you are asked to faktor this expression
you can do this! use your mathrobatics
to conquer all that lies before you
x2 +
3x +
2
1×2
21 2
that means to break it up into two parenthesis
this is called a "quadratic expression"
and it needs to be broken up into two
groups of parenthesis, each that has only
two things inside it - something with only
two things, or two "terms", is call a
"binomial"
x2 +
3x +
2
1×2
21 2
well, how are you supposed to do that?
with your brain ... or the internet,
whichever is more convenient
x2 + 3x + 2 ???
!!!
step one is to look at the first, and last, coefficients
in this case, they are one and two; the coefficient
of just "x" is one, because there is a, sort of,
invisible "times one" infront of the x
1x2 +
3x +
2
1×2
21 2
we multiply them together
1x2 +
3x +
2
1×2
21 2
one times two is not simple, but it equals two
1x2 +
3x +
2
1×2
21 2
now, think about the faktors of two ...
1x2 +
3x +
2
1×2
21 2
are there faktors of two we can combine to make three?
x2 +
3x +
2
1×2
21 2
we only have one option, but those actually werk!
x2 +
3x +
2
1×2
21 2
we're going to replace that middle term ...
x
2 +
3x
+ 2
re
mem
ber
1 & 2
... with the sum of the faktors: one-plus-two
then drop those parenthesis
now, you're thinking "how does THIS help?!!"
don't worry - things are going to get
much worse
x2 + 1x + 2x + 2
why
???!!!
group them into two different groups
would it be possible to faktor the left group?
x2
x, x
+
1x
1, x
+
2x + 2
yes, they each have an x in common
x2
x, x
+
1x
1, x
+
2x + 2
move that equis out to the left
x
x2
x, x
+
1x
1, x
+
2x + 2
multiply by x infront, and restore the remaining faktors
x
x + 1
+ 2x + 2
now, focus on the right group of parenthesis
what are the faktors of its elements?
do they have any common faktors? yes, a two
pull that two out, and to the left, of the parenthesis
x
x + 1 +
2
2x
2 x
+
2
1 2
multiply the two infront, and restore inside the
parenthesis
ok, this is somehow much worse
x
x + 1
+
2
x + 1
!!!
???
!!!
think of this all as one big group
then think of (x+1) as a single term
as though each term has (x+1) as a faktor
x·x + 1
x, (x+1)
+
2·x + 1
2, (x+1)
pull that (x+1) out to the left and see what happens
(x+1)
x·x + 1
x, (x+1)
+
2·x + 1
2, (x+1)
multiply by (x+1) in front, and restore the faktors
x+1
x + 2
you can convert those brakettes into parenthesis
x+1
[(
x + 2
])
wow - look at that! it finally seems better
x+1 x+2
now we just need to check if this is correct
...