Year 10A Polynomials, Factor and Remainder Theorems Worksheets & Practice

Worked examples

Year 10A polynomials and factor theorem: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Remainder theorem

   Find the remainder when $P(x) = 2x^3 - 5x^2 + x + 3$ is divided by $(x - 2)$.

   Show worked solution

   Key idea: The remainder theorem states that when $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.

   1. The divisor is $(x - 2)$, so $a = 2$.
   2. Substitute $x = 2$ into $P(x)$:
      $P(2) = 2(2)^3 - 5(2)^2 + (2) + 3$
   3. Evaluate each term:
      $\begin{aligned} P(2) &= 2(8) - 5(4) + 2 + 3 \\ &= 16 - 20 + 2 + 3 \end{aligned}$
   4. Simplify:
      $P(2) = 1$

   Answer: The remainder is $1$.

2. Example 2 · Factor theorem and full factorisation of a cubic

   Show that $(x + 1)$ is a factor of $P(x) = x^3 + 2x^2 - x - 2$, then fully factorise $P(x)$.

   Show worked solution

   Key idea: By the factor theorem, $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. Once one factor is found, use polynomial long division to find the remaining quadratic, then factorise it.

   1. Test $x = -1$ (since the factor is $(x + 1)$, so $a = -1$):
      $P(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0$ Since $P(-1) = 0$, $(x + 1)$ is a factor.
   2. Divide $P(x)$ by $(x + 1)$ using long division:
      $\begin{aligned} x^3 + 2x^2 - x - 2 &\div (x + 1) \\ x^3 + 2x^2 - x - 2 &= (x+1)(x^2 + x - 2) \end{aligned}$
   3. Factorise the quadratic $x^2 + x - 2$:
      $x^2 + x - 2 = (x + 2)(x - 1)$
   4. Write the full factorisation:
      $P(x) = (x + 1)(x + 2)(x - 1)$

   Answer: $P(x) = (x+1)(x+2)(x-1)$

3. Example 3 · Solving a cubic equation in context

   A storage company designs an open-top rectangular box by cutting squares of side length $x$ cm from each corner of a 12 cm by 18 cm rectangular sheet of cardboard, then folding up the sides. The volume of the box is given by $V(x) = x(12 - 2x)(18 - 2x)$. The company needs a volume of 160 cm³. Show that this requirement leads to the equation $x^3 - 15x^2 + 54x - 40 = 0$, then find all valid values of $x$.

   Show worked solution

   Key idea: Expand the volume expression and rearrange to form a cubic equation equal to zero, then use the factor theorem and long division to fully factorise and solve.

   1. Expand $V(x) = x(12 - 2x)(18 - 2x)$ and set it equal to 160:
      $\begin{aligned} (12 - 2x)(18 - 2x) &= 216 - 24x - 36x + 4x^2 = 4x^2 - 60x + 216 \\ V(x) &= x(4x^2 - 60x + 216) = 4x^3 - 60x^2 + 216x \end{aligned}$ Setting $V(x) = 160$: $4x^3 - 60x^2 + 216x = 160$ Divide everything by 4: $x^3 - 15x^2 + 54x - 40 = 0$ ✓
   2. Test integer factors of 40. Try $x = 1$:
      $1 - 15 + 54 - 40 = 0 \checkmark$ So $(x - 1)$ is a factor.
   3. Divide $x^3 - 15x^2 + 54x - 40$ by $(x - 1)$:
      $x^3 - 15x^2 + 54x - 40 = (x - 1)(x^2 - 14x + 40)$
   4. Factorise $x^2 - 14x + 40 = (x - 4)(x - 10)$, so:
      $(x-1)(x-4)(x-10) = 0 \implies x = 1, \; x = 4, \; x = 10$
   5. Check validity: the shorter side of the sheet is 12 cm, so the cut must satisfy $2x < 12$, meaning $x < 6$. Also $x > 0$. So $x = 10$ is rejected.

   Answer: The valid values are $x = 1$ cm and $x = 4$ cm.

Common mistakes

Where students usually lose marks

Try one yourself

Your turn

Got a different polynomials and factor theorem question wrong? Upload it to Tuterly and get the worked solution and a fresh one to practise, free.