A Cubic with Two Unknown Coefficients

Two unknowns need two equations. Each factor of a polynomial gives one equation via the factor theorem ($f(k) = 0$), so two factors give two simultaneous equations in the unknowns.

The polynomial $f(x) = x^3 + ax^2 + bx - 6$ has $(x - 1)$ and $(x + 2)$ as factors. Find the values of $a$ and $b$. [5]

The working

Step 1, apply the factor theorem to $(x - 1)$, so $f(1) = 0$: $f(1) = 1 + a + b - 6 = 0 \;\Rightarrow\; a + b = 5 \quad \text{(M1, A1)}$

Step 2, apply it to $(x + 2)$, so $f(-2) = 0$: $f(-2) = (-2)^3 + a(-2)^2 + b(-2) - 6 = -8 + 4a - 2b - 6 = 0 \;\Rightarrow\; 4a - 2b = 14 \;\Rightarrow\; 2a - b = 7 \quad \text{(M1, A1)}$

Step 3, solve simultaneously. Add $a + b = 5$ and $2a - b = 7$ (the $b$ cancels): $3a = 12 \;\Rightarrow\; a = 4, \qquad \text{then } b = 5 - 4 = 1 \quad \text{(A1)}$

So $a = 4$ and $b = 1$.

Where the marks are won and lost

• Two factors, two equations. Each factor $(x - k)$ gives $f(k) = 0$; use both, not just one.
• Signs in $f(-2)$: $(-2)^3 = -8$ and $a(-2)^2 = 4a$ (positive), $b(-2) = -2b$. A sign slip here corrupts the second equation.
• Solve the pair by elimination (adding cancels $b$ here), cleaner than substitution.

Common mistakes

• Using only one factor and trying to guess the second unknown.
• Sign errors evaluating $f(-2)$.
• Arithmetic slips solving the simultaneous equations.

Full method: Factor Theorem notes. Topic home: Factors of Polynomials pillar.