Using Polynomial Synthetic Division When the Divisor is in the Wrong Form

Description/Explanation/Highlights

Video Description

This video explains how to adjust a polynomial division problem to use synthetic division when the coefficient of the variable in the divisor is not 1 and is not in the form of \(x-r\)

Steps and Key Points to Remember

To adjust the divisor and use synthetic division when the coefficient in the divisor is not equal to one, 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 \(x-1\) would be in the correct form, as would \(x+1\) although in the later example realize that the value of \(r\) is -1. \(3x+1\) would not qualify for synthetic division as the coefficient of x is not equal to 1.
  3. Divisors that have a variable with a coefficient other than one but otherwise fit the required \(x-r\) form for a divisor can be adjusted to allow synthetic division to be used.
  4. To use synthetic division in this case, every term of the dividend and divisor must be divided by the coefficient of x in the divisor. For example, to adjust \((3x^4-5x^3+x^2+7x)\div(3x+1)\) to allow synthetic division, divide each term by 3 which is the coefficient of x in the divisor as follows. \((\frac{3x^4}{3}-\frac{5x^3}{3}+\frac{x^2}{3}+\frac{7x}{3})\div(\frac{3x}{3}+\frac{1}{3})\)
  5. Simplify the problem and do synthetic division as usual since the coefficient of x in the divisor is now equal to one. \((x^4-\frac{5x^3}{3}+\frac{x^2}{3}+\frac{7x}{3})\div(x+\frac{1}{3})\)
  6. Remember, to begin a polynomial division problem using synthetic division, start by verifying that the divisor is now in the correct form and find the value of \(r\). In the above example the divisor, \(x+\frac{1}{3}\) is in the form of \(x-r\) and the value of \(r\) is \(-\frac{1}{3}\).
  7. Write the value of \(r\) and the coefficients in front of the variables as follows. If a variable is missing, such as the constant in the above example, put a zero (0) in its place. \(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{8 pt}0\\\)
  8. 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.\(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{8 pt}0\\\hspace{27 pt}\underline{0\hspace{62 pt}}\\\hspace{27 pt}1\\\)
  9. 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. \(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{8 pt}0\\\hspace{27 pt}\underline{0\hspace{8 pt}\text{–}\frac{1}{3}\hspace{54 pt}}\\\hspace{27 pt}1\hspace{9 pt}\text{–}2\\\)
  10. 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. \(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{8 pt}0\\\hspace{27 pt}\underline{0\hspace{8 pt}\text{–}\frac{1}{3}\hspace{8 pt}\frac{2}{3}\hspace{54 pt}}\\\hspace{27 pt}1\hspace{9 pt}\text{–}2\hspace{10 pt}1\\\)
  11. Repeat for the remaining columns until all are complete. \(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{11 pt}0\\\hspace{27 pt}\underline{0\hspace{8 pt}\text{–}\frac{1}{3}\hspace{8 pt}\frac{2}{3}\hspace{3 pt}\text{–}\frac{1}{3}\hspace{5 pt}\text{–}\frac{2}{3}}\\\hspace{27 pt}1\hspace{9 pt}\text{–}2\hspace{10 pt}1\hspace{10 pt}2\hspace{4 pt}|\underline{\text{–}\frac{2}{3}}\\\)
  12. 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.  \(\underline{\text{–}\frac{1}{3}\hspace{4 pt}}|\hspace{8 pt}1\hspace{8 pt} \text{–}\frac{5}{3}\hspace{8 pt} \frac{1}{3}\hspace{8 pt} \frac{7}{3}\hspace{11 pt}0\\\hspace{27 pt}\underline{0\hspace{8 pt}\text{–}\frac{1}{3}\hspace{8 pt}\frac{2}{3}\hspace{3 pt}\text{–}\frac{1}{3}\hspace{5 pt}\text{–}\frac{2}{3}}\\\hspace{27 pt}1\hspace{9 pt}\text{–}2\hspace{10 pt}1\hspace{10 pt}2\hspace{4 pt}|\underline{\text{–}\frac{2}{3}}\\\hspace{26 pt}x^3\hspace{7 pt}x^2\hspace{7 pt}x\hspace{11 pt}c\hspace{6 pt}\text{remainder}\\\)
  13. Before writing the final answer, simplify the fraction that will make up the remainder. Remember write the remainder over the divisor. In this case it is the original divisor that we divided by 3 to get in the correct form for synthetic division \(-\frac{\textstyle\frac{2}{3}}{\textstyle\frac{(3x+1)}{3}}\) By flipping the bottom fraction and multiplying, this simplifies to \(-\frac{\textstyle2}{\textstyle3x+1}\\\)
  14. 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. In the example presented here, the final polynomial would be: \(x^3-2x^2+x+2-\frac{2}{3x+1}\)

Here are some key points to keep in mind when using synthetic division to divide polynomials that are not in the correct form.

  • 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.
  • Problems that have an x coefficient of the divisor other than one can often be adjusted to be in the correct form of \(x-r\) that will allow synthetic division to be used
  • To adjust the problem, divide every term in the dividend and divisor by the coefficient of x.
  • If fractions become too complex, it may be easier to do long division instead of adjusting the problem.
  • After adjusting to get the problem in the right form, do synthetic division in the normal way.
  • To divide using synthetic division, first find the new value of the \(r\) constant after adjustment taking care to use the opposite sign when looking at the binomial divisor.
  • 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 the remainder as the numerator and the original divisor as the denominator.
  • When fractional coefficients are used, the fraction may have to be simplified to avoid complex fractions.

Video Highlights

  • 00:00 Introduction
  • 00:15 \((3x^4-5x^3+x^2+7x)\div(3x+1)\) example of a polynomial that must be adjusted to meet the criteria for synthetic division and the steps to solve
  • 04:08 Conclusion
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