Question #141156
Given the cubic equation 2z^3-z^2+4z+p=0, peR( p is an element of the set of real numbers)
z=1+2i (i) Find the other solution (ii) Determine the value of P
1
Expert's answer
2020-11-13T14:03:03-0500

f(z)=2z3z2+4z+p=0By the factor theorem,f(1+2i)=02(1+2i)3(1+2i)2+4(1+2i)+p=02(1+6i128i)(1+4i4)+4+8i+p=02(2i11)(3+4i)+4+8i+p=04i22+34i+4+8i+p=0p=15By the Fundamental theorem of Algebrathe conjugate of1+2iis also a root off(z)(z(1+2i))(z(12i))=0z2(1+2i+12i)z+(1+4)=0z22z+5=0By Long division, we obtain the last factor2z+3z22z+52z3z2+4z+15(2z34z2+10z)3z26z+15(3z26z+15)=0    (z22z+5)(2z+3)=02z+3=0    z=32z=32is another solution.(i)32(ii)p=15\displaystyle f(z) = 2z^3-z^2+4z+p = 0\\ \textsf{By the factor theorem,}\, f(1 + 2i) = 0\\ 2(1 + 2i)³ - (1 + 2i)² + 4(1 + 2i) + p = 0\\ 2(1 + 6i - 12 - 8i) - (1 + 4i - 4) + 4 + 8i + p = 0\\ 2(-2i - 11) - (-3 + 4i) + 4 + 8i + p = 0\\ -4i - 22 + 3 - 4i + 4 + 8i + p = 0\\ \therefore p = 15\\ \textsf{By the Fundamental theorem of Algebra}\\ \textsf{the conjugate of}\, 1 + 2i \, \textsf{is also a root of}\, f(z)\\ \therefore (z - (1 + 2i))(z - (1 - 2i)) = 0\\ z^2 - (1 + 2i + 1 - 2i)z + (1 + 4) = 0\\ z^2 - 2z + 5 = 0\\ \textsf{By Long division, we obtain the last factor}\\ 2z + 3\\ z^2 - 2z + 5|2z^3-z^2+4z+15\\ -(2z^3 - 4z^2 + 10z)\\ 3z^2 - 6z + 15\\ -(3z^2 - 6z + 15) = 0\\ \implies(z^2 - 2z + 5)(2z + 3) = 0\\ 2z + 3 = 0 \implies z = -\frac{3}{2}\\ \therefore z = -\frac{3}{2}\, \textsf{is another solution.}\\ (i)\\ -\frac{3}{2}\\ (ii)\\ p = 15


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