# How to factorize quadratic equations quickly?

It takes me more than a minute to quickly factorise this kind of quadratic expression. $$3n^2 -53n + 232$$

I need to solve them in less than $$10$$-$$15$$ seconds. Please tell me a way I can solve them.

• After using the prime factorisation of $3$ and $232$, it just comes down to observing that $3\times 8+1\times 29=53$.
– JC12
Commented Oct 24, 2020 at 6:38
• Since you are asking for this kind of equation in general, you are out of luck. Use the quadratic formula and a calculator, if you want to get a result in under 15 seconds.
– vvg
Commented Oct 24, 2020 at 6:41
• I have an exam to give without using calculators.
– user821898
Commented Oct 24, 2020 at 6:42
• @JC12 I hope prime factorisation doesn’t take much of my time .I will try it.Thnx
– user821898
Commented Oct 24, 2020 at 6:44
• Please use MathJax for all equations on this site. There's a tutorial with lots of examples at math.meta.stackexchange.com/questions/5020 Commented Oct 24, 2020 at 7:17

$$3n^2 -53n + 232$$

I use a process called "A-C Method", "Grouping", or "Split the Middle Term".

$$3*232=696$$

Now, what two numbers add to -53 but multiply to 696?

$$-23 * -30$$? That's $$690$$. Close.

Now, keep in mind, (a+b)(a-b) is a maximum when b=0. Therefore, since we are at $$690$$, and wish a higher result, the numbers need to get closer, so we just move by 1, since $$690$$ was so close to $$696$$.

$$-24 * -29$$? That's it! Second guess.

$$3n^2-24n-29n+232$$ (we split the middle term so the 24 is a nice multiple of 3)

$$(3n^2-24n)-(29n-232)$$ (Now, we group, and be mindful of that minus)

$$3n(n-8)-29(n-8)$$

$$(3n-29)(n-8)$$

Will update/edit if OP needs clarification.

• You don’t which way to go then.You may think 65*12 as well and go the other way.So to not do that, I wanted a quicker trick and not make mistake in the way you’re going.
– user821898
Commented Oct 24, 2020 at 16:02
• @robertpatrick - of course I do. (a+b)(a-b) gets larger as b goes to zero. If I failed to articulate that, I should edit it in. But the process itself is 2-3 guesses, tops. (Update, I edited a note for clarity) Commented Oct 24, 2020 at 16:10
• @PM2Ring - Yes. I am 58 and work in a HS, a second chapter, post retirement. As a student, it was guess and check. In my second year in the HS, a student showed me this, and I have used it ever since. 2/3 of students embrace it, and about 1/3 reject it respectfully, as Robert seems to have done. My approach is that there are many ways to solve a problem. Find the method you are most comfortable and successful with. Commented Oct 24, 2020 at 16:16
• I didn’t understand what you meant there.I want to know how were so sure of using the numbers the way you did and not the way I did.I use the same method but am not getting it when I go the wrong way.🙂
– user821898
Commented Oct 24, 2020 at 16:41
• The two numbers must sum to a known value, here, -53. Whatever 2 numbers you pick, you have a product as well. To get the product to move higher, bring the numbers closer, to go lower, move the numbers further away. The truth is, once you understand the process, with a bit of practice, the result comes very fast. Commented Oct 24, 2020 at 16:48

As Raffaele says, it's tricky when the leading coefficient isn't $$1$$, or a perfect square. Especially if you don't know if the expression is reducible or not. You can tell that by looking at the discriminant, $$b^2-4ac$$, but if you've gone that far, you might as well eliminate all the guesswork and just use the quadratic formula. ;)

But anyway, assuming it is factorisable, because $$3$$ is prime, $$-53<0$$ and $$232>0$$, we know the factors must be of the form $$(3n-u)(n-v)$$ with $$3v+u=53$$ and $$uv=232$$ for $$u,v>0$$.

Now $$232=8\cdot29$$ and any multiple of $$29>53$$, which eliminates all possibilities except $$3v+29=53$$. Thus $$u=29$$ and $$v=8$$, so the desired factorisation is $$(3n-29)(n-8)$$

Another approach is to complete the square, but it is a bit tedious, and the numbers may get too large for rapid mental calculation.

We need the leading coeffient to be a square, and the coefficient of the $$n$$ term to be even. So we have to multiply this expression by $$12$$. Thus $$36n^2 - 12\cdot53n + 12\cdot8\cdot29$$ $$(6n)^2 - 2\cdot6\cdot53n + 53^2 - 53^2 + 12\cdot8\cdot29$$ $$(6n-53)^2 - (53^2 - 12\cdot8\cdot29)$$ That constant term looks pretty bad, until we notice that $$53=29+24$$ and $$4\cdot24=12\cdot8$$ $$(6n-53)^2 - ((29+24)^2 - 4\cdot24\cdot29)$$ $$(6n-53)^2 - (29-24)^2$$ $$(6n-53)^2 - 5^2$$ $$(6n-53+5)(6n-53-5)$$ $$(6n-48)(6n-58)$$ $$12(n-8)(3n-29)$$ And now we can drop that multiplier of $$12$$ $$(n-8)(3n-29)$$

As I said, it's a bit tedious, but we got there eventually. ;)

$$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where$$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a};\;x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$

• I didn’t understand it.How would you use it for my questions.
– user821898
Commented Oct 24, 2020 at 6:42
• @robertpatrick You didn't say what level are you in math. Another tricky way is $$3 n^2 - 29 n - 24 n + 232=3n(n-8)-29(n-8)=(n-8)(3n-29)$$ Commented Oct 24, 2020 at 6:47
• How did you find that 24 and 29 are those numbers sir.
– user821898
Commented Oct 24, 2020 at 6:48
• @robertpatrick I told you it was tricky. Use the quadratic formula. Commented Oct 24, 2020 at 6:50
• @Raffaele - I can't help but wonder. And this comment is really with MathEducators.SE in mind. Would a teacher be okay with this? We teach quadratic equation. We teach factoring, by a couple different methods. But, if a student was asked to factor, and 'show work', this might not be accepted. Commented Oct 24, 2020 at 17:43

I am not sure if this helps you. If $$ax^2+bx+c=0$$ has a rational solution $$\frac p q$$ then $$p\mid c$$ and $$q\mid a$$. So if $$ax^2+bx+c=0$$ then the only integer solutions are divisors of $$232$$ because $$3$$ is prime. We have $$232=2^3\cdot 29$$. So the divisors are $$\{\pm 1, \pm2, \pm4, \pm 8\, \pm 29, \pm2 \cdot 29, \pm4 \cdot 29, \pm 8 \cdot 29\}$$

If we have an integer solution $$x_1$$, the numerator $$p$$ of the other solution $$x_2=\frac p q$$ is a divisor of $$232$$, too, and it satisfies $$53-3\cdot x_1=p\in D$$. From this we see that 53-3*8=29$$asns so$$8\$ is the solution.

But all in all I think using the formula

$$x_1=\frac{-b-\sqrt{b^2-4ac}}{2};\;x_2=\frac{-b+\sqrt{b^2-4ac}}{2}$$

is the most efficient way. Calculating $$b^2-4ac$$ shouldn't be a problem, and calculating the square root of a 4 digit number is also rather simple. It is a two digit number. The left digit can be estimated by comparing the size of the numbr to $$10^2,20^2,...$$ and the right digit can be estimated by comparing the squares modulo $$10$$.