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 ...
x2 + 3x + 2 re
mem
ber
1 & 2
... with the sum of the faktors: one-plus-two
x2 + 3x1x+2x + 2
then drop those parenthesis
x2 + 3x 1x+2x + 2
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
x2 + 1x + 2x + 2
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
xx + 1 + 2x + 2
what are the faktors of its elements?
xx + 1 + 2x2 x + 21 2
do they have any common faktors? yes, a two
xx + 1 + 2x2 x + 21 2
pull that two out, and to the left, of the parenthesis
xx + 1 + 2 2x 2 x + 2 1 2
multiply the two infront, and restore inside the parenthesis
xx + 1 + 2 (x + 1)
ok, this is somehow much worse
xx + 1 + 2x + 1 !!!
???
!!!
think of this all as one big group
xx + 1 + 2x + 1
then think of (x+1) as a single term
xx + 1 + 2x + 1
as though each term has (x+1) as a faktor
x + 1 x, (x+1) + x + 1 2, (x+1)
pull that (x+1) out to the left and see what happens
(x+1) x + 1 x, (x+1) + 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
...