Question #117071
Find the real root of the equation z3 + z + 10 = 0 given that one root is 1 − 2i.
(b) Given that 3 + i is a root of the equation z3 − 3z2 − 8z + 30 = 0, find the remaining roots.
(c) Given that 1 + i is a root of the equation z3 − 2z + k = 0, find the other two roots and the value of the real constant k.
(d) Given that 2−3i is a root of the equation z3 +pz2 +qz +13 = 0, find the other two roots and the values of the real constants p and q.
(e) Show that z = i is a root of the equation z4 + z3 + z − 1 = 0. Find the three other roots.
(f) Show that z = −1+i is a root of the equation z4 −2z3 −z2 +2z +10 = 0. Find the remaining roots.
(g) Solve the equation 6z4 − 47z3 + 148z2 − 167z + 52 = 0
1
Expert's answer
2020-05-24T17:52:00-0400

a)

z1=12i,z2=1+2iz_1=1-2i, z_2=1+2i

z3+z+10=(z(12i))(z(1+2i))(Az+B)z^3+z+10=(z-(1-2i))(z-(1+2i))(Az+B)

z3+z+10=((z1)2+4)(Az+B)z^3+z+10=((z-1)^2+4)(Az+B)

z3+z+10=(z22z+5)(Az+B)z^3+z+10=(z^2-2z+5)(Az+B)

z3+z+10=Az3+Bz22Az22Bz+5Az+5Bz^3+z+10=Az^3+Bz^2-2Az^2-2Bz+5Az+5B

z3+z+10=Az3+(B2A)z2+(5A2B)z+5Bz^3+z+10=Az^3+(B-2A)z^2+(5A-2B)z+5B

A=1,B=2A=1, B=2

Az+B=0Az+B=0

z+2=0z+2=0

Answer:

z3=2z_3=-2


b)

z1=3+i,z2=3iz_1=3+i, z_2=3-i

z33z28z+30=(z(3+i))(z(3i))(Az+B)z^3 − 3z^2 − 8z + 30=(z-(3+i))(z-(3-i))(Az+B)

z33z28z+30=((z3)2+1)(Az+B)z^3 − 3z^2 − 8z + 30=((z-3)^2+1)(Az+B)

z33z28z+30=(z26z+10)(Az+B)z^3 − 3z^2 − 8z + 30=(z^2-6z+10)(Az+B)

z33z28z+30=Az3+Bz26Az26Bz+10Az+10Bz^3 − 3z^2 − 8z + 30=Az^3+Bz^2-6Az^2-6Bz+10Az+10B

z33z28z+30=Az3+(B6A)z2+(10A6B)z+10Bz^3 − 3z^2 − 8z + 30=Az^3+(B-6A)z^2+(10A-6B)z+10B

A=1,B=3A=1, B=3

Az+B=0Az+B=0

z+3=0z+3=0

Answer:

z3=3z_3=-3


c)

z1=1+i,z2=1iz_1=1+i, z_2=1-i

z32z+k=(z(1+i))(z(1i))(Az+B)z^3 − 2z + k=(z-(1+i))(z-(1-i))(Az+B)

z32z+k=((z1)2+1)(Az+B)z^3 − 2z + k=((z-1)^2+1)(Az+B)

z32z+k=(z22z+2)(Az+B)z^3 − 2z + k=(z^2-2z+2)(Az+B)

z32z+k=Az3+Bz22Az22Bz+2Az+2Bz^3 − 2z + k=Az^3+Bz^2-2Az^2-2Bz+2Az+2B

z32z+k=Az3+(B2A)z2+(2A2B)z+2Bz^3 − 2z + k=Az^3+(B-2A)z^2+(2A-2B)z+2B

A=1,B=2A=1, B=2

2B=k2B=k

Answer:

z3=2,k=4z_3=-2, k=4


d)

z1=23i.z2=2+3iz_1=2-3i. z_2=2+3i

z3+pz2+qz+13=(z(23i))(z(2+3i))(Az+B)z^3 +pz^2 +qz +13=(z-(2-3i))(z-(2+3i))(Az+B)

z3+pz2+qz+13=((z2)2+9)(Az+B)z^3 +pz^2 +qz +13=((z-2)^2+9)(Az+B)

z3+pz2+qz+13=(z24z+13)(Az+B)z^3 +pz^2 +qz +13=(z^2-4z+13)(Az+B)

z3+pz2+qz+13=Az3+Bz24Az24Bz+13Az+13Bz^3 +pz^2 +qz +13=Az^3+Bz^2-4Az^2-4Bz+13Az+13B

