Factoring a Polynomial W/O Quadratic formula : GMAT Quantitative Section
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# Factoring a Polynomial W/O Quadratic formula

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08 Jun 2013, 17:41
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
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08 Jun 2013, 18:46
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 1852 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.

Last edited by lchen on 08 Jun 2013, 19:29, edited 3 times in total.
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08 Jun 2013, 19:01
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The example problem that you posted took me a little less than 10 seconds to do in my head, without writing anything down.

What two numbers, when multiplied, are equal to + 9, but when added together (and one of the two numbers is multiplied by 2) equals -9?
Well the only things that multiply out to 9 are 3x3, -3x-3, 9x1, and -9x-1.

Then I have to find out which of these add up to -9 if one of them is multiplied by 2 (I get the 2 from 2x^2.. if the question was 5x^2 i would look for something multiplied by 5)

only thing that holds up here is -3, -3

These aren't the factors though. I now have to determine which number was multiplied by two when I added them up to get -9. In this case, it doesn't matter since they're both -3... (-3) + (2*-3) = -9

Now I divide the one that was multiplied by two and get (-3/2)

Then I flip the signs, and there are my factors. It's a very fast process.

If you're looking at

Ax^2 + Bx + C

It's really easy to find the factors if A is one... just figure out what two numbers add up to B and multiply out to C

If A is a prime number, then you have twice as many possibilities: one of the two numbers has to be multiplied by A when figuring out B, but C is left alone...

If A isn't prime, then it gets a bit more complicated but the same theory holds true.. just use your head and practice

EX: 20x^2+102x+10

since the adding part (b) is so much higher than the multiplying part (c), we know that the 20 is probably not evenly distributed..

(20x + 2) (x + 5)
=
20(x+.1)(x+5)

x= -.1, -5
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09 Jun 2013, 02:02
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

Hope it helps.
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10 Jun 2013, 20:49
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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

$$f(x) = Ax^2 + Bx + C = 2x^2 - 9x + 9$$

Sum of factors $$= B = -9$$
Product of factors $$= AC = 2*9 = 18$$

Two numbers that multiply to give 18 and add up to -9 are -3 and -6 (break down 18 into 2*3*3 and combine them in various ways to get a sum of 9)

Factors are:

$$A(x - 3/A)(x - 6/A) = 2(x - 3/2)(x - 6/2)$$

x can take 2 values: 3/2 and 3
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27 Jun 2013, 21:52
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

For any polynomial, one needs to check whether the polynomial can be factorized and further solved. The possible answer choices could be integers,fractions or imaginary values(i).

Here,

2x^2-9X+9 as ax^2+bx+c

Now simply multilpy a and c .. i.e 18.
factors of 18 are 2,3,3. i.e 2*3*3=> 18

Now, try to form combinations from 2,3 and 3 to get the value of b as -9 and the product of terms as +18.
i.e -6 and -3, now 6*3=>18 and (-6) + (-3) => -9

Therefore, we can rewrite 2x^2-9x+9 as 2x^2 -6x-3x +9

2x^2-6x-3x+9
2x(x-3) -3(x-3)
(2x-3)(x-3)

x can be 3 or 3/2.
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22 May 2016, 07:57
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

An alternative approach: just factor out the coefficient of x^2 and solve as normal.

f(x) = 2(x^2 - 4.5x + 4.5)

Even though 4.5 isn't an integer, it still shouldn't take too much energy to work out that -3-1.5 = -4.5, and (-3)(-1.5) = 4.5.
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Re: Factoring a Polynomial W/O Quadratic formula   [#permalink] 22 May 2016, 07:57
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