<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Synthetic Division of Polynomials

## Concise method of dividing polynomials when the divisor is x minus a constant.

0%
Progress
Practice Synthetic Division of Polynomials
Progress
0%
Synthetic Division of Polynomials

The volume of a rectangular prism is $2x^3 + 5x^2 - x - 6$ . Determine if $2x + 3$ is the length of one of the prism's sides.

### Guidance

Synthetic division is an alternative to long division from the previous concept. It can also be used to divide a polynomial by a possible factor, $x - k$ . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

#### Example A

Divide $2x^4-5x^3-14x^2-37x-30$ by $x - 2$ .

Solution: Using synthetic division, the setup is as follows:

Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is $2x^3-x^2-16x+15$ . Notice that when we synthetically divide by $k$ , the “leftover” polynomial is one degree less than the original. We could also write $(x-2)(2x^3-x^2-16x+15)=2x^4-5x^3-14x^2+47x-30$ .

#### Example B

Determine if 4 is a solution to $f(x)=5x^3+6x^2-24x-16$ .

Using synthetic division, we have:

The remainder is 304, so 4 is not a solution. Notice if we substitute in $x = 4$ , also written $f(4)$ , we would have $f(4)=5(4)^3+6(4)^2-24(4)-16=304$ . This leads us to the Remainder Theorem.

Remainder Theorem: If $f(k) = r$ , then $r$ is also the remainder when dividing by $(x - k)$ .

This means that if you substitute in $x = k$ or divide by $k$ , what comes out of $f(x)$ is the same. $r$ is the remainder, but also is the corresponding $y-$ value. Therefore, the point $(k, r)$ would be on the graph of $f(x)$ .

#### Example C

Determine if $(2x - 5)$ is a factor of $4x^4-9x^2-100$ .

Solution: If you use synthetic division, the factor is not in the form $(x - k)$ . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put $\frac{5}{2}$ up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this example, we need zeros for the $x^3-$ term and the $x-$ term.

This means that $\frac{5}{2}$ is a zero and its corresponding binomial, $(2x - 5)$ , is a factor.

Intro Problem Revisit

If $2x + 3$ divides evenly into $2x^3 + 5x^2 - x - 6$ then it is the length of one of the prism's sides.

If we use synthetic division, the factor is not in the form $(x - k)$ . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put $\frac{-3}{2}$ up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of 0. This means that $(2x + 3)$ is a factor of the volume. Therefore, it is also the length of one of the sides of the rectangular prism.

### Guided Practice

1. Divide $x^3-9x^2+12x-27$ by $(x + 3)$ . Write the resulting polynomial with the remainder (if there is one).

2. Divide $2x^4-11x^3+12x^2+9x-2$ by $(2x + 1)$ . Write the resulting polynomial with the remainder (if there is one).

3. Is 6 a solution for $f(x)=x^3-8x^2+72$ ? If so, find the real-number zeros (solutions) of the resulting polynomial.

1. Using synthetic division, divide by -3.

The answer is $x^2+6x-6-\frac{9}{x+3}$ .

2. Using synthetic division, divide by $-\frac{1}{2}$ .

The answer is $2x^3-12x^2+18x-\frac{2}{2x+1}$ .

3. Put a zero placeholder for the $x-$ term. Divide by 6.

The resulting polynomial is $x^2-2x-12$ . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

$x=\frac{2 \pm \sqrt{2^2-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}$

The solutions to this polynomial are 6, $1+\sqrt{13} \approx 4.61$ and $1-\sqrt{13} \approx -2.61$ .

### Vocabulary

Synthetic Division
An alternative to long division for dividing $f(x)$ by $k$ where only the coefficients of $f(x)$ are used.
Remainder Theorem
If $f(k) = r$ , then $r$ is also the remainder when dividing by $(x - k)$ .

### Practice

Use synthetic division to divide the following polynomials. Write out the remaining polynomial.

1. $(x^3+6x^2+7x+10) \div (x+2)$
2. $(4x^3-15x^2-120x-128) \div (x-8)$
3. $(4x^2-5) \div (2x+1)$
4. $(2x^4-15x^3-30x^2-20x+42) \div (x+9)$
5. $(x^3-3x^2-11x+5) \div (x-5)$
6. $(3x^5+4x^3-x-2) \div (x-1)$
7. Which of the division problems above generate no remainder? What does that mean?
8. What is the difference between a zero and a factor?
9. Find $f(-2)$ if $f(x)=2x^4-5x^3-10x^2+21x-4$ .
10. Now, divide $2x^4-5x^3-10x^2+21x-4$ by $(x + 2)$ synthetically. What do you notice?

Find all real zeros of the following polynomials, given one zero.

1. $12x^3+76x^2+107x-20; -4$
2. $x^3-5x^2-2x+10; -2$
3. $6x^3-17x^2+11x-2; 2$

Find all real zeros of the following polynomials, given two zeros.

1. $x^4+7x^3+6x^2-32x-32; -4, -1$
2. $6x^4+19x^3+11x^2-6x; 0, -2$

### Vocabulary Language: English

Oblique Asymptote

Oblique Asymptote

An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes

Oblique Asymptotes

An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Remainder Theorem

Remainder Theorem

The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division

Synthetic Division

Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.