Hint
Try factors with leading terms \(3x\) and \(2x\). Cancellation does not remove a restriction from the original denominator.
Worked solution
(a)
Split \(-x=8x-9x\). Then \(6x^2+8x-9x-12=2x(3x+4)-3(3x+4)=(3x+4)(2x-3)\).
Answer: \((3x+4)(2x-3)\)
(b)
Set each factor equal to zero: \(3x+4=0\) or \(2x-3=0\).
Answer: \(\displaystyle x=-\frac43\) or \(\displaystyle x=\frac32\).
(c)
The original denominator requires \(\displaystyle x\ne - \frac{4}{3}\). For allowed values, the equation becomes \(2x-3=1\), so \(x=2\), which is allowed.
Answer: \(\displaystyle x\ne-\frac43\); the solution is \(x=2\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Use a valid middle-term split or consistent pair of linear factors.
- 1 mark: Obtain \((3x+4)(2x-3)\).
Alternative methods: Any correct factorisation method, including finding roots then checking expansion.
Common error: An incorrect sign in a constant factor.
(b) · 2 marks
- 1 mark: Apply the zero-product rule to both factors.
- 1 mark: State both roots \(\displaystyle - \frac{4}{3}, \frac{3}{2}\).
Alternative methods: Quadratic formula with both correct roots also earns full credit.
Common error: Dividing by one factor and losing its root.
(c) · 3 marks
- 1 mark: Exclude \(\displaystyle x= - \frac{4}{3}\).
- 1 mark: Reduce to \(2x-3=1\) on the permitted domain.
- 1 mark: Obtain and accept \(x=2\) only.
Alternative methods: Multiplication followed by factorisation is valid if the excluded root is rejected.
Common error: Accepting the excluded denominator root.
Total: 7 marks.