Find all zeros of F(x) = x^3 - x^2 - x - 2.?

Question

Hi, I am having trouble solving this problem I am not sure if I am supposed to solve it graphically or a different way. Thanks for any help

Find all zeros of F(x) = x^3 - x^2 - x - 2.

Answer 1

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x=2,8-4-2-2=0So,(x-2)(x^2+x+1)=0, x=2,x=(-1+3i)/2

& x=(-1-3i)/2.

Where i=(-1)^1/2

Answer 2

Well, graphing is usually fine but graphing doesn't show irrational roots very nicely. So we have to resort to p/q and then using synthetic division.

Here p/q or ±2/1 gives us +2 and -2 to check.

Checking +2: (2)^3-(2)^2-2-2=0

0=0

So x=2, or (x-2) is in fact a rational root.

Now using x-2 and dividing into x^3-x^2-x-2 reveals the quadratic

x^2+x+1

And this does indeed have irrational roots that graphing would not have easily revealed.

Using the quadractic formula on this we find

x = -1/2±1/2√(3)i

So there's the three roots or zeros of the cube function, one real and two immaginary.

Have a good day!

Answer 3

Trying x=2 (a divisor of 2)

F(2) = 2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0

F(x) = x^3 - x^2 - x - 2 = x^3 - 2x^2 + x^2 - 2x + x - 2 =

= x^2 (x - 2) + x(x - 2) + (x - 2) = (x - 2)(x^2 + x + 1)

x^2 + x + 1 cannot be further factored in R, because its discriminant

(-1)^2 - 4*1*1 = 1 - 4 = -3 < 0

Thus, the only zero is 2.

Answer 4

rational zeros?

+/-2 or +/-1

try 2:

8-4-2-2 = 0 , so 2 is a zero, or (x-2) is a factor

do synthetic division to get it reduced to a quadratic

2 L 1 -1 -1 -2

.........2....2...2

......1....1....1.0

F(x) = x^3 - x^2 - x - 2 = (x-2) ( x^2+x+1)

imaginary solutions?? depends on your level?

if only real solutions: x = 2