How to refactor a polynomial?

Question

Might be a bit basic, but I'm more or less starting from scratch. :-)

We start with the following inequality:

x² + x + 1 > 7

From what I can remember from my high school days, we solve this by refactoring it, and from what I can find in my old textbooks, the solution can be found like this:

$x² - sx + p = (x + x_1) (x - x_2)$

Where:
$x_1 + x_2 = s$
$x_1 \times x_2 = p$

My question... I can find solutions within my head, but I was wondering... how do we solve those 2 equations?

$x_1 + x_2 = s$
$x_1 \times x_2 = p$

Answer 1

In the general case? Poorly.

Usually when encountering polynomials of this type, you're dealing with integers, so polynomials of the types $$\begin{align*} &x^2 -x + 2 \\ &2x^2 - 3x + 1 \\ &4x^2 - 9 \end{align*}$$ and so on. When you're dealing with non-integers, it begins to get rougher.

If your polynomials have rational coefficients, say $$\begin{align*} &\frac 1 2 x^2 - \frac 9 5 \\ &x^2 + \frac 3 4 x - \frac 1 7 \end{align*}$$ and so on, you can reduce to the first case by multiplying by the lowest common denominator for the fractions, i.e. the lowest common multiple of the denominators. Hence in both of these cases, you can "do it in your head" by checking factorizations of the constant term, essentially.

If you have irrational coefficients, you're stuck. You might get lucky (certain polynomials may have a form you can exploit, e.g. from the difference of squares formula, you can get by easily if you don't have a linear coefficient), but in general you have little choice but to use the quadratic formula. For the quadratic $ax^2+bx+c$, you will have two roots, $$ r_{+,-} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ and then $ax^2+bx + c = a(x-r_+)(x-r_-)$.

Answer 2

Trying to get back to the essence of the original question. I think the quadratic formula is the best approach, but the question specifically mentioned the sum and product of the roots, so...

There is a technique I read about known as Loh's method which says this:

$x^2+bx+c=0$ has roots $r_1$ and $r_2$ where the average of the roots is $\frac{-b}{2}$ and the product $r_1\times r_2=c$

In your example $x^2+x-6=0$ so $b=1, c=-6$ and so the average of $r_1$ and $r_2$ will be $\frac{-1}{2}$

Since the roots are equidistant from $\frac{-1}{2}$ we can write $r_1=\frac{-1}{2}-s$ and $r_2=\frac{-1}{2}+s$ where $s\geq0$

Since $r_1r_2=c=-6$ we can write $(\frac{-1}{2}+s)(\frac{-1}{2}+s)=-6$ which can be expanded to give $\frac{1}{4}-s^2=-6$ and therefore $s^2=\frac{25}{4}$ and since $s\geq0$ we can conclude $s=\frac{5}{2}$.

From there it is a simple matter to determine $r_1=\frac{-1}{2}-\frac{5}{2}=-3$ and $r_2=\frac{-1}{2}+\frac{5}{2}=2$.

Loh's method can easily be generalised to non-monic quadratics as well. Is it easier than the quadratic formula? In general probably not. If you know the roots are rational... maybe.

Answer 3

I know the quadratic formula, but personally, I'd prefer to derive it every time I need to solve a quadratic:

$$ \begin{align} {x}^{2} - 2 a x + {g}^{2} & = 0, \\ {x}^{2} - 2 a x + {a}^{2} + {g}^{2} & = {a}^{2}, \\ {x}^{2} - 2 a x + {a}^{2} & = {a}^{2} - {g}^{2}, \\ {\left( x - a \right)}^{2} & = {a}^{2} - {g}^{2}, \\ x - a & = \pm \sqrt {{a}^{2} - {g}^{2}}, \\ x & = a \pm \sqrt {{a}^{2} - {g}^{2}}. \end{align} $$

Let $\overline {\text {AC}} = {x}_{1}$ and $\overline {\text {AD}} = \overline {\text {BC}} = {x}_{2}$. Then $\overline {\text {OP}} = \overline {\text {OA}} = \overline {\text {OB}} = a = \left( {x}_{1} + {x}_{2} \right) / \, 2$ and $\overline {\text {CP}} = g = \sqrt {{x}_{1} {x}_{2}}$, so $\overline {\text {OC}} = \sqrt {{a}^{2} - {g}^{2}}$. Think of $\text {A}$ as the origin.