Worked Solutions
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.
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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; $\alpha^2 + \beta^2 + \gamma^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$.