Expert: Sherman D. - 2/20/2005

Question
I understand what you did to the previous cubed power with grouping, but you can't use that for factoring:
x^3+x^2+x-3.
So how would you factor something like that.
I appreciate your time that you take out to answer these questions.Thank you so much.

Answer
yeah, this ones a different one

off and i can tell that x = 1, because
1^3 + 1^2 + 1 - 3 = 1 + 1 + 1 - 3 = 3 - 3 = 0

so using (x - 1) as a factor

1'|'1'1'1'|'-3
''|'''1'2'|''3
----------------
''|'1'2'3'|''0

so this gives you

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

to factor further, i would use the quadratic formula to find if there are any other x-intercepts

x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-(2) ± sqrt((2)^2 - 4(1)(3)))/2(1)
x = (-2 ± sqrt(4 - 12))/2
x = (-2 ± sqrt(-8))/2
x = (-2 ± sqrt(-4 * 2))/2
x = (-2 ± 2isqrt(2))/2
x = -1 ± isqrt(2)
x = -1 + isqrt(2) or -1 - isqrt(2)

so unless you want it looking like

(x - 1)(x - (1 - isqrt(2)))*(x - (1 + sqrt(2)))

i would say that your answer is (x - 1)(x^2 + 2x + 3)

going to www.quickmath.com to check my work confirms that (x - 1)(x^2 + 2x + 3) is the answer.

anymore you would like help on, bring them on, i like trying to factor problems.