## Question 9:

$a = 0$ (is one possibility) | **B1** (1.1) |

$1^4 + 1^3 + 3\times1^2 - 5\times1 = 0$ so $a = 1$ (is another possibility) | **B1** (3.1a) | or e.g. $f(1) = 0$ if intent is clear but must be some justification.

$x^4 + x^3 + 3x^2 - 5x = x(x^2(x-1) + 2x(x-1) + 5(x-1)) = x(x-1)(x^2+2x+5)$ and discriminant of quadratic $= 2^2 - 4\times1\times5 = -16 < 0$ so no further real roots. | **B1** (2.3) | Some justification must be given that there are no more real roots. Allow for finding the two complex roots $(-1\pm2i)$ - must have seen the correct $x^2+2x+5$. If **B1B0B0** or **B0B0B0** then **SC1** for "$a=1$ and no others" or "$a=1$, $(-1\pm2i)$" without justification.

$a = 0 \Rightarrow 2+3i$ is a root of $z^4 - 2z^3 - 10z^2 + 86z - 195\ [=0]$ so $2-3i$ is also a root. OR $a=1 \Rightarrow 3+3i$ is a root of $z^4 - 4z^3 + 11z^2 + 6z + 90\ [=0]$ so $3-3i$ is also a root | **M1** (3.1a) | Condone small errors in calculation of coefficients in equation. Need both the pair of complex roots SOI and the quartic shown (allow sign slips). Only need one case for the M1.

$2+3i + 2-3i = 4$ and $(2+3i)(2-3i) = 13$ so $z^2 - 4z + 13$ is a factor. OR $3+3i + 3-3i = 6$ and $(3+3i)(3-3i) = 18$ so $z^2 - 6z + 18$ is a factor | **M1** (1.1) | oe e.g. expanding $(z-(2+3i))(z-(2-3i))$ or $(z-(3+3i))(z-(3-3i))$. Attempt to find quadratic factor from the complex roots. Only one case needed for M1. Allow with no working.

$z^4 - 2z^3 - 10z^2 + 86z - 195 = (z^2-4z+13)(z^2+2z-15)$ OR $z^4 - 4z^3 + 11z^2 + 6z + 90 = (z^2-6z+18)(z^2+2z+5)$ | **M1** (1.1) | Attempt to factorise their quartic with their quadratic factor (at least $z^3$ and constant terms consistent). Only one case needed. MUST see some evidence of factorisation here.

$z^2 + 2z - 15 = 0 \Rightarrow z = -5,\ z = 3$ and $2\pm3i$ stated as roots (possibly earlier). | **A1** (2.2a) | All four roots SOI for the $a=0$ case. **DR** so need to see evidence of where the roots came from i.e. factorisation into two quadratics.

$z^2 + 2z + 5 = 0 \Rightarrow z = -1\pm2i$ and $3\pm3i$ stated as roots (possibly earlier). | **A1** (2.2a) | All four roots SOI for the $a=1$ case. If extra values of $z$ found (from complex $a$ or incorrect $a$ values) then A0.

# Mark Scheme Extraction

## Question (Quartic/Complex Roots Problem):

### Alternate Method 1 (for 1st and 2nd M marks):

| Working | Mark | Guidance |
|---------|------|----------|
| $(a+2+3i)+(a+2-3i)=2a+4$ and $(a+2+3i)(a+2-3i)=(a+2)^2+9$ so $z^2-(2a+4)z+[a^2+4a+13]$ is a factor | **M1** | Quadratic factor found in general case. Can also be found by expanding $[z-(a+2+3i)][z-(a+2-3i)]$. Constant term can be either $(a+2)^2+9$ or $[a^2+4a+13]$ |
| $a=0 \Rightarrow z^2-4z+13$ is a factor OR $a=1 \Rightarrow z^2-6z+18$ is a factor | **M1** | |

### Alternate Method 2 (quartic in $z$ with $a=0$, via linear factors):

| Working | Mark | Guidance |
|---------|------|----------|
| $a=0$ is one possibility | **B1** | |
| $1^4+1^3+3\times1^2-5\times1=0$ so $a=1$ is another possibility | **B1** | Or e.g. $f(1)=0$ if intent is clear, but must be some justification |
| $x^4+x^3+3x^2-5x=x(x^2(x-1)+2x(x-1)+5(x-1))=x(x-1)(x^2+2x+5)$ and discriminant of quadratic $=2^2-4\times1\times5=-16<0$ so no further real roots | **B1** | Some justification must be given that there are no more real roots. Allow for finding the two complex roots $(-1\pm2i)$ (must have seen correct quadratic). If **B1B0B0** or **B0B0B0** then **SC1** for "$a=1$ and no others" or "$a=1$, $(-1\pm2i)$" without justification |

### If $a=0$ (quartic route):

| Working | Mark | Guidance |
|---------|------|----------|
| If $a=0$ quartic is $z^4-2z^3-10z^2+86z-195$; $f(3)=0$ so $(z-3)$ is a factor; $f(z)=(z-3)(z^3+z^2-7z+65)$ | **M1** | Identifying linear factor and factorising |
| $f(-5)=0$ so $(z+5)$ is a factor; $f(z)=(z-3)(z+5)(z^2-4z+13)$ | **M1** | |
| So the roots are $3,\ -5,\ 2+3i,\ 2-3i$ | **A1** | Identifying 4 roots. Final 2 marks unavailable |