SOLUTION: Factor completely
2x^2 + 16x + 32
x^3 + 3x^2 + x + 3
3x^2 + 6x - 24
Algebra.Com
Question 30879: Factor completely
2x^2 + 16x + 32
x^3 + 3x^2 + x + 3
3x^2 + 6x - 24
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
You need to find a factor of this ie a value of x that makes the whole thing zero. Since all the terms are positive, then x must be a negative value, to make the and x negative.
As to the value of x, well this is trial and error to find, but looking at the equation, the value -3 does jump out. Put x=-3 into the expression and you do get zero. So, x+3 is a factor.
You now need to divide this into the cubic expression and then factorise the resulting quadratic, if possible. Having done this for you, the final answer is . So try it.
Actually, there is a simpler method for this cubic: written as . Look the 2 sets of 2 terms and factorise each of those:
You can see that both halves have , so we can factorise again, as
Third example looks to be like the first.
jon.
RELATED QUESTIONS
factor completely... (answered by akmb1215)
Factor completely,
11x(4x-3)-6(4x-3)=
2x(x-5)-(x-5)
3x^3-4x^2+6x-8=... (answered by KMST)
factor completely... (answered by vleith)
Factor Completely.
3x^3-6x^2-12x+24
Thank... (answered by dfrazzetto,ewatrrr)
Factor Completely
6x^3 - 16x^2 -... (answered by MathLover1,josmiceli)
factor completely:
X^4-2X^2-16X +32 = 0
(answered by rapaljer)
Factor completely.
-3x^3 - 6x^2 +... (answered by mananth)
Factor completely... (answered by mananth)
Factor completely:... (answered by lynnlo)