Problem 25 - Algebra and Trigonometry

Question

25-38 . Find the quotient and remainder using synthetic division. $$ \frac{x^{2}-5 x+4}{x-3} $$

Step-by-Step Solution

Answer

Quotient is \(x-2\) and remainder is \(-2\).

Step 1: Set up the synthetic division

To perform synthetic division, we write down the divisor, which in this case is \(x - 3\). Here, we only use the number 3, ignoring the sign. We then list the coefficients of the dividend \((x^2 - 5x + 4)\) as 1, -5, and 4. Place these numbers in a row: "3 | 1 -5 4".

Step 2: Bring down the leading coefficient

The first number below the line is always the same as the first number above the line. So, directly write down the 1 below the line, which is the leading coefficient of the dividend. This number represents the leading coefficient of the quotient.

Step 3: Multiply and add

Multiply the number you just brought down \((1)\) by the divisor \((3)\) and write the result \((3)\) under the next coefficient. Then add this result to the second coefficient \((-5)\) to get a new number \((-2)\) below the line.

Step 4: Repeat multiply and add process

Now, multiply the new number \((-2)\) by the divisor \((3)\) to get \((-6)\). Write this below the last coefficient \((4)\) and add them to get \((-2)\). This is the remainder of the division.

Step 5: Write the quotient

Using the numbers from the synthetic division, the quotient is represented by the coefficient line which is "1, -2". This translates to the polynomial \(x - 2\). The remainder is \(-2\). Hence, \(\frac{x^2 - 5x + 4}{x - 3} = x - 2\) with remainder \(-2\).

Key Concepts

Polynomial Division

When you're dividing polynomials, the process can be similar to long division, but with algebraic expressions. In the given example, we have \(\frac{x^2 - 5x + 4}{x - 3}\). This process lets us find out how many times the divisor \(x - 3\) fits into the dividend \(x^2 - 5x + 4\).

There are several methods to perform polynomial division:

• Long Division
• Synthetic Division

Here, we focus on synthetic division, which streamlines the process when you're dividing by a linear polynomial like \(x - 3\). It involves working with only the coefficients of the polynomial, which makes calculations faster and more straightforward.This approach is efficient because it reduces the complexity of operations by keeping track of only the essential parts of division.

By practicing synthetic division, you will enhance your ability to handle polynomial expressions more flexibly and efficiently.

Remainder Theorem

The Remainder Theorem connects division and evaluation in a fascinating way. If you have a polynomial \(f(x)\) and divide it by a linear divisor \(x - c\), the remainder you get is actually the same as the value of \(f(c)\).

In our exercise, we divided \(x^2 - 5x + 4\) by \(x - 3\) and found that the remainder was \(-2\). According to the Remainder Theorem:

• Take \(c = 3\) from \(x - 3\)
• Substitute it into the polynomial: \(f(3) = 3^2 - 5 \times 3 + 4\)
• Simplifying gives: \(9 - 15 + 4 = -2\)

Notice how the remainder matches \(f(3)\)? This theorem provides a quick check for your division results, ensuring you can verify outcomes rapidly by simply substituting values.

Quotient and Remainder

After dividing polynomials, what you get are the quotient and remainder. Understanding both is crucial to the division process because they provide the components of the original expression.

In synthetic division, as in our example:

• The numbers on the top row under the synthetic division bar become the coefficients of the quotient.
• Our coefficients “1, -2” correspond to the quotient polynomial \(x - 2\).
• The last number from the process is the remainder.

Therefore, in the division \(\frac{x^2 - 5x + 4}{x - 3}\), the quotient is \(x - 2\) and the remainder is \(-2\). This means you can express the relationship between dividend, quotient, divisor, and remainder as:\[x^2 - 5x + 4 = (x - 3)(x - 2) + (-2)\]
By grasping the concept of quotient and remainder, you're essentially mastering how polynomials decompose under division, which is foundational in many areas of algebra.