# Why not always use the quadratic equation

The is a very simple question, but I have just started studying quadratics. I understand how to factor them using different methods and also understand solving a quadratic using the formula, but my question is why bother learning to factorise when the quadratic equation always allows you to solve a quadratic? Is there a point to factorising that I am missing? Thanks

• Perhaps the point is that if $f$ is a polynomial with coefficients in a field $F$, then $\alpha \in F$ is a root of $f$ if and only if $x - \alpha$ divides $f$. This approach helps in understanding, among others, what it means for a polynomial to have a multiple root. – Andreas Caranti Jan 24 '16 at 11:18

If you are only interested in solving an equation like $$x^2+3x+2=0$$ you are of course free to use any method you like (completing the square, use any formula, try to factorize...). But there is more to that than just solving equations.

If you solve the equation above you will get $x_1=-2,x_2=-1$. The fact that based on these solutions you can now write $x^2+3x+2=(x-x_1)(x-x_2)=(x+2)(x+1)$ holds for polynomials with higher degree. So if you solved the equation $$x^3+6x^2+11x+6=0$$ (which is a little bit trickier and would likely involve long division of polynomials), you'd get $x_1=-3,x_2=-2,x_1=-3$ and hence could write $$x^3+6x^2+11x+6=(x+3)(x+2)(x+1).$$ Just to outline some uses:

1. Solving polynomial inequalities. Assume you are not interested in the quadratic equation $$x^2+3x+2=0$$ but in the quadratic inequation $$x^2+3x+2\geq 0.$$ There are no formulas to apply directly, so you have to use something different. Factorizing comes in handy as we can write: $$x^2+3x+2\geq 0 \Leftrightarrow (x+2)(x+1)\geq 0.$$ Now we know that the product of two (real) numbers is $\geq 0$, if and only if both numbers are $\geq 0$ or both numbers are $\leq 0$. Using this we can break that inequation down to two systems of linear inequations $$x+2\geq 0,~x+1\geq 0\quad \text{or}\quad x+2\leq0,~x+1\leq 0$$ which are easier to solve. Of course, this method can be refined and made more efficient, but it still needs you to factorize the given polynomial expression.
2. Determining the type of a singularity. Let $$f:D\rightarrow\mathbb R,~f(x)=\frac{x^2+3x+2}{x^3+6x^2+11x+6}$$ where $D$ denotes the maximal domain of the function. As seen above we have $$x^3+6x^2+11x+6=0 \Leftrightarrow x\in\{-3,-2,-1\},$$ therefore we have $D=\mathbb R\setminus\{-3,-2,-1\}$. Now one can ask if there exists a contiuous extension of the function in $x=-3,-2,-1$. Using the factorizations from above we can write (for $x\neq -3,-2,-1$): $$f(x)=\frac{(x+2)(x+1)}{(x+3)(x+2)(x+1)}=\frac{1}{x+3}.$$ Thus we can say that there exists a continuous extension of $f$ in $x=-2$ and $x=-1$, but not in $x=-3$.
3. Simplyfing expressions. More general than what I did in point 2, this can be used to simply expressions of the form $\frac{P(x)}{Q(x)}$ where $P,Q$ are polynomials. Uses of this are very wide (types of singularties of a function, calculating the limit of a sequence, haveing a "nicer" expression to differentiate...)

So it is not just another method to solve a equation in yet another, maybe time saving way, but it actually has some relevance for more advanced problems.

• Hirshy, nice points, but all of them can be done using factorisation by quadratic formula. the only caveat would be that the'll be too complex and cumbersome. – Max Payne Jan 24 '16 at 11:22
• I was aiming for the question "why bother factorizing". Sure if one only talks about solving equations there is no need to factorize expect for less time if you can immediately see the solution. And of course there are polynomial expressions that can't be factorized over $\mathbb R$ (which in itself is an important information!) or where you cant directly get to the factorization. But even then you can use a method to solve $x^2+5x-18=0$ which has really ugly solutions to factorize $x^2+5x-18$ and keep on working (e.g. solve $x^2+5x-18\geq 0$ which again is straightforward after you factorize) – Hirshy Jan 24 '16 at 11:35

Its basically about knowing many different methods so that we can apply them suitably. Methods like splitting the middle term, or completing the square are often times a lot more efficient and time saving than blindly applying the quadratic formula.

Example

suppose you factor the following:

$$3x^2+28x+9=0$$

Using the quadratic formula, you end up in a mess:

$$\dfrac{-28\pm\sqrt{28^2 - (4\times3\times9)}}{2\times3}$$, but by observing the ratio, and splitting the $x$ term so that the ratio is constant, we get $$3x(x+9)+1(x+9) = 0$$ or $$(3x+1)(x+9)=0$$

giving the roots $\frac{-1}{3} \text{ and } -9$

Its derivation involves completing square (i.e. $a^2 \pm 2ab + b^2 = (a \pm b)^2$). Note that the right side of the identity is actually factorization.
• Can you solve $5x^2+1=0 \pmod{10}$ with the quadratic formula? :-p. – YoTengoUnLCD Jan 25 '16 at 6:39