2.1, 3.1 - Dividing polynomials

2.1, 3.1 - Dividing polynomials

Dividing a polynomial produces a quotient and, sometimes, something left over. In this shared part of 2.1 and 3.1, you will organise division by linear and quadratic expressions, including missing powers and non-unit leading coefficients. Multiplication will provide a complete check of every result.

Quotient and remainder

Can a polynomial division leave an expression containing xx, or must the remainder be a number? Compare ordinary division: 17=5(3)+217=5(3)+2, where the remainder is too small for another whole group of 55. For polynomials, “too small” refers to degree rather than numerical value.

Polynomial division

A polynomial is a finite sum of terms in non-negative integer powers of xx. Dividing a polynomial f(x)f(x) by a nonzero polynomial d(x)d(x) means finding a quotient q(x)q(x) and remainder r(x)r(x) such that

f(x)=d(x)q(x)+r(x).f(x)=d(x)q(x)+r(x).

The remainder is zero or has degree less than the divisor. The degree is the largest power with a nonzero coefficient.

A linear divisor leaves a constant remainder, possibly zero. A quadratic divisor can leave Ax+BAx+B, including a constant or zero. If the supposed remainder still has the divisor's degree, division has not finished.

Worked example 1: a nonzero remainder

Question. Find the quotient and remainder when x2+3x+5x^2+3x+5 is divided by x+1x+1.

Plan. Match the squared term first, then express the dividend as a multiple of the divisor plus a constant.

x2+3x+5=(x+1)x+(2x+5)=(x+1)x+2(x+1)+3=(x+1)(x+2)+3\begin{aligned} x^2+3x+5&=(x+1)x+(2x+5)\\ &=(x+1)x+2(x+1)+3\\ &=(x+1)(x+2)+3 \end{aligned}

Answer. Quotient x+2x+2, remainder 33.

Check. Multiplying (x+1)(x+2)(x+1)(x+2) gives x2+3x+2x^2+3x+2; adding 33 recovers the dividend.

Change only the constant term from 55 to 22. The quotient stays the same, but now nothing remains.

Worked example 2: exact division

Question. Find the quotient and remainder when x2+3x+2x^2+3x+2 is divided by x+1x+1.

Plan. Match the same leading terms and examine what is left.

x2+3x+2=(x+1)x+2x+2=(x+1)(x+2).x^2+3x+2=(x+1)x+2x+2=(x+1)(x+2).

Answer. Quotient x+2x+2, remainder 00. The divisor is a factor, since it divides exactly.

Check. The product expands to all three terms of the dividend, with no correction needed.

The identity is valid for all real xx. When writing a fraction, division additionally requires d(x)≠0d(x)\ne0: here

x2+3x+5x+1=x+2+3x+1,x≠−1.\frac{x^2+3x+5}{x+1}=x+2+\frac3{x+1},\qquad x\ne-1.

Watch out

The remainder is 33, not 3/(x+1)3/(x+1). The latter is the extra fraction in the division result; the polynomial identity uses +3+3.

Recognise both outputs from another identity.

A linear divisor

Long division repeats one decision: which multiple of the divisor cancels the highest remaining power? Divide leading terms, multiply the whole divisor, subtract the whole product, then repeat. Write zero coefficients for missing powers so that like terms stay aligned.

For example, 2x3+3x2+52x^3+3x^2+5 should be organised as 2x3+3x2+0x+52x^3+3x^2+0x+5. The zero is a place-holder, not a new term.

Worked example 3: a missing linear term

Question. Divide 2x3+3x2+52x^3+3x^2+5 by x+1x+1.

Plan. Cancel the cubic, quadratic and linear terms in turn, subtracting a complete multiple each time.

The first quotient term is 2x3/x=2x22x^3/x=2x^2:

(2x3+3x2+0x+5)−2x2(x+1)=x2+0x+5.(2x^3+3x^2+0x+5)-2x^2(x+1)=x^2+0x+5.

Next use x2/x=xx^2/x=x:

(x2+0x+5)−x(x+1)=−x+5.(x^2+0x+5)-x(x+1)=-x+5.

Finally use −x/x=−1-x/x=-1:

(−x+5)−(−1)(x+1)=6.(-x+5)-(-1)(x+1)=6.

Answer. Quotient 2x2+x−12x^2+x-1, remainder 66.

Check. (x+1)(2x2+x−1)+6=2x3+3x2+0x+5(x+1)(2x^2+x-1)+6=2x^3+3x^2+0x+5.

Keep the dividend but change the divisor's leading coefficient. Divide by that coefficient at every stage, even when a fraction appears.

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