Problem 30
Question
25-38 . Find the quotient and remainder using synthetic division. $$ \frac{3 x^{3}-12 x^{2}-9 x+1}{x-5} $$
Step-by-Step Solution
Verified Answer
Quotient is \(3x^2 + 3x + 6\), and the remainder is \(31\).
1Step 1: Setup for Synthetic Division
Identify the coefficients of the dividend polynomial \(3x^3 - 12x^2 - 9x + 1\). The coefficients are \(3, -12, -9,\) and \(1\). The divisor is \(x - 5\); hence, use \(5\) as the number to the left of the grid.
2Step 2: Write Down the Initial Coefficient
Carry down the first coefficient (\(3\)) of the dividend polynomial as it is to the bottom row of your synthetic division setup.
3Step 3: Perform Synthetic Division
1. Multiply the carried down number by \(5\) (the divisor root) and write it under the next coefficient, i.e., \(-12+15=3\).2. Add the numbers in this column and repeat until done: - Multiply 3 by 5, add result to next coefficient: \(-9 + 15 = 6\). - Multiply 6 by 5, add result to next coefficient: \(1 + 30 = 31\).
4Step 4: Identify Quotient and Remainder
The last number is the remainder. The numbers at the bottom make up the coefficients of the quotient polynomial, \(3x^2 + 3x + 6\), with \(31\) being the remainder.
5Step 5: Write the Division Result
The quotient is \(3x^2 + 3x + 6\) and the remainder is \(31\). Therefore, the division results in: \(3x^2 + 3x + 6 + \frac{31}{x-5}\).
Key Concepts
Algebraic DivisionPolynomial Long DivisionRemainder Theorem
Algebraic Division
Algebraic division is a method used to divide a polynomial by another polynomial. It can be likened to long division with numbers, but instead, it involves polynomials. This process helps in simplifying expressions or solving higher-order polynomial equations by breaking them down into simpler terms.
This technique is foundational, and understanding it is key to mastering more complex algebraic operations, especially when dealing with polynomial expressions.
- The divisor in algebraic division is usually of a lower degree than the dividend for easier calculation.
- The main goal of algebraic division is to obtain a quotient and a remainder.
This technique is foundational, and understanding it is key to mastering more complex algebraic operations, especially when dealing with polynomial expressions.
Polynomial Long Division
Polynomial long division is very similar to the long division approach applied when dividing two numerical values. However, here, instead of numbers, polynomials are involved.
The process involves:
This remainder can be left as a fraction where it is divided by the original divisor. In our example exercise, polynomial long division could break down the equation manually for full understanding, if synthetic division isn't utilized. But overall, it shares the end goal - obtaining the quotient and remainder.
- Dividing the highest degree term of the dividend by the highest degree term of the divisor. The result becomes the first term of the quotient.
- Multiplying the entire divisor by this new term of the quotient to get an intermediate polynomial, which is then subtracted from the original polynomial.
- Repeating the process with the new, smaller dividend until a term lesser than the divisor's degree is reached.
This remainder can be left as a fraction where it is divided by the original divisor. In our example exercise, polynomial long division could break down the equation manually for full understanding, if synthetic division isn't utilized. But overall, it shares the end goal - obtaining the quotient and remainder.
Remainder Theorem
The Remainder Theorem is an important and helpful principle when dealing with polynomials and division. It states that if a polynomial, say \( P(x) \), is divided by \( x-c \), the remainder of such division is \( P(c) \).
In simpler terms, it means that if you want to find the remainder of a polynomial divided by a linear binomial, you simply substitute the root of the divisor into the polynomial.
In our problem, the remainder is given as \(31\), which matches the calculation from synthetic division. This theorem simplifies checks and can save time in larger computations.
In simpler terms, it means that if you want to find the remainder of a polynomial divided by a linear binomial, you simply substitute the root of the divisor into the polynomial.
- For the given polynomial \(3x^3 - 12x^2 - 9x + 1\) and division by \(x-5\), the root would be \(c=5\).
- Substitute \(5\) into the polynomial to compute \(3(5)^3 - 12(5)^2 - 9(5) + 1\).
- The result of this calculation tells you precisely what the remainder is without performing the actual division.
In our problem, the remainder is given as \(31\), which matches the calculation from synthetic division. This theorem simplifies checks and can save time in larger computations.
Other exercises in this chapter
Problem 30
Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=2 x^{4}-x^{3}-19 x^{2}+9 x+9 $$
View solution Problem 30
Find all horizontal and vertical asymptotes (if any). $$ r(x)=\frac{5 x^{3}}{x^{3}+2 x^{2}+5 x} $$
View solution Problem 30
Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(Q(x)=x^{4}+10 x^{2}+25\)
View solution Problem 30
\(27-40\) Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$ P(x)=-2 x^{3}-x^{2}+x $$
View solution