Polynomials — Worked Solutions (Preliminary Maths Extension 1)
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Worked examples for Preliminary Maths Extension 1 polynomials. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Extension 1 — Polynomials. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Remainder and factor theorem
Question
The polynomial $P(x) = x^3 + ax^2 - 5x + b$ has a factor of $(x - 1)$ and leaves a remainder of $-18$ when divided by $(x + 2)$. Find $a$ and $b$.
Solution
Two conditions, two unknowns. Use the factor theorem and the remainder theorem.
Factor $(x-1)$ means $P(1) = 0$: $1 + a - 5 + b = 0 \Rightarrow a + b = 4$.
Remainder $-18$ on division by $(x+2)$ means $P(-2) = -18$: $-8 + 4a + 10 + b = -18 \Rightarrow 4a + b = -20$.
Subtract the equations: $3a = -24 \Rightarrow a = -8$, then $b = 4 - a = 12$.
So $a = -8$, $b = 12$. Always evaluate $P$ at the right value — $(x+2)$ means substitute $x = -2$, not $+2$.
We have two pieces of information, so we'll build two equations. The factor theorem says that if $(x-1)$ is a factor, then substituting $x = 1$ makes the polynomial zero: $P(1) = 1 + a - 5 + b = 0$, which tidies to $a + b = 4$.
The remainder theorem says the remainder when dividing by $(x + 2)$ equals $P(-2)$ — the value that makes $x + 2$ zero. So $P(-2) = (-2)^3 + a(-2)^2 - 5(-2) + b = -8 + 4a + 10 + b = -18$, giving $4a + b = -20$.
Now we solve the pair. Subtracting the first from the second eliminates $b$: $3a = -24$, so $a = -8$, and then $b = 4 - (-8) = 12$.
The reason both theorems work is that division leaves $P(x) = (x - c)Q(x) + R$, so plugging in $x = c$ wipes out the quotient and leaves just the remainder.
Two unknowns → two equations.
- Factor: $P(1) = 0 \Rightarrow 1 + a - 5 + b = 0 \Rightarrow a + b = 4$
- Remainder: $P(-2) = -18 \Rightarrow -8 + 4a + 10 + b = -18 \Rightarrow 4a + b = -20$
- Subtract: $3a = -24 \Rightarrow a = -8$, $b = 12$
$a = -8$, $b = 12$.
Where the marks go
- 1 mark: Uses factor theorem: $P(1) = 0$ giving $a + b = 4$
- 1 mark: Uses remainder theorem: $P(-2) = -18$ giving $4a + b = -20$
- 1 mark: Solves the simultaneous equations
- 1 mark: Correct values $a = -8$ and $b = 12$
Key idea
Factor theorem: $(x-c)$ is a factor iff $P(c) = 0$. Remainder theorem: the remainder on division by $(x-c)$ is $P(c)$.
Example 2 — Roots and coefficients
Question
The cubic $2x^3 - 3x^2 + 4x - 1 = 0$ has roots $\alpha$, $\beta$ and $\gamma$. Find the value of $\alpha^2 + \beta^2 + \gamma^2$.
Solution
Use the relationships between roots and coefficients, then the algebraic identity.
For $2x^3 - 3x^2 + 4x - 1 = 0$: $\sum \alpha = \tfrac{3}{2}$ and $\sum \alpha\beta = \tfrac{4}{2} = 2$.
The identity is $\alpha^2 + \beta^2 + \gamma^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta$.
So $= \left(\tfrac{3}{2}\right)^2 - 2(2) = \tfrac{9}{4} - 4 = -\tfrac{7}{4}$.
Don't try to solve the cubic — the symmetric-function shortcut is the point of the question.
We never actually find the roots here. Instead we use the sum and product relationships that come straight from the coefficients. For $ax^3 + bx^2 + cx + d = 0$, the sum of roots is $-\tfrac{b}{a}$ and the sum of products in pairs is $\tfrac{c}{a}$.
Here $\sum \alpha = -\tfrac{-3}{2} = \tfrac{3}{2}$ and $\sum \alpha\beta = \tfrac{4}{2} = 2$.
Now the clever bit: squaring the sum of roots gives $\left(\sum \alpha\right)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2\sum \alpha\beta$. Rearranging gives exactly what we want: $\alpha^2 + \beta^2 + \gamma^2 = \left(\tfrac{3}{2}\right)^2 - 2(2) = \tfrac{9}{4} - 4 = -\tfrac{7}{4}$.
A negative answer is fine — these are squares of complex or irrational roots, not real lengths.
Roots–coefficients, then identity.
- $\sum \alpha = \tfrac{3}{2}$, $\sum \alpha\beta = 2$
- $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$
- $= \tfrac{9}{4} - 4 = -\tfrac{7}{4}$
Answer: $-\tfrac{7}{4}$.
Where the marks go
- 1 mark: Correct $\sum \alpha = \tfrac{3}{2}$
- 1 mark: Correct $\sum \alpha\beta = 2$ and uses the identity $(\sum\alpha)^2 - 2\sum\alpha\beta$
- 1 mark: Correct value $-\tfrac{7}{4}$
Key idea
Sums and products of roots come directly from the coefficients; $\alpha^2 + \beta^2 + \gamma^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$.
Frequently asked questions
Step-by-step solutions to Polynomials questions in Preliminary Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Extension 1, using the methods and notation expected in exams and assessments.