How to do factoring faster?

Question

Hello,

This is more a general question, not directly related to a specific question of the GMAT.

I need a lot of time for factoring, for example here:

(t^2)+4t-672=0 --> factor
(t-24)(t+28)=0

How do you do that? What is your approach?
What I would do is, split 672 up into its factors, which are 2^5*3*7 ... and then I try every single calculation to find the right figures.
In some cases this takes me around 2 minutes or so...

I also know another approach from a YouTube video, where you draw a big X. On top you write the product of a and c (here it is 1*(-672)), and b on the bottom (here 4).
The left and right side will be your two values of x, and their product will match the top value and their sum will match the bottom value.
But still, you have to come up with the two values of x first, and that's where I need too much time. How to improve?

Thanks,
Tom

Bunuel · Answer

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm Hope it helps.

Blackbox · Answer

Finding factor pairs for a big number like 672 in the quadratic equation is time consuming. At the same time, with prime factors, finding the right combination that would yield the exact co-efficient is also time consuming. What  Bunuel  recommended as solution is very resourceful but, at the same time, does not address situations where you have big numbers like 672. I wish there were a quicker method. Nevertheless, thank you Bunuel.

broall · Answer

Generally, we can't factoring quadratic equation just by Prime Factorization. The best way is to solve that equation.

However, I would like to introduce a better solution.

$x^2+4x-672=0$

Note that $(a+b)^2=a^2 + 2ab + b^2$

Hence $x^2+4x-672 = 0 \implies x^2 + 2 \times x \times 2 + 2^2 = 676 \implies (x+2)^2 = 676$

Now make Prime Factorization: $676 = 2 \times 338 = 2^2 \times 169 = 2^2 \times 13^2 = 26^2$

Hence $(x+2)^2 = 26^2 \implies (x+2)^2 - 26^2 =   =(x-24)(x+28)=0$

ccooley · Answer

I'd use the Quadratic Formula to factor that equation. It's often faster for equations where it isn't immediately clear how to factor. If you look at the equation like this:

ax^2 + bx + c = 0

The formula to find the values of x is as follows:

first value: x = (-b + sqrt(b^2 - 4ac))/2a
second value: x = (-b - sqrt(b^2 - 4ac))/2a

Blackbox · Answer

Thanks Cooley. I stumbled on this formula (^^) while reading OG. However, can one apply the same formula for this equation below?
9 x^2  - 25

When I applied the formula (assuming b=0), I am ending up with a negative number under the root which I believe is an imaginary number and is not tested in GMAT. Innit?

broall · Answer

Solving math through many ways, not just by applying formula.

$9x^2-25=0 \implies 9x^2=25 \implies x^2 = \frac{25}{9} \implies x = \pm \frac{5}{3}$

Blackbox · Answer

nguyendinhtuong - Yes, I am aware that it could be solved like that. But I was trying to apply the formula and see if it could be solved using the formula as well.

broall · Answer

$9x^2-25=0$

$a=9$
$b=0$
$c=-25$

The roots are $x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{\pm \sqrt{4 \times 9 \times 25}}{2 \times 9}=\frac{\pm 2 \times 3 \times 5}{2 \times 9}=\pm \frac{5}{3}$

Blackbox · Answer

Wait...how come there is no "minus" before 4ac in your solution?

broall · Answer

$b^2-4ac=0^2-4(9)(-25)=4 \times 9 \times 25$

Hope this helps.

Blackbox · Answer

Aha! Totally overlooked the preceeding sign for 'c'. Thanks for the breakdown.

JeffYin · Answer

Good question, Tom. Once you have the prime factorization, instead of trying every single calculation to find the right pair, you could use the rest of the information provided to be strategic about testing pairs of factors that multiply to give you 672. In this example, you know that the factors have to differ by 4. This tells you that:
  Because at least one of them needs to be even (since there are powers of 2 in 672), both need to be even. 
 The two factors need to be pretty close together, so don't start with one as a small factor.

Worst case scenario, perhaps you start by choosing  2^{4} = 16  and  2*3*7 = 42 . Since these are too far apart, try swapping one of the factors of 2 in the first factor with one of the factors of 3 in the second factor, giving you  2^{3}*3 = 24  and  2^{2}*7 = 28 , which is the pair you're looking for.

I hope that using this type of reasoning can help you to get the pair in 2 or 3 tries, rather than trying all possible pairs of factors!

Cheers,
Jeff

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How to do factoring faster?

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