WMA12/01 Jun 2024, Question 4: Cubic Roots after Division

Part (a) · 1 mark

Give the remainder obtained on dividing f(x) by (x − 2).

1. 1f(2) = (0)(…) + 21
2. 2Remainder = 21

Answer21

Part (b) · 2 marks

Using the fact that (2x − 1) divides f(x) exactly, demonstrate that k = 11.

1. 1(2x − 1) is a factor, so f(½) = 0
2. 2(½ − 2)(2(¼) + 5(½) + k) + 21 = 0
3. 3(−3/2)(3 + k) + 21 = 0
4. 43 + k = 14, so k = 11

AnswerShown: f(½) = (−3/2)(3 + k) + 21 = 0 gives 3 + k = 14, so k = 11

Part (ci) · 3 marks

Write f(x) as a product of linear factors in full.

1. 1f(x) = (x − 2)(2x² + 5x + 11) + 21
2. 2= 2x³ + 5x² + 11x − 4x² − 10x − 22 + 21 = 2x³ + x² + x − 1
3. 3Divide by (2x − 1): 2x³ + x² + x − 1 = (2x − 1)(x² + x + 1)

Answerf(x) = (2x − 1)(x² + x + 1)

Part (cii) · 2 marks

Determine how many real values satisfy f(x) = 0, justifying your count.

1. 1For x² + x + 1: b² − 4ac = 1² − 4(1)(1) = −3
2. 2−3 < 0, so x² + x + 1 = 0 has no real roots
3. 3Only 2x − 1 = 0 gives a root, so f(x) = 0 has one real solution, x = ½

AnswerOne real solution (x = ½), because x² + x + 1 = 0 has discriminant 1 − 4 = −3 < 0 so it has no real roots

More past-paper questions on cubic roots after division

• WMA12/01 Jan 2023 Q5 · 9 marks
• WMA12/01 Jun 2023 Q2 · 6 marks
• WMA12/01 Jun 2021 Q7 · 10 marks
• WMA12/01 Oct 2021 Q4 · 8 marks