Synthetic Division for Polynomial Factors

Why Use Synthetic Division?

When a polynomial is divided by a linear expression \(x-a\), ordinary polynomial long division works, but it can involve several lines of algebra. Synthetic division organizes the same coefficient calculations into a compact table. It is useful for finding both the quotient and the remainder while making the arithmetic easy to check.

The method applies directly to division by a monic linear factor: the coefficient of \(x\) in the divisor must be \(1\), as in \(x-3\) or \(x+2\). The number placed into the synthetic-division calculation is \(a\), where the divisor is written as \(x-a\). In particular, the divisor \(x+2\) has \(a=-2\).

The Coefficient Procedure

Write the polynomial in descending powers, from \(x^n\) down to the constant term. Record a coefficient for every power, inserting \(0\) wherever a power is missing. Write \(a\) at the left of the coefficient row. Then bring down the leading coefficient. Repeatedly multiply the most recent bottom-row entry by \(a\), place the product under the next coefficient, and add.

All bottom-row entries except the last are the coefficients of the quotient, in descending powers. The last entry is the remainder. Thus, if the entries are \(b_{n-1}, b_{n-2},\ldots,b_0,r\), then \(Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_0\).

Worked Example: A Cubic with Zero Remainder

Why the Procedure Works

Synthetic division is not a shortcut based on a guess. Its steps are the coefficient equations required by the division identity. To see this, write a degree-\(n\) polynomial as \(P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\), where \(c_n\ne 0\). Suppose its quotient has the form \(Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+b_0\).

Multiplying \(Q(x)\) by \(x-a\) shifts each quotient term up by one power when multiplied by \(x\), and subtracts \(a\) times each quotient term. Therefore, matching coefficients from highest degree downward gives the following equations:

The first equation says to bring down the leading coefficient. Each later equation rearranges to “next quotient coefficient equals current dividend coefficient plus \(a\) times the preceding quotient coefficient.” That is exactly the multiply-and-add step in the table. After the coefficient of \(x\) has been matched, the constant term left over is \(c_0+ab_0\). It is the remainder \(r\). This proves that the recurrence used in synthetic division produces \(P(x)=(x-a)Q(x)+r\).

Worked Example: Include a Missing Power

Worked Example: A Nonzero Remainder

Remainder, Factors, and a Common Sign Error

The remainder is not an extra quotient coefficient. It is the constant \(r\) in the identity \(P(x)=(x-a)Q(x)+r\). If \(r=0\), the identity says \(P(x)=(x-a)Q(x)\), so \(x-a\) is a factor of \(P(x)\). This is consistent with the Factor Theorem: the remainder is also \(P(a)\). If \(r\ne 0\), the divisor is not a factor.

A frequent error is entering the opposite sign from the one required. For division by \(x+2\), using \(2\) instead of \(-2\) is incorrect. For the polynomial \(x^2+1\), the mistaken calculation would be:

This calculation uses \(a=2\), so it divides by \(x-2\), not \(x+2\). The correction is to use \(a=-2\): bring down \(1\), multiply by \(-2\), and add to the next coefficient, giving \(-2\); multiply \(-2\) by \(-2\), giving \(4\), and add to \(1\), giving \(5\). Thus the correct quotient is \(x-2\) and remainder \(5\). The check confirms it:

For a reliable calculation, write the divisor as \(x-a\) before beginning, include every coefficient position, and verify the final identity. Synthetic division is designed for divisors of the form \(x-a\); when the linear divisor has a leading coefficient other than \(1\), first account for that coefficient rather than applying this table unchanged.