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I'm quite eager to learn what is the difference between factoring quadratics (the $(x + a)(x + b)$ method), and using the typical formula (where $x = (-b \pm \sqrt{b^2 - 4ac})/2a$), and in what situations should I use either of them.

Take this for a problem (I didn't make it myself)

enter image description here

I know that the quadratic formula is required to solve the answer, but why not factor?

I previously asked this on Math Overflow, but this is the right place.

Thanks in advance.

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    $\begingroup$ What may be worth doing in this case is cancelling a constant factor rather than a linear one, to make the numbers more manageable in the quadratic formula, so we go from $16t^2-164t+92=0$ to $4t^2-41t+23=0$. However, you'll probably need a calculator either way, but it's worth bearing in mind for other quadratics later. $\endgroup$ – J.G. Mar 21 at 11:14
  • $\begingroup$ See also math.stackexchange.com/questions/3154893/… $\endgroup$ – Gerry Myerson Mar 21 at 11:31
  • $\begingroup$ Who said you cannot factor? $\endgroup$ – Taladris Mar 21 at 13:57
  • $\begingroup$ IIRC the primary value of factoring (wrt to particular solutions) is finding the zeroes (where the function crosses the x-axis). (x-3)(x+5)=88 (in fact, converting to quadratic, then using quadratic forms for closed form is what i'd do) isn't particularly easy to solve for an exact solution, but (x-3)(x+5) = 0 can be solved by inspection (x = 3 or x = -5). $\endgroup$ – Kristian H Mar 21 at 19:06
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Which is appropriate to use when is simply a matter of one's computational speed.

Factoring, for instance, is something we often think about only being doable when we have rational or integer coefficients, and generally smaller ones. For the most part, that's true - if you don't want to spend a painstaking amount of work, you want to focus on those cases. For instance, consider your equation:

$$h(t) = -16t^2 + 164t - 92 = 0$$

To properly factor this, you need to figure out which factors of $92\cdot -16=-1472$ have a sum of $164$. You could sit there all day enumerating the various factors and checking the sum, or you can just go "screw this" and decide to just plug-and-chug using the quadratic formula, and probably solve it faster. Really, try it - see which one you solve faster!

(Anecdote: when you see a "round to the nearest hundredth" tenet in a question, a lot of the time that means the question will have not-so-nice roots, i.e. they'll be irrational probably. Which touches on another issue: if the square root in the quadratic formula is not a perfect square, then you definitely cannot factor it just by using integers, you might have to reach for square root factors instead ... which makes this even more of a mess and is why factoring isn't always optimal. Let's nail that point home...)

Okay, so forget the big numbers and such - let's say you're great at factorizing even large numbers. But then there's a problem: factoring doesn't necessitate integers/rationals. They are easier to work with like that, but consider:

$$(x+4\sqrt 2)(x + 2\sqrt 3) = x^2 + (4 \sqrt 2 + 2 \sqrt 3) x + 8 \sqrt 6$$

At a glance, how would you even factor that latter expression, without knowing what led up to it? Or perhaps another, nastier, crueler one:

$$(x + 4 \sqrt 6)(x - 3 \sqrt 6) = x^2 + \sqrt 6 x - 72$$

At least if they're "nice enough" you can guess perhaps a means to factorize it, but I see no obvious way for the above polynomial. How would you even begin to approach this a priori?

I'd rather sit there and just plug-and-chug, than try to guess how to factor some weird expression. Simpler to do and process.


That's not to say either method is "invalid" in either situation. Factoring takes advantage of the fact that, if a polynomial of degree $n$ has roots $r_1,r_2,...,r_n$, then it can be written in the form

$$a(x-r_1)(x-r_2)...(x-r_n) =0$$

where $a$ is its leading coefficient. And it so happens that for integers and similar-enough polynomials, factoring can be relatively easy. But that's simply not always the case, which is why it's nice that the quadratic formula provides a quick, general solution just in case the expressions aren't nice enough - which is something we can't say for all polynomials!

In short - use whichever method you choose, both are valid. But always be cognizant of the time it takes for you to get the solution any one way, as opposed to another way. No sense in wasting time and energy when the end result is the same.

