Question

Write the polynomial equation of lowest degree with real coefficients having 2 and (1-3i) as two of its roots.

Answer 1

if a+bi is a root, then a-bi is also a root of that equation. (with real co-efficients)

so,if here 1-3i is a root, whats the other one?

Answer 2

@hartnn 1+3i.

Answer 3

correct!
now if a is the root of f(x), then (x-a) is one of the factor.

so when '2' iss the root, the factor is ?

Answer 4

@hartnn x-2.

Answer 5

correct!
now 1-3i is also a root,
so (x-(1-3i)) = (x-1+3i) is also a factor,
lastly, since 1+3i is also a root, which is the last factor of f(x) ?

Answer 6

@raechelvictoria do u get it

Answer 7

@hartnn (x-(1-3i))

Answer 8

@nirmalnema Isn't the other factor supposed to be (x-(1+3i))?

Answer 9

yup its correct...

Answer 10

x-(1-3i) = x-1+3i is what i showed
and other will be x-(1+3i) = x-1-3i
ok?

Answer 11

(x-2)(x-(1+3i))(x-(1-3i))
(x-2)(x^2-2x-9i^2)
x^3-2x^2-9i^2x-2x^2+4x-18i^2
now i^2=(-1)
so you will get
x^3-4x^2+13x+18=0

Answer 12

i put the wrong sign...sorry

Answer 13

there is an error on your second step, you left out the '+1'

Answer 14

and @raechelvictoria , which step are you on ? stuck or trying ?

Answer 15

@hartnn I'm on the step you just showed :)

Answer 16

ok, so we have 3 factors as x-2, x-1+3i, x-1-3i

so, your f(x) = (x-2)(x-1+3i)(x-1-3i) = .... ?
can you simplify?

Answer 17

@hartnn the part where i'm having trouble is the part where you simplify the complex numbers :(

Answer 18

ok, we use the fact that (a+b)(a-b) = a^2-b^2

so, we take, (x-1+3i)(x-1-3i) = ...
take a= x-1 and b = 3i
\(\large [(x-1)+(3i)][(x-1)-(3i)]=.....? \\knowing, (a-b)(a+b)=a^2-b^2\)

Answer 19

\[(x-1)^{2} + 9\] ?

Answer 20

correct! good :)
f(x) = (x-2)(x^2-2x+10)
right ? simplify this further, this part is easy :)

Answer 21

\[x ^{3}-4x ^{2}+14x-20?\]

Answer 22

yes! absolutely correct :)

Answer 23

@hartnn thank you so much! :)

Answer 24

you're welcome ^_^