Question

Determine where the real zeros of f(x) = x4 - 2x3 + x - 2 are located.

Answer 1

Answer:

a. at -1 & 2

Step-by-step explanation:

= x^3(x -2) + 1(x - 2) = 0

= (x^3 + 1) + (x - 2) = 0

x^3 + 1 = 0

x^3 = - 1

Taking the square root on both sides, we get:

x = - 1

x - 2 = 0

x = 2

So x = - 1, 2 (option a)

Hope that helps!

Answer 2

Step-by-step explanation:

Step 1:  Factor by grouping

$x^4 - 2x^3 + x - 2$

$x^3(x - 2) + 1(x - 2)$

$(x^3 + 1)(x - 2)$

Step 2:  Solve for x

$x^3 + 1 = 0$

$x^3 = -1$

$\sqrt[3]{x^3} = \sqrt[3]{-1}$

$x = -1$

$x - 2 = 0$

$x - 2 + 2= 0 + 2$

$x = 2$

Answer: $Option\ A, \ at\ -1 \ \& \ 2$