Synthetic Division is one of three common techniques for dividing one polynomial by another. The other two are long division and the box method.
Of the three methods, students generally prefer synthetic division. It has the advantage of having minimal writing and being strictly algorithmic: Add, multiply, add, multiply….
First, let’s see how it works. Second, let’s see why it works.
How It Works
Let’s say we want to evaluate \[\frac{2x^3-5x^2-6x+9}{x-3}\]
We would set up to synthetic division for this as shown:
The number to the left, 3, is the value that makes the denominator (\[(x-3)\]) equal to zero. The other numbers are the coefficients of the numerator polynomial.
We bring the first coefficient down:
We then multiply that coefficient by the value to the right and place it on the second row:
From here, we add the column, then multiply by the value to the right, repeating until the chart is filled:
The number in the box to the left is the remainder. If this remainder is zero, then the denominator is a factor of the numerator.
The other numbers are the coefficients of the resulting quotient polynomial. That is, in this case: \[\frac{2x^3-5x^2-6x+9}{x-3}=2x^2-x-3\]
Synthetic division can also be used to evaluate functions at specific values. For instance, assume \(f(x)=2x^3-5x^2-6x+9\) and we want to know \(f(5)\). Synthetic division gives us:
which tells us that \(f(5)=104\). Of course, we could evaluate more directly by calculating \(f(5)=2(5)^3-5(5)^2-6(5)+9=2(125)+5(25)-6(5)+9=250-125-30+9=104\), but this can be more prone to student error. Also, having two different methods for evaluating a function is useful for verifying work.
Finally, using synthetic division to evaluate functions means that, when a root is found, the factor polynomial has also been found.
Why It Works
Synthetic division relies on an opaque algorithm: We strip the variable, perform prescribed steps, and then reintroduce the variable. Let’s look at what’s really happening under the hood.
Here is an example with parametric variables instead of numeric constants: \[\frac{ax^3+bx^2+cx+d}{x-n}\]
The corresponding synthetic division is set up like this:
and performed like this:
Given \(f(x)=ax^3+bx^2+cx+d\), this yields \(f(n)=((an+b)n+c)n+d\)\(=(an+b)n^2+cn+d=an^3+bn^2+cn+d\). This is \(f(x)\) evaluated when \(x=n\).
But why are the other values the coefficients of the quotient polynomial? What this is saying is:
\[\frac{ax^3+bx^2+cx+d}{x-n}=ax^2+(an+b)x+(an^2+bn+c)+\frac{an^3+bn^2+cn+d}{x-n}\]
Multiplying all but the last term of the right hand side by \(\frac{x-n}{x-n}\) yields: \[\frac{(ax^2+(an+b)x+(an^2+bn+c))(x-n)}{x-n}+\frac{an^3+bn^2+cn+d}{x-n}\\=\frac{ax^3+(an+b)x^2+(an^2+bn+c)x-anx^2-(an+b)nx-(an^2+bn+c)n}{x-n}\\+\frac{an^3+bn^2+cn+d}{x-n}\\=\frac{ax^3+anx^2+bx^2+an^2x+bnx+cx-anx^2-an^2x-bnx-an^3-bn^2-cn}{x-n}\\+\frac{an^3+bn^2+cn+d}{x-n}\\=\frac{ax^3+bx^2+cx-an^3-bn^2-cn}{x-n}+\frac{an^3+bn^2+cn+d}{x-n}\\=\frac{ax^3+bx^2+cx-an^3-bn^2-cn+an^3+bn^2+cn+d}{x-n}\\=\frac{ax^3+bx^2+cx+d}{x-n}\]
This is what we started with.
Extension (for Advanced Readers)
We could use the results of this to avoid (direct use of) synthetic division. For instance, if \(r\) is a root of \(f(x)=ax^3+bx^2+cx+d\), then \(\frac{f(x)}{x-r}=ax^2+(ar+b)x+ar^2+br+c\). Even more generally, for any value \(n\), \[\frac{f(x)}{x-n}=ax^2+(an+b)x+an^2+bn+c+\frac{f(n)}{x-n}\]
Breaking this down into steps: \[f(x)=ax^3+bx^2+cx+d\\A=a\\B=an+b\\C=an^2+bn+c\\\frac{f(x)}{x-n}=Ax^2+Bx+C+\frac{f(n)}{x-n}\]
This can be generalized for any degree polynomial. Given \(g(x)=ax^4+bx^3+cx^2+dx+f\): \[\frac{g(x)}{x-n}=ax^3+(an+b)x^2+(an^2+bn+c)x+an^3+bn^2+cn+d+\frac{g(n)}{x-n}\]
This looks daunting, but the pattern uses the coefficients of the dividend polynomial to create lower degree polynomials using the same coefficients. In the quartic example, \(a\) is constant, \(an+b\) is linear, \(an^2+bn+c\) is quadratic, and \(an^3+bn^2+cn+d\) is cubic. The general rule, then, is: Given a nth degree polynomial dividend with a divisor of the form \(x-n\), use the coefficients of the dividend to build polynomials of every degree from one to \(n-1\).
