3A - Dividing polynomials

Can a cubic divided by a linear expression give an exact quadratic answer? You will divide polynomials by monic and non-monic linear expressions, keep track of missing powers and describe any remainder. Multiplying back will give you a reliable way to check every division.

Division by a monic linear factor

What must multiply $x-2$ to give $x^3+2x^2-5x-6$ ? Expanding a product is familiar; division reverses that process, one leading term at a time.

Polynomial division

A polynomial is a finite sum of terms with nonnegative integer powers of the variable. Its degree is its highest power with a nonzero coefficient. A monic polynomial has leading coefficient $1$ , so $x-2$ is monic but $2x-1$ is not.

The expression you divide by is the divisor. The polynomial answer is the quotient.

In long division, divide the leading term of what remains by the leading term of the divisor. Multiply the whole divisor by that new quotient term, subtract the whole product, and repeat. Each subtraction removes the current highest power.

Worked example 1: removing one power at a time

Question. Divide $x^3+2x^2-5x-6$ by $x-2$ .

Plan. Choose successive quotient terms that cancel the cubic, quadratic and linear terms.

First, $x^3\div x=x^2$ . Multiply and subtract:

\begin{aligned} (x^3+2x^2-5x-6)-(x^3-2x^2)&=4x^2-5x-6 \end{aligned}

Next, $4x^2\div x=4x$ :

(4x^2-5x-6)-(4x^2-8x)=3x-6.

Finally, $3x\div x=3$ :

(3x-6)-(3x-6)=0.

The quotient terms are $x^2$ , $4x$ and $3$ .

Answer. The quotient is $x^2+4x+3$ , with no remainder.

Check. Expand the proposed product:

(x-2)(x^2+4x+3)=x^3+2x^2-5x-6.

It reconstructs every coefficient of the original polynomial.

The next dividend has no quadratic term. The method stays the same, but writing its coefficient as zero keeps the powers aligned.

Worked example 2: a missing power

Question. Divide $x^3-7x+6$ by $x-2$ .

Plan. Write $x^3+0x^2-7x+6$ , then use the same long-division steps.

\begin{aligned} x^3\div x&=x^2\\ (x^3+0x^2-7x+6)-(x^3-2x^2)&=2x^2-7x+6\\[2pt] 2x^2\div x&=2x\\ (2x^2-7x+6)-(2x^2-4x)&=-3x+6\\[2pt] -3x\div x&=-3\\ (-3x+6)-(-3x+6)&=0 \end{aligned}

Answer. The quotient is $x^2+2x-3$ , with no remainder.

Check. Use coefficient matching as an alternative check. If the quotient is $Ax^2+Bx+C$ , then

(x-2)(Ax^2+Bx+C)=Ax^3+(B-2A)x^2+(C-2B)x-2C.

Matching coefficients gives $A=1$ , $B-2A=0$ , $C-2B=-7$ and $-2C=6$ . These give $A=1$ , $B=2$ , $C=-3$ , agreeing with the division.

Coefficient matching works because equal polynomials have equal coefficients of each power. It can find a quotient directly when a factor is known: write an unknown quadratic quotient, expand, then compare coefficients. Include the zero coefficients in this method too.

Watch out

Subtract the whole product. For instance,

(2x^2-7x+6)-(2x^2-4x)=-3x+6,

because subtracting $-4x$ adds $4x$ . Losing this sign changes every later step.

Keep a place for the missing quadratic term in your own division.

Division by a non-monic linear factor

When the divisor starts with $2x$ , divide by $2x$ , rather than by $x$ , to choose each quotient term.

For a linear divisor $ax+b$ , the coefficient $a$ must be nonzero. A cubic divided by a linear polynomial has a quadratic quotient: the leading powers subtract, since $x^3\div x=x^2$ .

Worked example 3: a leading coefficient of two

Question. Divide $2x^3+5x^2-7x+2$ by $2x-1$ .

Plan. Divide each leading term by $2x$ , then multiply by the whole divisor before subtracting.

\begin{aligned} 2x^3\div2x&=x^2\\ (2x^3+5x^2-7x+2)-x^2(2x-1)&=6x^2-7x+2\\[2pt] 6x^2\div2x&=3x\\ (6x^2-7x+2)-3x(2x-1)&=-4x+2\\[2pt] -4x\div2x&=-2\\ (-4x+2)-(-2)(2x-1)&=0 \end{aligned}

Answer. The quotient is $x^2+3x-2$ , with no remainder.

Check. $(2x-1)(x^2+3x-2)=2x^3+5x^2-7x+2$ , so multiplying back recovers the dividend.

