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I've been working through some tutorials that show how to factorised in multiple ways, e.g.

$2x^2 + 11x +12$

Can be factorised to:

$(2x + 3)(x + 4)$

... and to:

$2[(x + \frac{11}{4})^2 - \frac{25}{16}]$

When is one technique preferred instead of the other? I haven't seen this advice yet in any of the books and tutorials I'm reading.

Chris Snow
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    There is no hard and fast rule. I'm assuming you're solving it equal to 0. If you can quickly notice the factorization, it is faster. Otherwise the quadratic formula (the generalization) $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ is the way to go. – Gregory May 01 '17 at 19:37
  • The thing does not factor unless $b^2 - 4ac$ is a perfect square. In that case, it is guaranteed to factor. I gave a recipe for when you cannot do it by eye, apparently I also posted a C++ program that does that recipe http://math.stackexchange.com/questions/1867805/formula-for-factorization-of-a-quadratic-equation/1868137#1868137 – Will Jagy May 01 '17 at 19:54

2 Answers2

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The first method is better when you are looking for the roots, the second -- for putting the expression into a specific form, usually done to understand its geometric shape. Like, from your example, it is an upward-open scaled parabola with vertex at $(-11/4,-25/16)$.

gt6989b
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Each has their own advantages. If you can spot the roots explicitly, the factorisation into a product is useful, and tells you the roots exactly. But it requires spotting the roots, unless one uses the quadratic formula. But where does the quadratic formula come from?

Completing the square tells you something different: in the form $$ a(x+p)^2+q, $$ it tells you that the minimum is unique, occurs when $x=-p$, and is $q$. And it is entirely algorithmic: $$ ax^2+2bx+c = a\left(x+\frac{b}{a}\right)^2 + c-\frac{b^2}{a}, $$ which one can do in one's head, unlike the quadratic formula in the general case. And from this, one derives the quadratic formula: if this is zero, $$ \left(x+\frac{b}{a}\right)^2 = \frac{b^2-ac}{a^2}, $$ so $$ x = \frac{-b \pm \sqrt{b^2-ac}}{a}. $$ This is the easiest way to get the quadratic formula: otherwise, one would be messing about with the rules about sums and products of roots.

Chappers
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