Question

The polynomial P(x) = x^3 - 2x^2 + kx + 24 has roots a , b and γ
(i) It is known that two of the roots are equal in magnitude but opposite in sign.
Find the third root and hence find the value of k

Can you please help .

Answer 1

:@

Answer 2

:@ ?? why yu angry ? LOL

Answer 3

3rd root is 2

Answer 4

not angry

Answer 5

just having a bad day

Answer 6

how is 3rd root 2 ?

Answer 7

lols , having a bad day ? why ?

Answer 8

I passed the entry test to uni, but apparently, I need to get minimum of 2 A levels to get into it. A levels take 2 years to get. I have 4 months.

Answer 9

ohh ok LOL

Answer 10

I got so far (for roots a,-a and c)

3a +c = 2 and
a^2 c = 24

U get that?

Answer 11

yeah

Answer 12

w8 , what's c ?

Answer 13

I don't mean understand, I mean u calculated same?

Answer 14

I said for roots, a,-a and c

Answer 15

ohh lols , i dnt even know how to do it

Answer 16

There is similar (not same) formula like Vieta for cubic as for quadratic....

Answer 17

What's Vieta ?

Answer 18

Formulae for sum and product of roots of quadratic.

Answer 19

ohh that

Answer 20

Let a be the first root >0. Then b=-a
Now since a cubic function with 3 real roots then must be of the form:
(x-a)(x-b)(x-v)= x^3 - 2x^2 + kx + 24
But b=-a so LHS is (x-a)(x+a)(x-v)=(x^2-a^2)(x-v)
Multiplying the left hand side out we get:
=x^3-v*x^2-a^2*x+v*a^2=P(x)
From this we see that v=2 which implies 24=2*a^2
then a^2=12 but -a^2*x=kx
thus k=-12

Answer 21

thank you (: