1. ## Factoring a Trinomial

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.

2. Originally Posted by JayRich88
There is this one trinomial that I can't seem to factor.

30x^2 -34x - 84
I get $30x^2 -34x - 84 = 2(15x^2 -17x - 42) = 2(5x+6)(3x-7)$

3. Originally Posted by JayRich88
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

$2(15x^2-17x-42)$

Check which numbers are factors of 42 and which make 15

$15 = 3 \times 5$
$42 = 7 \times 6$

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.

$2(5x+6)(3x-7) = 2(15x^2-35x+18x-42) = 2(15x^2-17x-42) = 30x^2-34x-84$

4. Originally Posted by e^(i*pi)
First of all I will factor out a 2 to make it easier

$2(15x^2-17x-42)$

Check which numbers are factors of 42 and which make 15

$15 = 3 \times 5$
$42 = 7 \times 6$

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.

$2(5x+6)(3x-7) = 2(15x^2-35x+18x-42) = 2(15x^2-17x-42) = 30x^2-34x-84$
I'm a little confused. I did not find any factors of 42 that added up to 15. Also, I was taught to find factors of the last number (in this case, 42) that added up to the middle number (in this case, -17).

5. Originally Posted by JayRich88
I'm a little confused. I did not find any factors of 42 that added up to 15. Also, I was taught to find factors of the last number (in this case, 42) that added up to the middle number (in this case, -17).
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 $(ax+b)(cx+d)$ we expand this to give $(ac)x^2 + (ad+bc)x + (bd)$.

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 $ad+bc = -17$ when it is known that $ac = 15$ and $bd =-42$

6. 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.