Moderate -0.3 This is a standard Further Maths question on complex roots with real coefficients. Once students recognize that 2-5i must also be a root (conjugate root theorem), they can find the quadratic factor and solve for the third root using routine algebraic methods. While it requires multiple steps, the approach is formulaic and well-practiced in FM courses.
3 In this question you must show detailed reasoning.
You are given that \(x = 2 + 5 \mathrm { i }\) is a root of the equation \(x ^ { 3 } - 2 x ^ { 2 } + 21 x + 58 = 0\).
Solve the equation.
Question 3:
3 | DR
Second root is the conjugate of2+5i, sox=2−5i soi
So the cubic x3−2x2 +21x+58=0 can be written
( x−a )( x−(2+5i) )( x−(2−5i) )=0
⇒−a(2+5i)(2−5i)=58
⇒a=−2
So the solution of the cubic is
⇒ x=−2, 2±5i | B1
M1
A1
A1
[4] | 2.2a
1.1
2.1
1.1 | Soi by correct quadratic
2
𝑥𝑥 −4𝑥𝑥+29
Attempt to factorise using
complex conjugate
Shown convincingly
oe (i.e. (x + 2) seen | Any valid method to find
real root by reasoning,
including division, or
listing or using the factor
theorem or sum of roots.
NB. A DR question so
full reasoning must be
shown.
3 In this question you must show detailed reasoning.\\
You are given that $x = 2 + 5 \mathrm { i }$ is a root of the equation $x ^ { 3 } - 2 x ^ { 2 } + 21 x + 58 = 0$.\\
Solve the equation.
\hfill \mbox{\textit{OCR Further Pure Core 1 2019 Q3 [4]}}