P2.2 - Algebra and functions

Learning objectives

P2.2.1Simple algebraic divisionSimple algebraic division; use of Only division by (ax + b) or (ax – b) will be required, e.g. the Factor Theorem and the b students should know that if f(x) = 0 when x =, Remainder Theorem. a then (ax – b) is a factor of f(x).; Students may be required to factorise cubic expressions such as x3 + 3x2 − 4 and 6x3 + 11x2 − x − 6.; Students should be familiar with the terms ‘quotient’ and ‘remainder’ and be able to determine the remainder when the polynomial f(x) is divided by (ax + b).

Dividing a polynomial $f(x)$ by a linear expression gives a quotient $Q(x)$ and constant remainder $R$ . One identity connects the calculation to both the Remainder and Factor Theorems.

$f(x)=(ax+b)Q(x)+R$

Divisor	Input that makes it zero	Remainder / factor test
$x-c$	$c$	remainder $f(c)$ ; factor iff $f(c)=0$
$ax-b$	$b/a$	remainder $f(b/a)$ ; factor iff $f(b/a)=0$
$ax+b$	$-b/a$	remainder $f(-b/a)$ ; factor iff $f(-b/a)=0$

For long division, order descending powers and insert zero coefficients for missing terms. Divide the leading terms, multiply the whole divisor by the new quotient term, subtract, and repeat until the remainder is constant. Verify with dividend = divisor × quotient + remainder.

For $f(x)=2x^3+x^2-7x+2$ , division by $x-2$ gives $f(x)=(x-2)(2x^2+5x+3)+8.$ The theorem check agrees: $f(2)=16+4-14+2=8$ , so $x-2$ is not a factor.

For $g(x)=2x^3-3x^2-8x+12$ , $g(2)=0$ , so $x-2$ is a factor. Division gives $2x^2+x-6=(2x-3)(x+2)$ ; hence $g(x)=(x-2)(2x-3)(x+2).$

Use the zero of the entire divisor: for $ax-b$ it is $b/a$ , not $b$ . A zero remainder proves a factor; a non-zero remainder does not. This P2 scope requires division only by linear expressions $ax\pm b$ .

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