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Question 9535: I was wondering how to find all zeros of the function....
h(x)=x^3-3x^2+4x-2....I could only get x=1
f(x)=x^4-10x^2+24......I got x=-2 but i can't get the other answers
THANKS SO MUCH!!!
Answer by mathmaven53(29)   (Show Source):
You can put this solution on YOUR website! Divide cubic polynomial by x - 1
The quotient is x^2 -2x + 2
The cubic equation factors into
(x-1)(x^2 -2x + 2) = 0
We already know that x=1 is a root
The other roots are obtained by solving
x^2 -2x + 2 = 0
This can be written
(x-1)^2 + 1 = 0
(x-1)^2 = -1
x-1 = i or x-1 = -i where i is square root of -1
x = 1 + i or x = 1 - i
The zeros of h(x)=x^3-3x^2+4x-2
are 1,1 + i, 1 - i
To find zeros of f(x) = x^4 -10x^2 + 24
We factor x^4 -10x^2 + 24
x^4 -10x^2 + 24 = (x^2 -4)(x^2-6)
= (x-2)(x+2)(x^2 - 6)
then solve
(x-2)(x+2)(x^2 - 6) = 0
By setting each factor equal to 0 and solving for x
x=2,or x=-2,or x=-sqrt 6,or x=sqrt 6
These are the zeros of f(x)
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