/*! 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 27 Use synthetic division to divide... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 27

Use synthetic division to divide. $$\left(3 x^{3}-17 x^{2}+15 x-25\right) \div(x-5)$$

Short Answer

Expert verified
The result of the division is \(3x^2 - 20x - 25 + \frac{100}{x-5}\). The division statement is \(3 x^{3}-17 x^{2}+15 x-25 = (x-5) \cdot (3x^2 - 20x - 25) + 100.

Step by step solution

01

Set Up

Synthetic division requires setting up a table with the coefficients of the polynomial along the top and the 0 of the divisor along the bottom left. Here are the setup materials: \nCoefficients from the polynomial: \(3, -17, 15, -25\) \nThe zero of the divisor (x - 5) is \(5\)
02

Begin Synthetic Division

We begin synthetic division by dropping the first coefficient (3 from the cubic term) straight down. This is the first coefficient of the quotient. Now the table looks like the following: \n 5| 3 -17 15 -25 \n --|------- \n 3
03

Continue Synthetic Division

The second step of synthetic division is to multiply the value just written with the divisor's 0 (5) and to place the result under the next coefficient (-17) and add the two together to get a new result (-20). This new result is written as the next coefficient of the quotient. Now multiply -20 with 5 and place the result under 15, add the two and write the new result (-25) as the next coefficient of the quotient. One last time, multiply -25 by 5, put the result under -25, and write the final result (100) on the bottom line. The updated table looks like this:\n 5| 3 -17 15 -25 \n --|------- \n-20 -25 100 \n3 -20 -25 100
04

Interpret Results

The last line of the table represents the coefficients of the quotient polynomial. The degree of this polynomial is one less than the degree of the original polynomial. Because the original was a third-degree polynomial, the quotient is a second-degree polynomial, and the coefficients correspond to: \(3x^2 - 20x - 25\). The final number, 100, is the remainder.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The coordinate system shown below is called the complex plane. In the complex plane, the point that corresponds to the complex number \(a+b i\) is \((a, b)\) (GRAPH CANNOT COPY) Match each complex number with its corresponding point. (i) 3 (ii) \(3 i\) (iii) \(4+2 i\) (iv) \(2-2 i\) (v) \(-3+3 i\) (vi) \(-1-4 i\)

Find the rational zeros of the polynomial function. $$P(x)=x^{4}-\frac{25}{4} x^{2}+9=\frac{1}{4}\left(4 x^{4}-25 x^{2}+36\right)$$

Determine (if possible) the zeros of the function \(g\) when the function \(f\) has zeros at \(x=r_{1}, x=r_{2},\) and \(x=r_{3}\) $$g(x)=-f(x)$$

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

Use the given zero to find all the zeros of the function. Function \(f(x)=x^{3}+4 x^{2}+14 x+20\) Zero \(-1-3 i\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.