For a linear divisor or factor \(ax\pm b\), assume \(a\ne0\). For the product \((ax-b)(cx-d)\), assume \(c\ne0\) and \(ad-bc\ne0\), so the two roots are distinct. If the roots coincide, divisibility by the repeated factor needs an additional multiplicity condition. A factor is a divisor that leaves remainder zero. Combining this with the remainder theorem gives a quick test for factors — and a strategy for factorising cubics with a known or discoverable root. Factor theorem StatementTest \(x-c\) is a factor of \(f(x)\)\(f(c)=0\), and vice versa \(ax-b\) is a factor of \(f(x)\)\(f\!\left(\dfrac{b}{a}\right)=0\), and vice versa \((ax-b)(cx-d)\) is a factor of \(f(x)\)\(ax-b\) and \(cx-d\) are each factors of \(f(x)\) Sum and difference of cubes \(a^3+b^3=(a+b)(a^2-ab+b^2)\) and \(a^3-b^3=(a-b)(a^2+ab+b^2)\) Method: solving a cubic equation Arrange the equation as \(f(x)=0\). Use a given root first. For an integer-coefficient polynomial, if the constant term is zero, first factor out \(x\). Otherwise any rational root in lowest terms is \(p/q\), where \(p\) divides the constant term and \(q\) divides the leading coefficient. Test these candidates; if \(f(k)=0\), then \(x-k\) is a factor. A cubic need not have a rational root, so this trial method is not guaranteed to find one. Write \(f(x)=(x-k)(ax^2+bx+c)\) and find the quadratic factor by comparing coefficients or by long division. Solve the quadratic factor: factorise, or use the formula or completing the square if it does not factorise. What the quadratic factor can give Quadratic factorWorkingRoots of the cubic Factorises\((x-1)(x^2-x-6)=0\) \((x-1)(x-3)(x+2)=0\)\(x=1,\ x=3,\ x=-2\) Does not factorise using integers, \(b^2-4ac>0\)\((x+2)(x^2-2x-1)=0\) \(x=-2\) or \(x=\dfrac{2\pm\sqrt{8}}{2}\)\(x=-2,\ x=1\pm\sqrt{2}\) \(b^2-4ac<0\)\((x-2)(x^2+x+5)=0\) \(1-4(1)(5)=-19<0\)\(x=2\) only Exam check When factorising over the real numbers, "factorise completely" means no factor can be broken down further: stop at linear factors and any quadratic factor with \(b^2-4ac<0\).

The Factor Theorem and Cubic Equations

Secondary 3

TGM Original Questions

For a linear divisor or factor \(ax\pm b\), assume \(a\ne0\). For the product \((ax-b)(cx-d)\), assume \(c\ne0\) and \(ad-bc\ne0\), so the two roots are distinct. If the roots coincide, divisibility by the repeated factor needs an additional multiplicity condition.

A factor is a divisor that leaves remainder zero. Combining this with the remainder theorem gives a quick test for factors — and a strategy for factorising cubics with a known or discoverable root.

Factor theorem

Statement	Test
\(x-c\) is a factor of \(f(x)\)	\(f(c)=0\), and vice versa
\(ax-b\) is a factor of \(f(x)\)	\(f\!\left(\dfrac{b}{a}\right)=0\), and vice versa
\((ax-b)(cx-d)\) is a factor of \(f(x)\)	\(ax-b\) and \(cx-d\) are each factors of \(f(x)\)

Sum and difference of cubes

\(a^3+b^3=(a+b)(a^2-ab+b^2)\) and \(a^3-b^3=(a-b)(a^2+ab+b^2)\)

Method: solving a cubic equation

Arrange the equation as \(f(x)=0\).
Use a given root first. For an integer-coefficient polynomial, if the constant term is zero, first factor out \(x\). Otherwise any rational root in lowest terms is \(p/q\), where \(p\) divides the constant term and \(q\) divides the leading coefficient. Test these candidates; if \(f(k)=0\), then \(x-k\) is a factor. A cubic need not have a rational root, so this trial method is not guaranteed to find one.
Write \(f(x)=(x-k)(ax^2+bx+c)\) and find the quadratic factor by comparing coefficients or by long division.
Solve the quadratic factor: factorise, or use the formula or completing the square if it does not factorise.

What the quadratic factor can give

Quadratic factor	Working	Roots of the cubic
Factorises	\((x-1)(x^2-x-6)=0\) \((x-1)(x-3)(x+2)=0\)	\(x=1,\ x=3,\ x=-2\)
Does not factorise using integers, \(b^2-4ac>0\)	\((x+2)(x^2-2x-1)=0\) \(x=-2\) or \(x=\dfrac{2\pm\sqrt{8}}{2}\)	\(x=-2,\ x=1\pm\sqrt{2}\)
\(b^2-4ac<0\)	\((x-2)(x^2+x+5)=0\) \(1-4(1)(5)=-19<0\)	\(x=2\) only

Exam check

When factorising over the real numbers, "factorise completely" means no factor can be broken down further: stop at linear factors and any quadratic factor with \(b^2-4ac<0\).

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