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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 .

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:@ ?? why yu angry ? LOL
3rd root is 2

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Other answers:

not angry
just having a bad day
how is 3rd root 2 ?
lols , having a bad day ? why ?
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.
ohh ok LOL
I got so far (for roots a,-a and c) 3a +c = 2 and a^2 c = 24 U get that?
w8 , what's c ?
I don't mean understand, I mean u calculated same?
I said for roots, a,-a and c
ohh lols , i dnt even know how to do it
There is similar (not same) formula like Vieta for cubic as for quadratic....
What's Vieta ?
Formulae for sum and product of roots of quadratic.
ohh that
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
thank you (:

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