3x²+5x+2
Multiply the 3 by the 2 ignoring signs. Get 6.
Write down all the ways to have two positive integers
which have product 6, starting with 6*1
6*1
3*2
Since the last sign in 3x²+5x+2 is +, ADD them,
and place the SUM out beside that:
6*1 6+1=7
3*2 3+2=5
Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 3x²+5x+2
which is 5
So we replace the number 5 by 3+2
3x²+5x+2
3x²+(3+2)x+2
Then we distribute to remove the parentheses:
3x²+3x+2x+2
Factor the first two terms 3x²+3x by taking out the
greatest common factor, getting 3x(x+1)
Factor the last two terms +2x+2 by taking out the
greatest common factor, +2, getting +2(x+1)
So we have
3x(x+1)+2(x+1)
Notice that there is a common factor, (x+1)
3x(x+1)+2(x+1)
which we can factor out leaving the 3x and the +2 to put
in parentheses:
(x+1)(3x+2)
------------------------------
5x²-14x-3
Multiply the 5 by the 3 ignoring signs. Get 15
Write down all the ways to have two positive integers
which have product 15, starting with 15*1
15*1
5*3
Since the last sign in 5x²-14x-3 is -, SUBTRACT them,
and place the DIFFERENCE out beside that:
15*1 15-1=14
5*3 5-3=2
Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 5x²-14x-3
So we replace the number 14 by 15-1
5x²-14x-3
5x²-(15-1)x-3
Then we distribute to remove the parentheses:
5x²-15x+1x-3
Factor the first two terms 5x²-15x by taking out the
greatest common factor, 5x, getting 5x(x-3)
Factor the last two terms +1x-3 by taking out the
greatest common factor, +1, getting +1(x-3)
So we have
5x(x-3)+1(x-3)
Notice that there is a common factor, (x-3)
5x(x-3)+1(x-3)
which we can factor out leaving the 5x and the +1 to put
in parentheses:
(x-3)(5x+1)
---------------------------
The others are done just like the first one.
I wrote a lesson on this method of factoring. Go here:
http://www.algebra.com/algebra/homework/Expressions-with-variables/change-this-name32371.lesson
Edwin
