For a linear divisor or factor \(ax\pm b\), assume \(a\ne0\). Long division of polynomials works exactly like long division of numbers: divide, multiply, subtract, bring down, repeat. Every division statement can be rewritten as a multiplication identity. The division algorithm FormNumbersPolynomials Division identity\(23=4\times 5+3\)\(P(x)=D(x)\,Q(x)+R(x)\) Fraction form\(\dfrac{23}{4}=5+\dfrac{3}{4}\)\(\dfrac{P(x)}{D(x)}=Q(x)+\dfrac{R(x)}{D(x)}\) Size of the remainder\(3<4\)\(R(x)=0\) or \(\deg R<\deg D\) Here \(P(x)\) is the dividend, \(D(x)\) the divisor, \(Q(x)\) the quotient and \(R(x)\) the remainder. Dividing \(x^3+2x^2-5\) by \(x-1\) gives quotient \(x^2+3x+3\) and remainder \(-2\), so \(x^3+2x^2-5=(x-1)(x^2+3x+3)-2\). Method: long division Arrange the dividend and the divisor in descending powers of \(x\). Insert any missing term with a zero coefficient, e.g. write \(x^3+2x^2+0x-5\). Divide the leading terms, multiply the divisor by the result, subtract. Repeat until nothing remains or the remaining polynomial has degree less than the divisor; this is the remainder. Remainder theorem: the remainder without dividing DivisorRemainderExample: \(f(x)=2x^3-x+3\) \(x-c\)\(f(c)\)Divided by \(x-2\): \(f(2)=16-2+3\) \(=17\) \(ax-b\)\(f\!\left(\dfrac{b}{a}\right)\)Divided by \(2x+1\): \(f\!\left(-\dfrac12\right)=-\dfrac14+\dfrac12+3\) \(=3\dfrac14\) The remainder theorem gives the remainder only. When a question also needs the quotient, use long division. Exam check Set the divisor equal to zero to find the value to substitute: dividing by \(2x+1\) means substituting \(x=-\dfrac12\), sign included. The divisor \(D\) must be a nonzero polynomial. The division identity holds for every \(x\); its fractional form is used only where \(D(x)\ne0\).

Long Division and the Remainder Theorem

Secondary 3

TGM Original Questions

For a linear divisor or factor \(ax\pm b\), assume \(a\ne0\).

Long division of polynomials works exactly like long division of numbers: divide, multiply, subtract, bring down, repeat. Every division statement can be rewritten as a multiplication identity.

The division algorithm

Form	Numbers	Polynomials
Division identity	\(23=4\times 5+3\)	\(P(x)=D(x)\,Q(x)+R(x)\)
Fraction form	\(\dfrac{23}{4}=5+\dfrac{3}{4}\)	\(\dfrac{P(x)}{D(x)}=Q(x)+\dfrac{R(x)}{D(x)}\)
Size of the remainder	\(3<4\)	\(R(x)=0\) or \(\deg R<\deg D\)

Here \(P(x)\) is the dividend, \(D(x)\) the divisor, \(Q(x)\) the quotient and \(R(x)\) the remainder. Dividing \(x^3+2x^2-5\) by \(x-1\) gives quotient \(x^2+3x+3\) and remainder \(-2\), so \(x^3+2x^2-5=(x-1)(x^2+3x+3)-2\).

Method: long division

Arrange the dividend and the divisor in descending powers of \(x\).
Insert any missing term with a zero coefficient, e.g. write \(x^3+2x^2+0x-5\).
Divide the leading terms, multiply the divisor by the result, subtract.
Repeat until nothing remains or the remaining polynomial has degree less than the divisor; this is the remainder.

Remainder theorem: the remainder without dividing

Divisor	Remainder	Example: \(f(x)=2x^3-x+3\)
\(x-c\)	\(f(c)\)	Divided by \(x-2\): \(f(2)=16-2+3\) \(=17\)
\(ax-b\)	\(f\!\left(\dfrac{b}{a}\right)\)	Divided by \(2x+1\): \(f\!\left(-\dfrac12\right)=-\dfrac14+\dfrac12+3\) \(=3\dfrac14\)

The remainder theorem gives the remainder only. When a question also needs the quotient, use long division.

Exam check

Set the divisor equal to zero to find the value to substitute: dividing by \(2x+1\) means substituting \(x=-\dfrac12\), sign included.

The divisor \(D\) must be a nonzero polynomial. The division identity holds for every \(x\); its fractional form is used only where \(D(x)\ne0\).

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