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    $\begingroup$ I think there's a typo in your first radical equation. $(x + 4\sqrt3)(x + 2\sqrt3) = x^2 + 6\sqrt{3}x + 24.$ $\endgroup$ – Bladewood Mar 21 at 9:44
  • $\begingroup$ Hm, thanks. I think I'll replace it with a different one that'll better emphasize my point, to be honest $\endgroup$ – Eevee Trainer Mar 21 at 9:46
  • $\begingroup$ $x^2 + (4 \sqrt 2 + 2 \sqrt 3) x + 8 \sqrt 6$ seems rather trivial to factor. Given that (x+a)(x+b)=x^2+(a+b)x+ab - it's quick to verify that $(4 \sqrt 2)(2 \sqrt 3)=8 \sqrt 6$ $\endgroup$ – Taemyr Mar 21 at 12:11
  • $\begingroup$ $x^2 + (4 \sqrt 2 + 2 \sqrt 3) x + 8 $ would be much harder. $\endgroup$ – Taemyr Mar 21 at 12:17
  • $\begingroup$ My point was that "okay, so even some of those with noninteger coefficients might be easy if you're observant, but then .. (much harder example here)". $\endgroup$ – Eevee Trainer Mar 21 at 20:15
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First of all, in the context of a mechanics problem like the one where this equation originates, often you'll only want an approximate decimal solution in the first place. That said, henceforth let's assume we really want an exact answer and that the coefficients of the quadratic equation are integers, or even rational numbers.

One can always in principle use either method, but the roots of $$a x^2 + b x + c = 0$$ often contain radical expressions, so it's often not easy to factor these. By inspecting the quadratic formula we can actually say when this will happen: The solutions of the equation will involve radicals (and factoring is, roughly speaking, hard) except when the quantity $$\boxed{\Delta := b^2 - 4 a c}$$ is a perfect square.

Example In the linked example, we're asked to find the solutions of $-16 t^2 + 164 t = 92$. Rearranging gives $$16 t^2 - 164 t + 92 = 0,$$ and computing with a pocket calculator shows that $\Delta = 21\,008$ and that $\Delta$ is not a perfect square, so the roots will involve $\sqrt\Delta = \sqrt{21\,008} = 4 \sqrt{1\,313}$---definitely not a quantity you want to deal with when factoring manually, if you have a choice.

Remark The quantity $\Delta$ is called the discriminant of the quadratic polynomial, and it gives us a good deal of information about it. For example, if it is negative, then the radical expression in the quadratic formula is an imaginary expression, and so the quadratic equation has no real roots at all.

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    $\begingroup$ Minor note: the polynomial in the question can't be meaningfully be factored that way. If you look closer, the question wants $t$ such that $$h(t) = 92 = -16t^2 + 164t$$ So in this scenario, you won't have that nice $(ax+b)x$ factoring $\endgroup$ – Eevee Trainer Mar 21 at 6:59
  • $\begingroup$ Oh, you're right. I just glanced at the polynomial and didn't see that we wanted to find the solutions of $h(t) = k$ rather than the roots of $h$. I'll adjust my answer---thanks for the catch! $\endgroup$ – Travis Mar 21 at 7:04
  • $\begingroup$ Perhaps the downvoter would explain their objection? $\endgroup$ – Travis Mar 21 at 19:03
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In fact there is no "by factoring" resolution method.

Unless the factorization is obvious (e.g. $3x^2-2x$ or, with more practice, $x^2+7x+12$), you cannot find the factorization "out of the blue" and you need to resort to... the quadratic formula.

E.g. try to factor

$$x^2-x-1$$

and

$$x^2-x+1.$$


An alternative approach known as "completing the square", with the step

$$ax^2+bx+c=a\left(\left(x+\frac b{2a}\right)^2+\cdots\right)=0$$

is indeed a resolution method. But looking closely, it is just a way to re-establish the quadratic formula.


A second alternative is to use the Vieta relations,

$$r+s=-\frac ba,\\rs=\frac ca$$

and solve for $r$ and $s$. There can be an obvious solution, and you are just in the case of factorization "at a glance". The second equation leads you to the "rational root theorem", which can be used when the coefficients are integers (factor $a$ and $c$ and try the ratios of the factors).

In the general case, analytical resolution of this system brings you back to… the quadratic formula.


As regards the way to present the answer, writing

$$r=\cdots,\\s=\cdots\ $$ or $$ax^2+bx+c=a(x-r)(x-s)$$ convey the same information.

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In fact, they are equivalent so it does not matter. $$(x+A)(x+B)=0$$ Expands to: $$x^2+(A+B)x+AB=0$$ Substituting in quadratic formula: $$x=\frac{-(A+B)\pm\sqrt{(A+B)^2-4AB}}{2}$$ $$x=\frac{-(A+B)\pm\sqrt{(A-B)^2}}{2}$$ Simplifies to: $$x=\frac{-(A+B)\pm(A-B)}{2}=-A,-B$$ which are exactly the roots found by factoring

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