Standard +0.3 This is a straightforward Further Pure question requiring evaluation of f(3/2) to identify a factor, then polynomial division and solving a quadratic. While it involves complex roots (typical for Further Maths), the method is completely standard with clear guidance provided. Slightly above average difficulty due to the fractional root and complex arithmetic, but remains a routine textbook exercise.
3 In this question you must show detailed reasoning.
The function \(\mathrm { f } ( \mathrm { z } )\) is given by \(\mathrm { f } ( \mathrm { z } ) = 2 \mathrm { z } ^ { 3 } - 7 \mathrm { z } ^ { 2 } + 16 \mathrm { z } - 15\).
By first evaluating \(\mathrm { f } \left( \frac { 3 } { 2 } \right)\), find the roots of \(\mathrm { f } ( \mathrm { z } ) = 0\).
f = − + 2 4 − 1 5 = − 9 + 9 = 0 [so 3/2 is a root]
Answer
Marks
Guidance
2 4 4
B1
1.1
must see some substitution, [so f = 0 alone is B0]
2
2z – 3 is a factor
f(z) = (2z – 3)(z2 – 2z + 5)
2 − 1 6
z =
2
3
z=1+2i,1−2i,
Answer
Marks
2
M1
M1
A1
M1
Answer
Marks
B1
2.2a
1.1
1.1
1.1
Answer
Marks
2.2a
or z − 3/2
attempt to factorise (oe, e.g. long division)
or (z – 3/2)(2z2 – 4z + 10)
or by completing the square or using sum and prod of roots
[z = a ib, 2a = 2, a2 + b2 = 5 a = 1, b = 2]
[3/2 may be stated as a root earlier]
If no working shown then award no marks
Alternative solution
Other roots are and where
3 7 3 3 3 1 5
, 8 , + + = + + = =
Answer
Marks
2 2 2 2 2 2
M1
A1
symmetric property of roots used (condone 7, 15, 16 or sign
errors) – must have at least 2 of the 3
Answer
Marks
M1
symmetric property of roots used (condone 7, 15, 16 or sign
errors) – must have at least 2 of the 3
A1
so = 5, + = 2
Answer
Marks
Guidance
2 − 2 + 5 = 0
2 − 2 + 5 = 0
A1
2 1 6 −
=
Answer
Marks
2
M1
M1
or by completing the square or using sum and prod of roots
[z = a ib, 2a = 2, a2 + b2 = 5 a = 1, b = 2]
3
z=1+2i,1−2i,
Answer
Marks
2
B1
[6]
B1
Question 3:
3 | DR
3 2 7 6 3
f = − + 2 4 − 1 5 = − 9 + 9 = 0 [so 3/2 is a root]
2 4 4 | B1 | 1.1 | 3
must see some substitution, [so f = 0 alone is B0]
2
2z – 3 is a factor
f(z) = (2z – 3)(z2 – 2z + 5)
2 − 1 6
z =
2
3
z=1+2i,1−2i,
2 | M1
M1
A1
M1
B1 | 2.2a
1.1
1.1
1.1
2.2a | or z − 3/2
attempt to factorise (oe, e.g. long division)
or (z – 3/2)(2z2 – 4z + 10)
or by completing the square or using sum and prod of roots
[z = a ib, 2a = 2, a2 + b2 = 5 a = 1, b = 2]
[3/2 may be stated as a root earlier]
If no working shown then award no marks
Alternative solution
Other roots are and where
3 7 3 3 3 1 5
, 8 , + + = + + = =
2 2 2 2 2 2 | M1
A1 | symmetric property of roots used (condone 7, 15, 16 or sign
errors) – must have at least 2 of the 3
M1 | symmetric property of roots used (condone 7, 15, 16 or sign
errors) – must have at least 2 of the 3
A1
so = 5, + = 2
2 − 2 + 5 = 0 | 2 − 2 + 5 = 0 | A1
2 1 6 −
=
2 | M1
M1 | or by completing the square or using sum and prod of roots
[z = a ib, 2a = 2, a2 + b2 = 5 a = 1, b = 2]
3
z=1+2i,1−2i,
2 | B1
[6]
B1
3 In this question you must show detailed reasoning.\\
The function $\mathrm { f } ( \mathrm { z } )$ is given by $\mathrm { f } ( \mathrm { z } ) = 2 \mathrm { z } ^ { 3 } - 7 \mathrm { z } ^ { 2 } + 16 \mathrm { z } - 15$.\\
By first evaluating $\mathrm { f } \left( \frac { 3 } { 2 } \right)$, find the roots of $\mathrm { f } ( \mathrm { z } ) = 0$.
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2023 Q3 [6]}}