Synthetic Division

Synthetic division is a shorthand method to find the quotient and remainder when dividing a polynomial by a monic linear binomial \((\)a polynomial of the form \(x-k).\)

\[\frac{x^3-3x^2+5x+6}{x+2} = x^2-5x+15 -\frac{24}{x+2} \\ \]

This process is equivalent to polynomial division, but it requires much less writing. In addition to this application, synthetic division can be used to evaluate a polynomial function at a certain value.

Synthetic division can be used whenever you are dividing a polynomial by a monic linear binomial. A "monic linear binomial" is simply a polynomial of the form \(x-k.\) Examples of monic linear binomials are \(x+2,\) \(x-4,\) and \(x+\frac{4}{3}.\)

Perform \( (x^2 - 3x + 2) \div (x + 1) .\)

The top three numbers are the coefficients of the polynomial, that is, 1, -3, and 2 in order. Our "side term" written to the left is based dividing by \( x - k,\) so we write negative rather than positive 1.

Most of the last line is the coefficients of the quotient (note the quotient has a degree 1 less than the original polynomial), except the last number which is the remainder.

Therefore, \( (x^2 - 3x + 2) \div (x + 1) = x - 4 + \frac{6}{x+1} .\)

Here's the general procedure; we'll follow this with another example.

• (Step 1) Write the coefficients of the polynomial as written in standard form, in order. With the example \( x^2 + 2x + 6 ,\) the coefficients are 1, 2, and 6. If there is a "missing term," then one or more 0s must be used. For example, \( 5x^4 + 2x^2 - 5 \) has the coefficients 5, 0, 2, 0, and -5. \(\big(\)Formally, a "missing term" means the polynomial can be written with at least one term \( 0x^k, \) where \(k\) is a non-negative integer less than the degree of the polynomial.\(\big)\)

• (Step 2) Draw a vertical line and bar, as shown below.

• (Step 3) Given that we are dividing by \( x-k ,\) to the left, write \(k.\) \((\)Notice the subtraction; it means if we are dividing by \(x+1,\) i.e. \(k = -1.)\)

• (Step 4) We start on the far left of the polynomial coefficients and "add" the first coefficient to the number below it—since there's no number, we're "adding" 0, and then write the same number below the line.

• (Step 5) We take our result and multiply it by the value \(k\) to the left. We then right result diagonally up and to the right from the last position.

• (Step 6) We continue like step 4 and add the next column (this time there are two numbers to add).

• (Step 7) We continue like step 5, taking the result of our sum, multiplying by \(k,\) and writing the result diagonally up and to the right.

• (Step 8) The steps 6 and 7 continue until the last column is reached. The final number written will be our remainder. \((\)If the remainder is 0, that means \(x-k\) was a factor of our original polynomial.\()\) The other numbers on the bottom row represent the result of multiplication; they are the coefficients of the quotient's polynomial, and the degree of the quotient's polynomial is 1 less than the degree of the original polynomial.

Do the division \((3x^4 - 2x^2 + 3) \div (x - 2).\)

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Note that \( 3x^4 - 2x^2 + 3 = 3x^4 + 0x^3 - 2x^2 + 0x + 3 ,\) and we need the "missing terms" when writing out the synthetic division. We are dividing by \( x - k = x - 2, \) so \( k = 2:\)

This means \( \frac{3x^4 - 2x^2 + 3}{x-2} = 3x^3 + 6x^2 + 10x + 20 + \frac{43}{x-2} .\) \(_\square\)

If you are curious why the algorithm works, compare the regular long division algorithm with the synthetic. Below the numbers that serve the same function are marked. The synthetic division diagram simply collapses the operations that would normally happen when dividing by \( (x-k) \) into a format simpler to write.

Note that we'd often do long division by multiplying and then subtracting, but because we've already reversed the sign of our divisor \(\big(\)by using \( (x-k) \) instead of \( (x+k)\big),\) we've accounted for this; in other words, multiplying the second term by -1 and adding is the same as subtracting. \(\big(\)In fact, it'd be possible to change the synthetic division algorithm to use subtraction and have the side term be \( (x+k),\) but the examples above follow the traditional form of the algorithm.\(\big)\)

Main Article: Remainder Factor Theorem

By applying the remainder factor theorem, we can evaluate polynomial functions with synthetic division.