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

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Brent Hanneson – Creator of greenlighttestprep.com

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