When and How to Use Polynomial Synthetic Division

Description/Explanation/Highlights

Video Description

This video explains when synthetic division can be used as a shorter and easier way to perform polynomial division and the steps used to do division of polynomials using synthetic division.

Steps and Key Points to Remember

To determine if synthetic division is appropriate to solve a polynomial division problem and to use synthetic division to solve a polynomial division problem, follow these steps:

  1. Synthetic division can be used when the divisor is in the form of \(x-r\) where \(x\) is a variable with a coefficient of 1 and \(r\) is a constant.
  2. A divisor of \(x-4\) would be in the correct form, as would \(x+4\) although in the later example realize that the value of \(r\) is -4. \(3x+2\) would not qualify for synthetic division as the lead coefficient is not 1 and \(x-2x\) would not work because \(r\) must be a constant term and \(2x\) contains a variable.
  3. To begin a polynomial division problem using synthetic division, start by verifying that the divisor is in the correct form and find the value of \(r\). In the example: \((4x^4+2x^2-4x+12)\div(x+2)\), the divisor, \(x+2\) is in the form of \(x-r\) and the value of \(r\) is \(-2\).
  4. Write the value of \(r\) and the coefficients in front of the variables as follows. If a variable is missing, such as \(x^3\) in the above example, put a zero (0) in its place. \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{8 pt} 2\hspace{4 pt} \text{–}4\hspace{8 pt}12\)
  5. After writing out the \(r\) value and the coefficients, put a zero (or just leave blank) under the first coefficient, draw a line, and combine the first coefficient and 0 (or just bring down the coefficient if you left it blank underneath) and write as follows. \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{8 pt} 2\hspace{4 pt} \text{–}4\hspace{8 pt}12\\\hspace{25 pt}\underline{0\hspace{62 pt}}\\\hspace{25 pt}4\\\)
  6. Multiply the \(r\) value to the left by the number you got when you combined the first coefficient and 0 and place the answer under the next coefficient. Combine the coefficient above with this answer as you did in the first column. \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{8 pt} 2\hspace{4 pt} \text{–}4\hspace{8 pt}12\\\hspace{25 pt}\underline{0\hspace{3 pt}\text{–}8\hspace{46 pt}}\\\hspace{25 pt}4\hspace{3 pt}\text{–}8\\\)
  7. Repeat the process with the next column. Multiply the number you got combining the second column by the value of \(r\) and combine as before. \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{10 pt} 2\hspace{4 pt} \text{–}4\hspace{8 pt}12\\\hspace{25 pt}\underline{0\hspace{3 pt}\text{–}8\hspace{5 pt}16\hspace{32 pt}}\\\hspace{25 pt}4\hspace{3 pt}\text{–}8\hspace{5 pt}18\\\)
  8. Repeat for the remaining columns until all are complete. \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{10 pt} 2\hspace{8 pt} \text{–}4\hspace{8 pt}12\\\hspace{25 pt}\underline{0\hspace{3 pt}\text{–}8\hspace{5 pt}16\hspace{2 pt}\text{–}36\hspace{8 pt}80}\\\hspace{25 pt}4\hspace{3 pt}\text{–}8\hspace{5 pt}18\hspace{3 pt}\text{–}40\hspace{4 pt}|\underline{92}\\\)
  9. Starting from the right, the last number represents the remainder. The number immediately to the left is the constant. The next number is the coefficient of \(x\); the next is the coefficient of \(x^2\); etc.  \(\underline{\text{–}2\hspace{4 pt}}|\hspace{8 pt}4\hspace{8 pt} 0\hspace{10 pt} 2\hspace{8 pt} \text{–}4\hspace{8 pt}12\\\hspace{25 pt}\underline{0\hspace{3 pt}\text{–}8\hspace{5 pt}16\hspace{2 pt}\text{–}36\hspace{8 pt}80}\\\hspace{25 pt}4\hspace{3 pt}\text{–}8\hspace{5 pt}18\hspace{3 pt}\text{–}40\hspace{4 pt}|\underline{92}\\\hspace{23 pt}x^3\hspace{2 pt}x^2\hspace{7 pt}x\hspace{13 pt}c\hspace{6 pt}\text{remainder}\\\)
  10. To write the final answer, add the variables after the constants, taking care to keep the correct sign, and write the remainder as a fraction with the remainder as the numerator and the original divisor as the denominator. In the example presented here, the final polynomial would be: \(4x^3-8x^2+18x-40+\frac{92}{x+2}\)

Here are some key points to keep in mind when using synthetic division to divide polynomials.

  • Synthetic division is a faster and easier alternative to polynomial long division.
  • Synthetic division can only be used if the divisor is in the form of \(x-r\). For example, \(x-7\) or \(x+4\).
  • To meet the \(x-r\) form the coefficient of \(x\) must be 1 and \(r\) must be a constant and contain no variable.
  • To divide using synthetic division, first find the value of the \(r\) constant taking care to use the opposite sign when looking at the binomial divisor. For example, when looking at \((x-7)\) as a divisor the value of \(r\) is 7, not -7.
  • When writing out the coefficients of the polynomial, be sure to add zeros from any “missing” variables.
  • When the problem is complete, read the answer from right to left with the first number on the right representing the remainder; the next to the left, the constant; next is the value of \(x\); then \(x^2\); etc.
  • Write the final answer by adding the variables after the coefficients and the remainder as a fraction with remainder as the numerator and the original divisor as the denominator.

Video Highlights

  • 00:00 Introduction
  • 00:16 \((2x^3-13x^2+26x-24)\div(x-4)\) example of a polynomial that meets the criteria for synthetic division and the steps to solve
  • 03:47 \((4x^4+2x^2-4x+12)\div(x+2)\) synthetic division example with a missing \(x^3\) term
  • 07:05 Conclusion
  •  
  • To watch this video on YouTube in a new window with clickable highlights, click here

Related Videos

Polynomials Definitions, Naming and Terms
Finding and Using the Degree of a Polynomial
Adding and Subtracting Polynomials
Multiplying Polynomials
Dividing a Polynomial by a Monomial
Using Polynomial Long Division Without Remainders
Using Polynomial Long Division Resulting in Remainders
Using Polynomial Synthetic Division When the Divisor is in the Wrong Form
Return to topics menu

Sponsored Links