Change the divisor from $2x-1$ to $2x+1$ . The leading-term decisions are unchanged, but every multiplication now uses a positive constant term.

Worked example 4: a positive constant in the divisor

Question. Divide $2x^3+7x^2-x-2$ by $2x+1$ .

Plan. Remove the leading powers in order, retaining the plus sign in $2x+1$ .

\begin{aligned} (2x^3+7x^2-x-2)-x^2(2x+1)&=6x^2-x-2\\ (6x^2-x-2)-3x(2x+1)&=-4x-2\\ (-4x-2)-(-2)(2x+1)&=0 \end{aligned}

The leading-term divisions are again $x^2$ , $3x$ and $-2$ .

Answer. The quotient is $x^2+3x-2$ , with no remainder.

Check. Expanding $(2x+1)(x^2+3x-2)$ gives $2x^3+7x^2-x-2$ . The same quotient can therefore occur with a different divisor and dividend.

The divisor's leading coefficient matters even when the final coefficients happen to be integers.

Watch out

Here $2x^3\div2x=x^2$ , not $2x^2$ . Check your first quotient term by multiplying it by the divisor's leading term.

Try the same process with a leading coefficient of three.

Quotient and remainder

Division need not end at zero: a constant left after dividing by a linear polynomial is the remainder.

For example, $17=5\times3+2$ records an ordinary quotient and remainder. Polynomial division records the same structure. The remainder must have lower degree than the divisor, so a nonzero remainder for a linear divisor is a constant. Stop there; a constant cannot be removed by another polynomial quotient term.

Division identity

For a polynomial $f(x)$ divided by $ax+b$ , where $a\ne0$ ,

f(x)=(ax+b)q(x)+r,

where $q(x)$ is the quotient and $r$ is a constant remainder.

Not on the formulae sheet: learn it.

The identity holds for every $x$ . Dividing both sides by $ax+b$ also gives $f(x)/(ax+b)=q(x)+r/(ax+b)$ , but that fraction statement excludes the value where $ax+b=0$ .

Worked example 5: a positive remainder

Question. Find the quotient and remainder when $2x^3+3x^2-5x+4$ is divided by $x+2$ .

Plan. Divide until only a constant remains, then record the division identity.

\begin{aligned} (2x^3+3x^2-5x+4)-2x^2(x+2)&=-x^2-5x+4\\ (-x^2-5x+4)-(-x)(x+2)&=-3x+4\\ (-3x+4)-(-3)(x+2)&=10 \end{aligned}

Answer. The quotient is $2x^2-x-3$ and the remainder is $10$ :

2x^3+3x^2-5x+4=(x+2)(2x^2-x-3)+10.

Check. The product expands to $2x^3+3x^2-5x-6$ . Adding $10$ gives the original constant $4$ , with all other coefficients unchanged.

Now change only the dividend's constant term. All quotient steps remain the same until the final subtraction, but the remainder can be negative.

Worked example 6: a negative remainder

Question. Find the quotient and remainder when $2x^3+3x^2-5x-9$ is divided by $x+2$ .

Plan. Repeat the leading-term cancellations and retain the sign of the final constant.

\begin{aligned} (2x^3+3x^2-5x-9)-2x^2(x+2)&=-x^2-5x-9\\ (-x^2-5x-9)-(-x)(x+2)&=-3x-9\\ (-3x-9)-(-3)(x+2)&=-3 \end{aligned}

Answer. The quotient is $2x^2-x-3$ and the remainder is $-3$ :

2x^3+3x^2-5x-9=(x+2)(2x^2-x-3)-3.

Check. Expanding the right side gives $2x^3+3x^2-5x-6-3$ , exactly the dividend.

The opening cubic divided exactly by $x-2$ . These examples show how to state an answer just as precisely when division leaves something over.

Key idea

Check a polynomial division by reconstructing the dividend: divisor times quotient, then add the remainder. A remainder of zero means the divisor is a factor.

Use the identity to make your final answer unambiguous.

Quick Recap

Divide leading terms, multiply the whole divisor, subtract and repeat.
Write zero coefficients for missing powers.
For $ax+b$ , divide leading terms by $ax$ .
Reconstruct with $f(x)=(ax+b)q(x)+r$ .

The opening division gives the exact quadratic $x^2+4x+3$ ; a nonzero constant left over would be recorded separately as a remainder.

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3A - Dividing polynomials

Division by a monic linear factor

Division by a non-monic linear factor

Quotient and remainder

Quick Recap

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