SOLUTION: Would you please show me the STEPS to factor the following: 6x^2+12x+6. I know the answer is 6(x+1)^2, I'm just confused as to how to get there.

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Would you please show me the STEPS to factor the following: 6x^2+12x+6. I know the answer is 6(x+1)^2, I'm just confused as to how to get there.      Log On


   

Question 443756: Would you please show me the STEPS to factor the following: 6x^2+12x+6. I know the answer is 6(x+1)^2, I'm just confused as to how to get there.
Found 3 solutions by rwm, swincher4391, josmiceli:
Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
factor out the obvious 6
6*(x^2+2x+1)
now we want factors of 1 that add up to 2
the only factors of one are 1 and 1 which add up to 2
6*(x+1)*(x+1) which is
6*(x+1)^2

Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Notice that each term has a common factor of 6.
Then you can factor it in this way 6%28x%5E2%2B2x%2B1%29
From there factor the quadratic you are presented: x%5E2%2B2x%2B1 Since 1*1 =1 and 1+1 =2, then we can factor this into %28x%2B1%29%28x%2B1%29+=+%28x%2B1%29%5E2
Then the total factorization is :
highlight%286%28x%2B1%29%5E2%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+6x%5E2%2B12x%2B6+
Divide both sides by 6
+x%5E2+%2B+2x+%2B+1+=+0+
Use the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+2
c+=+1
x+=+%28-2+%2B-+sqrt%28+2%5E2+-+4%2A1%2A1+%29%29%2F%282%2A1%29+
x+=+%28-2+%2B-+sqrt%28+4+-+4+%29%29+%2F+2+
x+=+%28-2%29+%2F+2+
+x+=+-1+
This is called a double root, since the
equation must have 2 roots, it has the same
root twice. Plug it back in for x to check.
Rewrite it as:
+x+%2B+1+=+0+ 1st factor
+x+%2B+1+=+0+ 2nd factor
+%28x+%2B+1%29%5E2+=+0+ answer
Note that after you multiply it out as
+x%5E2+%2B+2x+%2B+1+=+0
Multiplying both sides by 6 doesn't change anything