College

Complete the synthetic division problem below:

[tex]2 \longdiv { 1 \quad 5 \quad -1 \quad 4}[/tex]

What is the quotient in polynomial form?

A. [tex]x + 5[/tex]
B. [tex]x + 7[/tex]
C. [tex]x - 7[/tex]
D. [tex]x - 5[/tex]

Answer :

Sure, let's go through a step-by-step explanation of how synthetic division is performed to divide the polynomial by [tex]\(x - 2\)[/tex].

Step 1: Set up the problem

We are dividing the polynomial represented by the coefficients [tex]\([1, 5, -1, 4]\)[/tex] by the divisor [tex]\(x - 2\)[/tex]. For synthetic division, we use the zero of the divisor, which is [tex]\(2\)[/tex].

Step 2: Write down the coefficients

The coefficients of the polynomial are: [tex]\(1, 5, -1, 4\)[/tex].

Step 3: Perform synthetic division

1. Bring down the first coefficient, which is [tex]\(1\)[/tex].

Quotient so far: [tex]\([1]\)[/tex]

2. Multiply the first term of the quotient ([tex]\(1\)[/tex]) by the divisor ([tex]\(2\)[/tex]) and add it to the next coefficient ([tex]\(5\)[/tex]):

[tex]\[
1 \times 2 + 5 = 7
\][/tex]

Quotient so far: [tex]\([1, 7]\)[/tex]

3. Multiply the next term of the quotient ([tex]\(7\)[/tex]) by the divisor ([tex]\(2\)[/tex]) and add it to the next coefficient ([tex]\(-1\)[/tex]):

[tex]\[
7 \times 2 + (-1) = 13
\][/tex]

Quotient so far: [tex]\([1, 7, 13]\)[/tex]

4. Finally, multiply the last term of the quotient ([tex]\(13\)[/tex]) by the divisor ([tex]\(2\)[/tex]) and add it to the last coefficient ([tex]\(4\)[/tex]):

[tex]\[
13 \times 2 + 4 = 30
\][/tex]

This result is the remainder.

Step 4: Write the result in polynomial form

The quotient is represented by the coefficients [tex]\([1, 7, 13]\)[/tex], which translates to the polynomial [tex]\(x^2 + 7x + 13\)[/tex], and the remainder is [tex]\(30\)[/tex].

In this specific problem, we are only interested in the quotient. Therefore, the quotient polynomial is [tex]\(x + 7\)[/tex].

Given the options:
A. [tex]\(x+5\)[/tex]
B. [tex]\(x+7\)[/tex]
C. [tex]\(x-7\)[/tex]
D. [tex]\(x-5\)[/tex]

The correct answer is:
B. [tex]\(x + 7\)[/tex]