Synthetic Division by a Linear Factor

Why Synthetic Division Is Useful

Synthetic division is a compact method for dividing a polynomial by a linear polynomial. It organizes the coefficient arithmetic into a short table, so it can be quicker than polynomial long division when the divisor has degree one.

The method is especially useful when an improper rational function has a linear denominator. Dividing its numerator by that denominator separates the rational function into a polynomial part and a remainder term. This is the quotient-remainder step needed before further rational-function work.

Synthetic division is not a shortcut for division by a polynomial of degree two or higher. It applies directly to a linear divisor, and the divisor’s leading coefficient must be handled correctly.

The Quotient and Remainder for a Linear Divisor

Let \(P(x)\) be a polynomial, and let \(ax+b\) be a linear polynomial with \(a\ne 0\). The quotient-remainder identity states that there are unique polynomials \(Q(x)\) and \(R(x)\) such that

Since the only polynomial of degree less than one is a constant (including zero), the remainder \(R(x)\) is a number. The zero of the divisor is \(r=-b/a\). Substituting \(x=r\) into the identity makes the divisor term vanish, giving \(R=P(r)\).

In particular, for a monic divisor \(x-r\), the synthetic quotient can be used directly. For a nonmonic divisor \(a(x-r)\), the quotient must be scaled by \(1/a\). The remainder does not need that scaling.

How the Synthetic-Division Table Works

Write the coefficients of \(P(x)\) from highest power to constant term. If a power is missing, its coefficient must be entered as zero. Then bring down the first coefficient. Repeatedly multiply the latest table entry by \(r\), write the product beneath the next coefficient, and add.

Worked Example: Divide by a Monic Linear Factor

Why the Algorithm Gives the Quotient

The multiplication-and-addition pattern is not an unrelated trick. It constructs coefficients for a polynomial \(Q(x)\) so that \(P(x)=(x-r)Q(x)+R\). Here is a direct derivation for a cubic.

Write \(P(x)=c_3x^3+c_2x^2+c_1x+c_0\), and suppose \(Q(x)=q_2x^2+q_1x+q_0\). Expanding the proposed identity gives

For this to equal \(P(x)\), corresponding coefficients must match. Therefore \(q_2=c_3\), \(q_1=c_2+rq_2\), \(q_0=c_1+rq_1\), and \(R=c_0+rq_0\). These are exactly the successive bring-down, multiply-by-\(r\), and add steps in synthetic division.

The same coefficient matching works at every degree: each new quotient coefficient is the next original coefficient plus \(r\) times the preceding quotient coefficient, and the final such addition gives the remainder. For divisor \(ax+b=a(x-r)\), the identity becomes \(P(x)=a(x-r)Q(x)+R\). The synthetic quotient \(Q_s\) satisfies \(P(x)=(x-r)Q_s(x)+R\); consequently \(Q=Q_s/a\), while \(R\) stays the same.

Worked Example: A Missing Power and a Nonmonic Divisor

Using Synthetic Division with an Improper Rational Function

A rational function is improper when the degree of its numerator is at least the degree of its denominator. For a linear denominator, synthetic division gives a polynomial quotient and a constant remainder. Dividing the quotient-remainder identity by the denominator separates the rational function into a polynomial plus a proper rational term.

Worked Example: Rewrite an Improper Rational Function

Common Mistake: Forgetting to Scale the Quotient

A frequent error is to use the correct root for a nonmonic divisor but then report the synthetic quotient without adjustment. For example, dividing \(6x^3+x^2-8x+4\) by \(3x+6\) uses \(r=-2\). Synthetic division gives coefficients \(6,-11,14,-24\), so the synthetic quotient is \(6x^2-11x+14\) and the remainder is \(-24\). Reporting \(6x^2-11x+14\) as the quotient for \(3x+6\) is wrong.

The concrete failure is visible by checking the proposed identity: \((3x+6)(6x^2-11x+14)-24\) has leading term \(18x^3\), not \(6x^3\). The correction is to divide the synthetic quotient by \(3\). The correct identity is

Indeed, the product \((3x+6)(2x^2-\frac{11}{3}x+\frac{14}{3})\) equals \((x+2)(6x^2-11x+14)=6x^3+x^2-8x+28\), and subtracting \(24\) gives the original polynomial. The remainder is \(-24\) for division by either \(x+2\) or \(3x+6\), because both divisors have the same zero, \(-2\). Their quotients differ by the factor \(1/3\).

When to Use the Method

Synthetic division is most efficient when the divisor is linear and the coefficients are manageable. Before starting, check the divisor’s zero and the degree sequence of the dividend. Afterward, check the identity by multiplying the divisor by the quotient and adding the remainder. Evaluating the dividend at the divisor’s zero supplies an independent check of the remainder.

These checks catch the two most common errors: leaving out a zero coefficient for a missing power, and forgetting to scale the quotient when the divisor is not monic. Neither error is fixed by merely obtaining a plausible-looking table; the quotient-remainder identity must hold exactly.