Hic Sunt Dracones
Synthetic division has the limitation that it only works well when dividing by a linear expression. There are ways to use it for higher degree divisors, but those get very complicated quickly. In those cases, it is definitely better to use either long division or the box method.
But is there a direct route to dividing a quartic by a quadratic? There is.
Let’s set up the box method. We’re going to find the quotient \[\frac{a_1x^4+a_2x^3+a_3x^2+a_4x+a_5}{b_1x^2+b_2x+b_3}\]
To use the box method for division, write the divisor on the side of the box. The goal is to fill in the cells so that they add up to the dividend. The quotient will be on the top of the box: \[\begin{array}{c|c|c|c|} & & & \\\hline b_1x^2 & & & \\\hline b_2x & & & \\\hline b_3 & & & \\\hline \end{array}\]
The upper left cell will be the highest degree term: \[\begin{array}{c|c|c|c|} & & & \\\hline b_1x^2 & a_1x^4 & & \\\hline b_2x & & & \\\hline b_3 & & & \\\hline \end{array}\]
We then divide by the term on the left and put that on the top: \[\begin{array}{c|c|c|c|} & \frac{a_1}{b_1}x^2 & & \\\hline b_1x^2 & a_1x^4 & & \\\hline b_2x & & & \\\hline b_3 & & & \\\hline \end{array}\]
Complete the column by multiplying each term on the left by the term on the top: \[\begin{array}{c|c|c|c|} & \frac{a_1}{b_1}x^2 & & \\\hline b_1x^2 & a_1x^4 & & \\\hline b_2x & \frac{a_1b_2}{b_1}x^3 & & \\\hline b_3 & \frac{a_1b_3}{b_1}x^2 & & \\\hline \end{array}\]
We need the cubic terms to add to \(a_2x^3\), so the missing term will be \(A_2x^3=\frac{a_2b_1-a_1b_2}{b_1}x^3\). This yields: \[\begin{array}{c|c|c|c|} & \frac{a_1}{b_1}x^2 & & \\\hline b_1x^2 & a_1x^4 & A_2x^3 & \\\hline b_2x & \frac{a_1b_2}{b_1}x^3 & & \\\hline b_3 & \frac{a_1b_3}{b_1}x^2 & & \\\hline \end{array}\]
Completing the rest of this column yields: \[\begin{array}{c|c|c|c|} & \frac{a_1}{b_1}x^2 & \frac{A_2}{b_1}x & \\\hline b_1x^2 & a_1x^4 & A_2x^3 & \\\hline b_2x & \frac{a_1b_2}{b_1}x^3 & \frac{A_2b_2}{b_1}x^2 & \\\hline b_3 & \frac{a_1b_3}{b_1}x^2 & \frac{A_2b_3}{b_1}x & \\\hline \end{array}\]
We need the quadratic terms to add to \(a_3x^2\), so the missing term will be \(A_3x^2=\frac{a_3b_1-a_1b_3-A_2b_2}{b_1}x^2\). Completing the last column yields: \[\begin{array}{c|c|c|c|} & \frac{a_1}{b_1}x^2 & \frac{A_2}{b_1}x & \frac{A_3}{b_1} \\\hline b_1x^2 & a_1x^4 & A_2x^3 & A_3x^2 \\\hline b_2x & \frac{a_1b_2}{b_1}x^3 & \frac{A_2b_2}{b_1}x^2 & \frac{A_3b_2}{b_1}x \\\hline b_3 & \frac{a_1b_3}{b_1}x^2 & \frac{A_2b_3}{b_1}x & \frac{A_3b_3}{b_1} \\\hline \end{array}\]
The remainder for this will be \(A_4x+A_5\), where \(A_4=\frac{a_4b_1-A_2b_3-A_3b_2}{b_1}\) and \(A_5=\frac{a_5b_1-A_3b_3}{b_1}\). That is, \[\frac{a_1x^4+a_2x^3+a_3x^2+a_4x+a_5}{b_1x^2+b_2x+b_3}=\frac{a_1}{b_1}x^2 + \frac{A_2}{b_1}x + \frac{A_3}{b_1}+ \frac{A_4x+A_5}{b_1x^2+b_2x+b_3}\\=\frac{a_1}{b_1}x^2 + \frac{a_2b_1-a_1b_2}{b_1^2}x + \frac{a_1(b_2^2-b_1b_3)-a_2b_1b_2+a_3b_1^2}{b_1^3}\\+ \frac{(a_1b_2(2b_1b_3-b_2^2)+a_2b_1(b_2^2-b_1b_3)-a_3b_1^2b_2+a_4b_1^3)x}{b_1^3(b_1x^2+b_2x+b_3)}\\+ \frac{a_1b_3(b_1b_3-b_2^2)+a_2b_1b_2b_3+a_3b_1^2b_3+a_5b_1^3}{b_1^3(b_1x^2+b_2x+b_3)}\]