Question

Find all zeros of the polynomial P(x) = x3 − 2x2 − 3x + 10. Express any non-real roots in the form a + bi.

Answer 1

Answer:

$x = - 2 \: or \: x = 2 - i \: or \: x = 2 + i$

Step-by-step explanation:

The given equation is

$p(x) = {x}^{3} - 2 {x}^{2} - 3x + 10$

We can see that:

$p( - 2) = {( - 2)}^{3} - 2 {( -2) }^{2} - 3( - 2)+ 10$

$p( - 2) = - 8- 8 + 6+ 10 = 0$

This means x=-2 is a zero of p(x).

From the long division in the attachment,

We can rewrite the polynomial as:

${x}^{3} - 2 {x}^{2} - 3x + 10 = (x + 2)( {x}^{2} - 4x + 5)$

We now find solution to the quadratic part:

${x}^{2} - 4x + 5 = 0$

This is given by:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

We plug in the values to get:

$x = \frac{ - - 4 \pm \sqrt{ {( - 4)}^{2} - 4 \times 1 \times 5} }{2 \times 1}$

$x = \frac{ 4 \pm \sqrt{ 16 - 20} }{2 }$

$x = \frac{ 4 \pm \sqrt{ - 4} }{2 }$

$x = \frac{ 4 \pm2i}{2 } \\ x = 2 \pm \: i \\ x = 2 - i \: or \: x = 2 + i$

Therefore all solutions are:

$x = - 2 \: or \: x = 2 - i \: or \: x = 2 + i$