Can you illustrate in laymen's terms how I am supposed to factor this trinomial? 4x^2+11x-3
There are several simple methods to do this sort of problem; though I don't remember exactly how it goes, one called the 'diamond' or 'rectangle' method would probably work with this. Unfortunately since there's a constant in front of the x^2 term, everything becomes a lot more complicated.
i have a method that i do, and as long as it factors into rationals ... it works every time. the process is simple, but has a few steps
okay can you show me and help me learn for myself
my thought process goes: can the first term be factored out? if so, reduce, if not divide off and move on what are the operators of the middle and last terms? last term postive, same sign factors last term is negative, different sign factors middle term sign indicates the sign of the larger factor multiply first and last terms factor the results to find a set that matches the middle term write up factors, reduce fractions, and stick any bottoms up front
4x^2+11x-3 4 cannot be factored out, so divide it off (x /4) (x /4) 4x^2+11x-3 ; last term is negative, we have difference signs (x+ /4) (x- /4) 4x^2+11x-3 ; middle term is postive, largest factor is postive (x+ L/4) (x- S/4) now to find L and S, multiply first and last terms: 4*3 = 12 factors of 12 that "subtract" (since the signs are different) .. to get 11 12,1 ... well that worked nicely , use 12 and 1 L=12 S = 1 (x+ 12/4) (x- 1/4) , reduce the fractions (x+ 3) (x- 1/4) and fractions that remain, stick the bottom in front (x+ 3) (4x- 1) and its done
yes the first term factors from 4x^2 to (2x)(2x) or (2x)^2
if we were to do this with a more common method, the fractions would tend to get in the way, so i worked out this method :)
there is no fraction in this equation
\[4x ^{2}+11x-3\]
the common method is to first factor out the leading term and then work the problem \[4(x^2+\frac{11}{4}x-\frac34)\] if you want to deal with those fractions then do it the usual way ....
so, what 2 values multiply to get -3/4; and sum to get 11/4 ??
after about a dozen of these, i noticed a pattern than i used to develop the method i mentioned ....
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