Sample Answers with Examiner Commentary

Question 2(c) — Sample Answers

Grade A response. "For 2x² − 3x − 2 = 0, I use the quadratic formula with a = 2, b = −3, c = −2. The discriminant is b² − 4ac = (−3)² − 4(2)(−2) = 9 + 16 = 25. So x = [−(−3) ± √25] / (2 × 2) = (3 ± 5)/4. This gives x = (3 + 5)/4 = 8/4 = 2, or x = (3 − 5)/4 = −2/4 = −½. So the solutions are x = 2 and x = −½."

Mark: 4/4. Examiner commentary: A model solution. The student identifies the coefficients correctly — including the crucial sign of c = −2 — and computes the discriminant accurately, avoiding the very common sign error in −4(2)(−2) = +16. The formula is applied cleanly and both solutions are given exactly. Showing the discriminant as a separate step makes the work easy to follow and protects the method marks.

Grade C response. "a = 2, b = −3, c = −2. x = [3 ± √(9 − 16)] / 4. But 9 − 16 = −7, and you can't square root a negative, so there are no solutions."

Mark: 1/4. Examiner commentary: The student sets up the formula correctly and earns one method mark, but makes the classic sign error in the discriminant: −4ac = −4(2)(−2) should be +16, not −16, because c is negative. This gives 9 + 16 = 25, not 9 − 16 = −7, so there are in fact two real solutions. Taking extra care with the signs when substituting a negative value of c — writing −4 × 2 × (−2) = +16 explicitly — would have avoided the error and led to the correct answers x = 2 and x = −½.

Question 3(d) — Sample Answers

Grade A response. "To show (x − 1) is a factor, I use the Factor Theorem: if (x − 1) is a factor, then f(1) = 0. Substituting x = 1 into x³ + 2x² − 5x + 2 gives f(1) = 1 + 2 − 5 + 2 = 0. Since f(1) = 0, by the Factor Theorem (x − 1) is a factor of the polynomial. This also tells me that x = 1 is a root (or zero) of the polynomial — the value of the polynomial is zero there, so the graph of y = f(x) crosses the x-axis at x = 1."

Mark: 3/3. Examiner commentary: A complete answer. The student states the Factor Theorem, substitutes x = 1 correctly, shows the arithmetic equals zero, and draws the correct conclusion. Crucially, the final part connects the algebra to its meaning — that x = 1 is a root and the graph crosses the x-axis there — which is exactly what the 'explain what this tells you' part of the question is testing. This link between the Factor Theorem and the zeros of a polynomial is a key Common Core idea.

Grade C response. "f(1) = 1 + 2 − 5 + 2 = 0, so (x − 1) is a factor."

Mark: 2/3. Examiner commentary: The student correctly substitutes x = 1, shows the result is zero, and concludes that (x − 1) is a factor, earning two marks. The final mark is lost because the question also asks what this tells you about x = 1, and that part is not answered. The point is that x = 1 is a root/zero of the polynomial — where its value is zero and its graph meets the x-axis. Adding that interpretation would complete the answer.

Question 6 — Sample Answers

Grade A response. "Let the width be w meters. The length is 3 meters longer, so the length is (w + 3). The area of a rectangle is length × width, so the equation is w(w + 3) = 40. Expanding gives w² + 3w = 40, so w² + 3w − 40 = 0. This factors as (w + 8)(w − 5) = 0, giving w = −8 or w = 5. A width cannot be negative, so w = 5. Therefore the width is 5 m and the length is w + 3 = 8 m."

Mark: 8/8. Examiner commentary: A complete solution that models the situation correctly and solves it fully. The student defines the variable, forms the correct equation from the words, rearranges into standard form, factors accurately, and — crucially — rejects the negative solution with a reason ('a width cannot be negative') before stating both dimensions. This attention to the context of the problem, not just the algebra, is exactly what the 'creating equations' Common Core standard is assessing.

Grade C response. "w × (w + 3) = 40. So w² + 3w − 40 = 0. Using the factors (w + 8)(w − 5), w = 8 or w = 5. So the width could be 8 m or 5 m."

Mark: 6/8. Examiner commentary: The student sets up the equation correctly, rearranges it, and factors it accurately, which earns most of the marks. Two errors cost the rest: from (w + 8)(w − 5) = 0 the solution is w = −8 (not +8) or w = 5, so the sign is wrong; and the student then keeps both values as possible widths instead of rejecting the negative one. In a real-world problem, a length cannot be negative, so only w = 5 is valid, giving width 5 m and length 8 m. Reading the solutions off the factors with the correct signs, and checking them against the context, would gain full marks.