There is this one trinomial that I can't seem to factor.
30x^2 -34x - 84
When I checked my final answer (added the inner numbers and outer numbers of the final binomial), I did not get -34.
Help is greatly appreciated.
First of all I will factor out a 2 to make it easier
$\displaystyle 2(15x^2-17x-42)$
Check which numbers are factors of 42 and which make 15
$\displaystyle 15 = 3 \times 5$
$\displaystyle 42 = 7 \times 6 $
We know the signs in the brackets are opposite because we get -42 not 42.
Afterwards it's trial and error, keep going until you get the answer like mine below.
$\displaystyle 2(5x+6)(3x-7) = 2(15x^2-35x+18x-42) = 2(15x^2-17x-42) = 30x^2-34x-84$
Factors of 42 won't add up to 15. What you're looking for is two sets of numbers independent of each other.
The first set must multiply to make 15 (as in a and c in my example below).
The second set must multiply to make 42 (as in b and d below). This is what you're used to doing by the looks of things
Since a trinomial can be written as $\displaystyle (ax+b)(cx+d)$ we expand this to give $\displaystyle (ac)x^2 + (ad+bc)x + (bd)$.
Factors of 42: 1, 2, 6, 7 , 21, 42
Factors of 15: 1, 3, 5, 15
It's very unlikely to be any of the extrema so pick the middle pairs first.
For your sum you need to find $\displaystyle ad+bc = -17$ when it is known that $\displaystyle ac = 15$ and $\displaystyle bd =-42$
Hi JayRich88,
as e^(i*pi) showed,
you are looking for 4 values a, b, c and d.
But there are only 3 terms in the equation, so we don't have enough clues to solve
using a few simultaneous equations.
Hence it's necessary to examine how they combine.
You need to factorise -42 and 15 and combine these two sets of factors
within parentheses as already shown to achieve -17x.
This is because -42 and 15 themselves do not combine to give -17x.
Usually these will be the most obvious ones, like 5 and 3, 6 and 7.
and in this case they are!
It doesn't always happen to be the most expected ones,
but that's normally the best place to start.