alphabeta1234 wrote:

What's the quickest way to factor a polynomial that has a coefficient other than 1 on the x^2 term? i.e. without using the quadratic equation.

e.g.

f(x)=2x^2-9x+9

The best alternative way is to graph it out by plugging in numbers until you see what the graph looks like. Just start anywhere:

x=0: \(2(0^2)-9(0)+9\); y = 9

x=1: \(2(1^2)-9(1)+9\); y = 2

x=2: \(2(2^2)-9(2)+9\); y = -1

x=3: \(2(3^2)-9(3)+9\); y = 0

x=4: \(2(4^2)-9(4)+9\); y = 5

Attachment:

2x^2 - 9x + 9.jpg [ 32 KiB | Viewed 1893 times ]
So you know that one x-intercept is at 3 and the other is in between 1 and 2. From here, I'd just start guessing unless you know how to find the vertex of a parabola (which you

definitely do not need to know for the GMAT). If you can pull that out of your back pocket, the second x-intercept is equidistant from the vertex and the first x-intercept that you found. You will find that your other answer is \(\frac{3}{2}\). It will be much more difficult to plug in x to find y for multiple points if they are complex quadratic equations or their intercepts and vertex are far apart.

Is there a reason you do not want to solve it the usual way?

\(2x^2-9x+9 = (2x-3)(x-3)\)

x = \(\frac{3}{2}\) and 3

If it is because you don't understand it, it may be something you want to invest a little time into learning.

Just try all the combinations that have the product of the two integers equal the last number in the quadratic formula and find one that works. Obviously, it will get more complicated if the coefficient before \(x^2\) is not prime and/or there are multiple ways of dividing out the integer.

\((2x-9)(x-1) = 2x^2-11x+9\); no

\((2x+9)(x+1) = 2x^2+11x+9\); no

\((2x-1)(x-9) = 2x^2-19x+9\); no

\((2x+1)(x+9) = 2x^2+19x+9\); no

\((2x+3)(x+3) = 2x^2+9x+9\); no

\((2x-3)(x-3) = 2x^2-9x+9\); yes

You will get better at it over time. There is a reason that we haven't been taught other ways... this is the most efficient.

Kudos if this was helpful