Standard +0.3 This is a structured multi-part question on the Factor Theorem with guided steps. Part (a) involves routine verification and algebraic manipulation to find a repeated factor. Parts (b) and (c) apply the factorization through substitution (p² and cos θ), requiring understanding of domain restrictions but following a clear path. The question is slightly above average due to the multi-step nature and the need to count solutions in a specific interval, but the scaffolding and standard techniques keep it accessible.
6. \(\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 1\)
a. (i) Show that ( \(2 x - 1\) ) is a factor of \(\mathrm { f } ( x )\).
(ii) Express \(\mathrm { f } ( x )\) in the form \(( 2 x - 1 ) ( x + a ) ^ { 2 }\) where \(a\) is an integer.
Using the answer to part a) (ii)
b. show that the equation \(2 p ^ { 6 } + 3 p ^ { 4 } - 1\) has exactly two real solutions and state the values of these roots.
c. deduce the number of real solutions, for \(5 \pi \leq \theta \leq 8 \pi\), to the equation
$$2 \cos ^ { 3 } \theta + 3 \cos ^ { 2 } \theta - 1 = 0$$
Part (a)(i): Show \((2x-1)\) is a factor of \(f(x) = 2x^3+3x^2-1\)
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(f\!\left(\frac{1}{2}\right) = 2\!\left(\frac{1}{2}\right)^3 + 3\!\left(\frac{1}{2}\right)^2 - 1 = 0\), so \((2x-1)\) is a factor
B1
Correctly applies the factor theorem
Part (a)(ii): Express \(f(x)\) in the form \((2x-1)(x+a)^2\)
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Expand \((2x-1)(x^2+2ax+a^2)\) and equate coefficients; \((4a-1)=3 \Rightarrow a=1\)
M1
Attempts to find quadratic factor \((x^2 \pm ax \pm b)\) by long division or equating coefficients
Factorises to obtain \((x\pm a)^2\), e.g. \((x+1)^2\)
M1
Factorises 3-term quadratic to obtain \((x\pm a)^2\)
\((2x-1)(x+1)^2\)
A1
Correct answer only
(4 marks)
Part (b): Show \(2p^6+3p^4-1=0\) has exactly two real solutions
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Let \(x=p^2\): equation becomes \(2x^3+3x^2-1=0\); factorises as \((2x-1)(x+1)^2=0 \Rightarrow x=\frac{1}{2},\ x=-1\) (repeated)
M1
Uses part (ii) to factorise; gives partial explanation that \(p^2=-1\) has no real roots
\(x=\frac{1}{2} \Rightarrow p^2=\frac{1}{2} \Rightarrow p=\pm\frac{1}{\sqrt{2}}\); \(x=-1 \Rightarrow p^2=-1 \Rightarrow\) no real roots; therefore exactly two real solutions
A1
Complete proof that given equation has exactly two real solutions
(2 marks)
Part (c): Number of real solutions to \(2\cos^3\theta + 3\cos^2\theta - 1 = 0\) for \(5\pi \leq \theta \leq 8\pi\)
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Let \(y=\cos\theta \Rightarrow \cos\theta = -1\) or \(\cos\theta = \frac{1}{2}\); for \(5\pi \leq \theta \leq 8\pi\): \(\cos\theta=-1 \Rightarrow \theta=5\pi, 7\pi\); \(\cos\theta=\frac{1}{2} \Rightarrow \theta=\frac{17}{3}\pi, \frac{19}{3}\pi, \frac{23}{3}\pi\); therefore 5 solutions
B1
Deduces that there are 5 solutions
(1 mark)
## Question 6:
### Part (a)(i): Show $(2x-1)$ is a factor of $f(x) = 2x^3+3x^2-1$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(\frac{1}{2}\right) = 2\!\left(\frac{1}{2}\right)^3 + 3\!\left(\frac{1}{2}\right)^2 - 1 = 0$, so $(2x-1)$ is a factor | B1 | Correctly applies the factor theorem |
---
### Part (a)(ii): Express $f(x)$ in the form $(2x-1)(x+a)^2$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expand $(2x-1)(x^2+2ax+a^2)$ and equate coefficients; $(4a-1)=3 \Rightarrow a=1$ | M1 | Attempts to find quadratic factor $(x^2 \pm ax \pm b)$ by long division or equating coefficients |
| Factorises to obtain $(x\pm a)^2$, e.g. $(x+1)^2$ | M1 | Factorises 3-term quadratic to obtain $(x\pm a)^2$ |
| $(2x-1)(x+1)^2$ | A1 | Correct answer only |
**(4 marks)**
---
### Part (b): Show $2p^6+3p^4-1=0$ has exactly two real solutions
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $x=p^2$: equation becomes $2x^3+3x^2-1=0$; factorises as $(2x-1)(x+1)^2=0 \Rightarrow x=\frac{1}{2},\ x=-1$ (repeated) | M1 | Uses part (ii) to factorise; gives partial explanation that $p^2=-1$ has no real roots |
| $x=\frac{1}{2} \Rightarrow p^2=\frac{1}{2} \Rightarrow p=\pm\frac{1}{\sqrt{2}}$; $x=-1 \Rightarrow p^2=-1 \Rightarrow$ no real roots; therefore exactly two real solutions | A1 | Complete proof that given equation has exactly two real solutions |
**(2 marks)**
---
### Part (c): Number of real solutions to $2\cos^3\theta + 3\cos^2\theta - 1 = 0$ for $5\pi \leq \theta \leq 8\pi$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $y=\cos\theta \Rightarrow \cos\theta = -1$ or $\cos\theta = \frac{1}{2}$; for $5\pi \leq \theta \leq 8\pi$: $\cos\theta=-1 \Rightarrow \theta=5\pi, 7\pi$; $\cos\theta=\frac{1}{2} \Rightarrow \theta=\frac{17}{3}\pi, \frac{19}{3}\pi, \frac{23}{3}\pi$; therefore **5 solutions** | B1 | Deduces that there are 5 solutions |
**(1 mark)**
6. $\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 1$\\
a. (i) Show that ( $2 x - 1$ ) is a factor of $\mathrm { f } ( x )$.\\
(ii) Express $\mathrm { f } ( x )$ in the form $( 2 x - 1 ) ( x + a ) ^ { 2 }$ where $a$ is an integer.
Using the answer to part a) (ii)\\
b. show that the equation $2 p ^ { 6 } + 3 p ^ { 4 } - 1$ has exactly two real solutions and state the values of these roots.\\
c. deduce the number of real solutions, for $5 \pi \leq \theta \leq 8 \pi$, to the equation
$$2 \cos ^ { 3 } \theta + 3 \cos ^ { 2 } \theta - 1 = 0$$
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [7]}}