Roots of Polynomials. please help

Question

If one root of the equation is given, solve the equation for all roots.

 

23. x^3-4x^2+21x-34=0; x=2

I got up to (x-2) x(x^2-2x+17), but since (x^2-2x+17) isn\'t factorable, I\'m not sure where to go from there.

 

24. x^3+x+10=0; x=-2

Here, I got up to (x+2) (x^2-2x+5).

Answer 1

 

23.

when x² - 2x + 17 isn\'t factorable apply quadratic formula.

Roots of ax² + bx + c = 0 is x = (- b ± √(b² - 4 ac))/2a

 

Compare x² - 2x + 17 with ax² + bx + c and find the coefficients.

a = 1, b = -2 and c = 17

x = (2 ± √(2² - 4 * 1 * 17))/2 *1

x = (2 ± √(4 - 68))/2

x = (2 ± √(-64))/2

Substitute i² = -1

x = (2 ± √(64i²))/2

x = (2 ± 8i)/2

x = 1 ± 4i

Answer 2

 

24.

when x² - 2x + 5 isn\'t factorable apply quadratic formula.

quadratic formula:

Roots of ax² + bx + c = 0 is x = (- b ± √(b² - 4 ac))/2a

Comapre and find the coefficients.

a = 1, b = -2 and c = 5

x = (2 ± √(2² - 4 * 1 * 5))/2 *1

x = (2 ± √(4 - 20))/2

x = (2 ± √(-16))/2

Substitute i² = -1

x = (2 ± √(16i²))/2

x = (2 ± 4i)/2

x = 1 ± 2i