wingtipti

2022-01-20

Why are quadratics factored into 2 brackets?

$2{x}^{2}+8x+6$

$2x(x+4+3\times \frac{1}{x})$ or $2x(x+4+\frac{3}{x})?$

Appohhl

Beginner2022-01-21Added 11 answers

Step 1

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

$p\left(x\right)=2{x}^{2}+8x+6$ ,

$p\left(0\right)=6$ .

However if we call

$g\left(x\right)=2x(x+8+\frac{3}{x})$ ,

$g\left(0\right)$ is not defined.

Clearly$p\left(x\right)=g\left(x\right)\mathrm{\forall}x\ne 0$ , but we lose continuity and differentiability at $x=0$ for g.

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

Using your example

However if we call

Clearly

KickAntitte06

Beginner2022-01-22Added 11 answers

Step 1

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

$x+\frac{1}{x}$ e.g.

$a{x}^{4}+b{x}^{3}+2a{x}^{2}+bx+a$

$=(a{(x+\frac{1}{x})}^{2}+b(x+\frac{1}{x})+c){x}^{2}$

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.

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

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.

RizerMix

Expert2022-01-27Added 600 answers

Here's a few reasons: 1) No one likes division. It is difficult. 2) This factoring doesn't tell you what the roots might be. 3) Either way you have 4 ''terms.'' 4) It is not as neat. 5) Note that: $2x(x+8+3\times \frac{1}{x})=(2x)(x+8+3\times \frac{1}{x})$

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