Apply the remainder theorem and the factor theorem to find remainders, test for factors, and factorise a cubic polynomial

What this dot point is asking

SEAB wants you to use two closely linked results, the remainder theorem and the factor theorem, to work with polynomials quickly. The remainder theorem finds the remainder of a division by substituting a single value; the factor theorem tells you whether a linear expression is a factor. Together they let you factorise a cubic without slogging through long division every time. The key idea is that dividing a polynomial by $(x - a)$ and substituting $x = a$ are two faces of the same thing.

The answer

The remainder theorem

When a polynomial $f(x)$ is divided by $(x - a)$, the remainder is simply the value $f(a)$:

So instead of dividing, you substitute $x = a$. If the divisor is $(x + a)$, rewrite it as $(x - (-a))$ and use $f(-a)$. For a divisor $(bx - a)$, the remainder is $f\!\left(\dfrac{a}{b}\right)$.

The factor theorem

The factor theorem is the special case where the remainder is zero:

If substituting $x = a$ gives zero, the division is exact, so $(x - a)$ divides $f(x)$ with no remainder. This is the standard tool for finding a first factor of a cubic.

Factorising a cubic

To factorise a cubic $f(x)$ fully:

1. Find one factor by trying small values ($x = 1, -1, 2, -2, \ldots$, usually factors of the constant term) until $f(a) = 0$. Then $(x - a)$ is a factor.
2. Divide $f(x)$ by $(x - a)$ to get a quadratic, using long division or comparing coefficients.
3. Factorise the quadratic by the usual methods to get the remaining factors.

Finding an unknown coefficient

If a polynomial contains an unknown and you are told a remainder or a factor, set up an equation. "$(x - a)$ is a factor" gives $f(a) = 0$; "the remainder on division by $(x - a)$ is $r$" gives $f(a) = r$. Solve for the unknown.

Examples in context

Example 1. Solving a cubic equation. To solve $x^3 - 7x + 6 = 0$, factorise using the factor theorem to get $(x - 1)(x + 3)(x - 2) = 0$, so $x = 1, -3$ or $2$. The factor theorem is the standard route from a cubic equation to its three solutions.

Example 2. Preparing for partial fractions. Before splitting an algebraic fraction into partial fractions, the denominator must be factorised. The factor theorem provides the factors of a cubic denominator, which is why this topic feeds directly into the partial fractions work.

Try this

Q1. Find the remainder when $x^3 + x - 5$ is divided by $(x - 1)$. [2 marks]

• Cue. Remainder $= f(1) = 1 + 1 - 5 = -3$.

Q2. Show that $(x + 2)$ is a factor of $x^3 + 3x^2 - 4$. [2 marks]

• Cue. $f(-2) = -8 + 12 - 4 = 0$, so $(x + 2)$ is a factor.

Q3. Given that $(x - 3)$ is a factor of $x^3 - 2x^2 + kx - 3$, find $k$. [3 marks]

• Cue. $f(3) = 27 - 18 + 3k - 3 = 0$, so $3k + 6 = 0$, giving $k = -2$.