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Roots of Polynomials. please help

+3 votes

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).

asked Dec 25, 2012 in ALGEBRA 2 by alg2trig Rookie

2 Answers

+4 votes

 

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

answered Dec 26, 2012 by steve Scholar
+3 votes

 

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

answered Dec 26, 2012 by steve Scholar

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