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Why is it that no one seems to factor quadratics into just one bracket Eg: $$2x^2+8x+6$$ into $$2x\left(x+4+3\cdot\frac{1}{x}\right)\quad\text{or}\quad 2x\left(x+4+\frac{3}{x}\right)\quad ?$$

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Both of your given `factorizations' contain rational expressions, which are more difficult to find solutions of x. A polynomial that is factored gives you the roots of the equation. –  Joshua Shane Liberman Feb 9 '11 at 15:28
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up vote 11 down vote accepted

The natural domain of the function $f(x) = 2x^2 +8x+6$ is all real numbers (remember, the "natural domain" is the collection of all real numbers for which the formula 'makes sense', or yields a real number).

The natural domain of $g(x) = 2x\left(x + 4 + \frac{3}{x}\right)$ is all real numbers except $x=0$, because you cannot plug in $x=0$ into the formula (division by zero makes the universe explode, after all).

So one major problem with writing $2x^2 + 8x + 6 = 2x\left(x + 4 + \frac{3}{x}\right)$ is that it is not true. They are not equal! The left hand side makes sense at $x=0$, but the right hand side does not.

Another major problem is that polynomials are nice and easy, while rational functions are less nice and less easy (just wait until you get to integrals: if you have to integrate a rational function, you'll groan and settle yourself in for some strenuous work; if you have to integrate a polynomial, you'll smile at how easy your life is going to be). So you would much rather deal with polynomials than rational functions. Here you start with a polynomial, $2x^2+8x+6$, and end up with a product of a polynomial and a rational function, so you've made your life that much harder.

Another issue is that one is very interested in knowing when a function is equal to $0$. If you factor as usual, $2x^2+8x+6 = 2(x^2+4x+3) = 2(x+1)(x+3)$, then it is very easy to figure out where $2x^2+8x+6$ is zero, because a product is zero if and only if at least one factor is zero, so you would need either $2=0$ (impossible), $x+1=0$ (which means $x=-1$) or $x+3=0$ (which means $x=-3$). With a bit of practice you can basically just "read off" where the product is zero. If we factor like you suggest, $2x^2+8x+6 = 2x\left(x + 4 + \frac{3}{x}\right)$, first you might be misled into thinking that $x=0$ will make it zero, since you have the factor $2x$ (which would be zero at $x=0$; but unfortunately, you cannot plug in $x=0$ into the second factor, so that causes problems). And second, to figure out when $x+4+\frac{3}{x}$ is $0$, you would end up having to figure out where $x^2+4x+3=0$, that is, your original problem. So the factorization you propose doesn't simplify the problem, and introduces the possibility of error.

That's a couple of reasons, at any rate.

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Here's a few reasons:

  • No one likes division. It is difficult.

  • This factoring doesn't tell you what the roots might be.

  • Either way you have $4$ "terms."

  • It is not as neat.

  • Note that: $2x(x+8+3\cdot \frac{1}{x}) = (2x)(x+8+3\cdot \frac{1}{x})$

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In fact in some contexts it is fruitful to employ such rational factorizations. For example, see this question whose solution involves rewriting symmetric polynomials as polynomials in $\rm\ x + 1/x\ $ e.g.

$$\rm a\ x^4 + b\ x^3+ 2\:a\ x^2 + b\ x + a\ \ =\ \ \bigg(a\ \bigg(x+\frac{1}x\bigg)^2 + b\ \bigg(x+\frac{1}x\bigg)+c\bigg)\ x^2$$

Generally contexts enjoying some sort of innate rational structure or symmetry may similarly profit from factorizations or compositions involving rational function. You will discover some pretty examples of such if you study Galois theory.

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Factoring a constant out of quadratics in order to make it monic is fairly common. However, factoring it in the method that you have introduces a removable singularity at $x=0$. Using your example $p(x) = 2x^2+8x+6$, $p(0)=6$. However if we call $g(x) = 2x(x+8+\frac{3}{x})$, $g(0)$ is not defined. Clearly $p(x) = g(x) \forall x \neq 0$, but we lose continuity and differentiability at $x = 0$ for $g$.

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Why writing something in a complicated way, while you can write it in a simpler way?

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Because I find it easier to divide by x instead of using trial and error(effectively) on a non-calculator paper. –  Jonathan. Feb 9 '11 at 18:41
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