the polynomial as the product of linear factors and list all the zeros of the function. $$f(x)=x^{3}-x^{2}+x+39$$

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(s)=2 s^{3}-5 s^{2}+12 s-5$$

Write the polynomial as the product of linear factors and list all the zeros of the function. $$f(x)=x^{2}+36$$

Prove that the complex conjugate of the product of two complex numbers \(a_{1}+b_{1} i\) and \(a_{2}+b_{2} i\) is the product of their complex conjugates.
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