SOLUTION: Factor completely 2x^2 + 16x + 32 x^3 + 3x^2 + x + 3 3x^2 + 6x - 24

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Question 30879: Factor completely
2x^2 + 16x + 32
x^3 + 3x^2 + x + 3
3x^2 + 6x - 24

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+2x%5E2+%2B+16x+%2B+32+
+2%28x%5E2+%2B+8x+%2B+16%29+
+2%28x%2B4%29%28x%2B4%29+
+2%28x%2B4%29%5E2+

+x%5E3+%2B+3x%5E2+%2B+x+%2B+3+
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 x%5E3 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 %28x%2B3%29%28x%5E2%2B1%29. So try it.

Actually, there is a simpler method for this cubic: +x%5E3+%2B+3x%5E2+%2B+x+%2B+3+ written as +x%5E3+%2B+x+%2B+3x%5E2+%2B+3+. Look the 2 sets of 2 terms and factorise each of those:

+%28x%5E3+%2B+x%29+%2B+%283x%5E2+%2B+3%29+
+x%28x%5E2+%2B+1%29+%2B+3%28x%5E2+%2B+1%29+

You can see that both halves have x%5E2%2B1, so we can factorise again, as +%28x%5E2+%2B+1%29%28x+%2B+3%29+

Third example looks to be like the first.

jon.