Question 650112


what you are looking for is called {{{expansion}}} and {{{grouping}}} 

so, you have to to expand "{{{bx}}}" into {{{two}}} terms, such that one term will have a {{{common}}} {{{factor}}} with {{{a}}} and the other will have a {{{common}}} {{{factor}}} with {{{c}}}

for example:

{{{2x^2 - x - 3}}}

you are going find numbers that are factors of {{{-6}}} (these are {{{2 * -3}}}) ) that {{{add}}} {{{up}}} to {{{-1}}} ({{{-x = -1x}}})
these factors are {{{-3}}} and {{{2}}}

so you would rewrite the equation as:

{{{2x^2 + 2x - 3x - 3}}} (doing so,you didn't really change anything, because {{{2 - 3 = -1}}}, so the integrity of the expression is still there)

now you factor in parts. from the first part, take out {{{2x}}}, from the second, take out {{{-3}}}

so, you have:

{{{2x(x + 1) - 3(x + 1)}}} . . . .  (if you notice, both of the terms in the ()'s are the same, this will always happen if it doesn't, you have done something wrong

then you say:

{{{(2x - 3)(x + 1)}}}...that's factored 

if u multiply this out, you will get back to the original expression of {{{2x^2 - x - 3}}}