| (a) Attempt $f(3)$ or $f(-3)$; Use of long division is M0A0 as factor theorem was required | M1 |
|---|---|
| $f(-3)=162-63-120+21=0$ so $(x+3)$ is a factor | A1 |
| (b) **Way 1:** $f(x)=(x+3)(-6x^{2}+11x+7)$ | M1A1 |
| $=(x+3)(-3x-7)(2x+1)$ or $-(x+3)(3x-7)(2x+1)$ | M1A1 |
| **Way 2:** Uses trial or factor theorem to obtain $x=-\frac{1}{2}$ or $x=\frac{7}{3}$ | M1 |
| Uses trial or factor theorem to obtain both $x=-\frac{1}{2}$ **and** $x=\frac{7}{3}$ | A1 |
| Puts three factors together (see notes below) | M1 |
| Correct factorisation: $(x+3)(7-3x)(2x+1)$ or $-(x+3)(3x-7)(2x+1)$ oe | A1 |
| **Way 3:** No working three factors $(x+3)(-3x+7)(2x+1)$ otherwise need working | M1A1M1A1 |
| (c) $2^{y}=\frac{7}{3}$ $\rightarrow$ $\log\left(2^{y}\right)=\log\left(\frac{7}{3}\right)$ or $y=\log_{2}\left(\frac{7}{3}\right)$ or $y=\frac{\log(7/3)}{\log 2}$ | B1, M1 |
| $\{y=1.222392421\ldots\} \Rightarrow y=\text{awrt } 1.22$ | A1 |

**Notes:**
- (a) M1: for attempting either $f(3)$ or $f(-3)$ – with numbers substituted into expression; A1: for calculating $f(-3)$ correctly to 0, and they must state $(x+3)$ is a factor for A1 (or equivalent ie. QED, ☐ or tick)
- (b) **1st M1:** attempting to divide by $(x+3)$ leading to a 3TQ beginning with correct term, usually $-6x^{2}$. This may be done by variety of methods including long division, comparison of coefficients, inspection etc. Allow for work in part (a) if the result is used in (b); **1st A1:** usually for $(-6x^{2}+11x+7)\ldots$ Credit when seen and use isw if miscopied; **2nd M1:** for a valid attempt to factorise their quadratic (* see notes on page 6 - General Principles for Core Mathematics Marking section 1); **2nd A1** is cao and needs all three factors together fully factorised. Accept e.g. $-3(x+3)(x-\frac{7}{3})(2x+1)$ but $(x+3)(x-\frac{7}{3})(-6x-3)$ and $(x+3)(3x-7)(-2x-1)$ are A0 as not fully factorised
- **Ignore subsequent work (such as a solution to a quadratic equation.)**
- **Way 2:** The second M mark needs three roots together so $-6(x-\alpha)(\beta)(\gamma+\beta)$ or equivalent where they obtained $\alpha$ and $\beta$ by trial, so if correct roots identified, then $(x+3)(3x-7)(2x+1)$ can gain M1A1M1A0
- **N.B.** Replacing $(-6x^{2}+11x+7)$ (already awarded M1A1) by $(6x^{2}-11x-7)$ giving $(x+3)(3x-7)(2x+1)$ can have M1A0 for factorization so M1A1M1A0
- (c) **B1:** $2^{y}=\frac{7}{3}$; M1: Attempt to take logs to solve $2^{y}=\alpha$ or $2^{y}=1/\alpha$, where $\alpha > 0$ and $\alpha$ was a root of their factorization; **A1:** for an answer that rounds to 1.22. If other answers are included (and not "rejected") such as $\ln(-3)$ or $-1$ lose final A mark
- **Special case:** Those who deal **throughout** with $f(x)=6x^{3}+7x^{2}-40x-21$. They may have full credit in part (a). In part (b) they can achieve a maximum of M1A0M1A0 unless they return the negative sign to give the correct answer. This is then full marks. Part (c) is fine. So they could lose 2 marks on the factorisation. (Like a misread)