abbasiime wrote:
I quite didn't get how the expression 2x^2 + 3x - 27 = 0 was factorised
There's a formal technique for factoring quadratics that do NOT have a 1 as the coefficient of the x^2 term.
However, the GRE will never give us a super complicated expression to factor.
So, we can just play around with some numbers.
First, we know that the two x-terms must multiply together to give us 2x^2.
So, one term must be 2x and the other term must be x
So far, we have: 2x^2 + 3x - 27 = (2x +/- ??)(x +/- ??)
For the two missing terms, we know that they multiply to get -27
So, let's try some options.
How about 27 and -1?
We get: 2x^2 + 3x - 27 = (2x + 27)(x - 1)
To see if this works, we'll EXPAND (2x + 27)(x - 1) using the FOIL Method
(2x + 27)(x - 1) = 2x^2 - 2x + 27x - 27
= 2x^2 + 25x - 27
This doesn't equal 2x^2 + 3x - 27. So, our missing numbers are NOT 27 and -1
How about 9 and -3?
We get: 2x^2 + 3x - 27 = (2x + 9)(x - 3)
To see if this works, we'll EXPAND (2x + 9)(x - 3) using the FOIL Method
(2x + 9)(x - 3) = 2x^2 - 6x + 9x - 27
= 2x^2 + 3x - 27
Perfect - this equals the original expression!
So, our missing numbers are NOT 27 and -1
Go, given: 2x^2 + 3x - 27 = 0
Factor to get: (2x + 9)(x - 3) = 0
This means EITHER 2x + 9 = 0 OR x - 3 = 0
If 2x + 9 = 0, then x = -4.5
If x - 3 = 0, then x = 3
Our solutions are x = -4.5 and x = 3