A=1,B=1A=1, B=1

Answer:

z3=1z_3=-1

p=B4A=14=3p=B-4A=1-4=-3

q=13A4B=134=9q=13A-4B=13-4=9


e)

i4+i3+i1=1i+i1=0i^4 + i^3 + i −1=1-i+i-1=0

z1=i,z2=iz_1=i, z_2=-i

z4+z3+z1=(zi)(z+i)(Az2+Bz+C)z^4 + z^3 + z − 1=(z-i)(z+i)(Az^2+Bz+C)

z4+z3+z1=(z2+1)(Az2+Bz+C)z^4 + z^3 + z − 1=(z^2+1)(Az^2+Bz+C)

z4+z3+z1=Az4+Az2+Bz3+Bz+Cz2+Cz^4 + z^3 + z − 1=Az^4+Az^2+Bz^3+Bz+Cz^2+C

A=1,C=1,B=1A=1, C=-1, B=1

z2+z1=0z^2+z-1=0

Answer:

z3=11+42=152z_3=\frac {-1-\sqrt{1+4}}{2}=\frac {-1-\sqrt{5}}{2}

z4=1+52z_4=\frac {-1+\sqrt{5}}{2}


f)

(1+i)42(1+i)3(1+i)2+2(1+i)+10=(-1+i)^4 −2(-1+i)^3 −(-1+i)^2 +2(-1+i) +10=

=(12i1)22(12i1)(1+i)(12i1)2+2i+10==(1-2i-1)^2-2(1-2i-1)(-1+i)-(1-2i-1)-2+2i+10=

=4+4i(1+i)+4i+8=44i4+4i=0=-4+4i(-1+i)+4i+8=4-4i-4+4i=0

z1=1+i,z2=1iz_1=-1+i, z_2=-1-i

z42z3z2+2z+10=(z(1+i))(z(1i))(Az2+Bz+C)z^4 −2z^3 −z^2 +2z +10=(z-(-1+i))(z-(-1-i))(Az^2+Bz+C)

z42z3z2+2z+10=((z+1)2+1)(Az2+Bz+C)z^4 −2z^3 −z^2 +2z +10=((z+1)^2+1)(Az^2+Bz+C)

z42z3z2+2z+10=(z2+2z+2)(Az2+Bz+C)z^4 −2z^3 −z^2 +2z +10=(z^2+2z+2)(Az^2+Bz+C)

z42z3z2+2z+10=Az4+Bz3+Cz2+2Az3+2Bz2+2Cz+2Az2+2Bz+2Cz^4 −2z^3 −z^2 +2z +10=Az^4+Bz^3+Cz^2+2Az^3+2Bz^2+2Cz+2Az^2+2Bz+2C

A=1,C=5A=1, C=5

2C+2B=22C+2B=2

C+B=1C+B=1

B=C=5B=-C=-5

z25z+5=0z^2-5z+5=0

Answer:

z3=525202=552z_3=\frac {5-\sqrt{25-20}}{2}=\frac {5-\sqrt{5}}{2}

z4=5+52z_4=\frac {5+\sqrt{5}}{2}


g)

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient.

In this case, the Leading Coefficient is  6  and the Trailing Constant is  52.

The factor(s) are:

of the Leading Coefficient :  1,2 ,3 ,6

of the Trailing Constant :  1 ,2 ,4 ,13 ,26 ,52

After the test we have:

z1=12,z2=43z_1=\frac {1}{2}, z_2=\frac {4}{3}

Then:

6z447z3+148z2167z+52(2z1)(3z4)=z26z+13\frac {6z^4 − 47z^3 + 148z^2 − 167z + 52}{(2z-1)(3z-4)}=z^2-6z+13

z26z+13=0z^2-6z+13=0

z3=636522=32iz_3=\frac {6-\sqrt{36-52}}{2}=3-2i

z4=3+2iz_4=3+2i


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