Answer:
[tex]x = - 2 \: or \: x = 2 - i \: or \: x = 2 + i[/tex]
Step-by-step explanation:
The given equation is
[tex]p(x) = {x}^{3} - 2 {x}^{2} - 3x + 10[/tex]
We can see that:
[tex]p( - 2) = {( - 2)}^{3} - 2 {( -2) }^{2} - 3( - 2)+ 10[/tex]
[tex]p( - 2) = - 8- 8 + 6+ 10 = 0[/tex]
This means x=-2 is a zero of p(x).
From the long division in the attachment,
We can rewrite the polynomial as:
[tex] {x}^{3} - 2 {x}^{2} - 3x + 10 = (x + 2)( {x}^{2} - 4x + 5)[/tex]
We now find solution to the quadratic part:
[tex] {x}^{2} - 4x + 5 = 0[/tex]
This is given by:
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
We plug in the values to get:
[tex]x = \frac{ - - 4 \pm \sqrt{ {( - 4)}^{2} - 4 \times 1 \times 5} }{2 \times 1} [/tex]
[tex]x = \frac{ 4 \pm \sqrt{ 16 - 20} }{2 } [/tex]
[tex]x = \frac{ 4 \pm \sqrt{ - 4} }{2 } [/tex]
[tex] x = \frac{ 4 \pm2i}{2 } \\ x = 2 \pm \: i \\ x = 2 - i \: or \: x = 2 + i[/tex]
Therefore all solutions are:
[tex]x = - 2 \: or \: x = 2 - i \: or \: x = 2 + i[/tex]
