Question

Given that root 2 is a zero of the cubic polynomial 6x3+2x2-10x-4root2, find its other two zeroes

Answer 1

But √2 is <u>not</u> a root of

6x^3+2x^2-10x-4√2

6(√2)^3+2(√2)^2-10√2-4√2
= 6*2√2 + 2*2 - 10√2 - 4√2
= -2√2 + 4

Care to fix your mistake(s)?

Answer 2

We know that (x-sqrt 2) is a factor

6 x^3 + 2 x^2 - 10 x -4 sqrt 2 = 0

(x-sqrt 2)( a x^2 + b x + c) = 0

well, c better be 4
(x-sqrt 2)(a x^2 + b x + 4) = 0
now I will use distributive property to split this up
x (a x^2 + b x + 4)
-sqrt 2 (a x^2 + b x + 4)
= 0
x terms:
4 x - b sqrt 2 x = 0
4 = b sqrt 2
b = 4/sqrt 2 = 2 sqrt 2
so now I have
x ( a x^2 + 2 sqrt 2 x + 4 )
-sqrt 2 (a x^2 + 2 sqrt 2 x + 4 ) = 0
so
x^2 terms
2 sqrt 2 x^2 - sqrt 2 a x^2 = 0
a = 2
so now I have
x ( 2 x^2 + 2 sqrt 2 x + 4 )
-sqrt 2 (2 x^2 + 2 sqrt 2 x + 4 ) = 0
in other words
(x-sqrt2)(2 x^2+2 sqrt 2 x + 4 = 0
so find zeros of
x^2 + sqrt 2 x + 2 = 0
x = [ -sqrt 2 +/- sqrt(2-32)] /2
= -(1/2) sqrt 2 +/- (i/2)sqrt 30

check my arithmetic, use for method only.

Answer 3

LOL, should have checked that !