Factoring polynomials given one obvious factor

Question

Got a basic algebra question that I seem to have forgotten how to do and need in the context of a larger question.

Given a polynomial, say $x^2 + m^2(x+1)^2 = 1$, that is $(1+m^2)x^2 + 2m^2x + m^2 - 1 = 0$, how do I find the roots knowing one 'visible' solution. I know there is one obvious solution of $x = -1$ (and another $x_m$) but from this how do we obtain the roots $Q = (x_m, y_m) = (\frac{1-m^2}{1+m^2}, \frac{2m}{1+m^2})$. Namely how do I find the other root $x_m$ easily from knowing the one root $x = -1$.

Note $y_m = m(x_m+1)$.

Appreciate the help!

Answer 1

The product of roots in a quadratic equation $$ ax^2 + bx +c =0$$

is $c/a$

For your equation $$(1+m^2)x^2 + 2m^2x + m^2 - 1 = 0$$

The product is $$ \frac {m^2 - 1}{ 1+m^2}.$$

Since you know one root is $-1$, the other is $$ \frac {1-m^2 }{ 1+m^2}.$$

Answer 2

The bulky, traditional way is to use the formula for the roots of the quadratic: $$\begin{align} x_{1,2} &=\frac{-2m^2 \mp \sqrt{4m^4-4(m^2+1)(m^2-1)}}{2(m^2+1)}\\ &=\frac{-2m^2 \mp \sqrt{4m^4-4(m^4-1)}}{2(m^2+1)}\\ &=\frac{-2m^2 \mp \sqrt{4}}{2(m^2+1)}\\ &=\frac{-2m^2 \mp 2}{2(m^2+1)}\\ &=\frac{-m^2 \mp 1}{m^2+1}.\end{align}$$ So $x_1=-1$ and $x_2=-\frac{m^2-1}{m^2+1}=-1+\frac{2}{m^2+1}$.

Answer 3

The answer is polynomial division by, in this case, $x+1$. (In general, $x-r$, where $r$ is the root.)

We have \begin{align} \frac{(1+m^2)x^2 + 2m^2x + m^2-1}{x+1} &= \frac{(1+m^2)x^2 \color{red}{{} + (1+m^2)x - (1+m^2)x} + 2m^2x + m^2-1}{x+1} \\ &= \frac{(1+m^2)(x^2+x) + (m^2-1)x + m^2-1}{x+1} \\ &= (1+m^2)x + \frac{(m^2-1)x + m^2-1}{x+1} \\ &= (1+m^2)x + m^2-1. \end{align} So the remaining root can be found by solving $(1+m^2)x + m^2-1 = 0$.

In the case of a quadratic polynomial, there are all kinds of shortcuts, but this is the approach you take for a polynomial of arbitrary degree.

Answer 4

The sum of the roots in $$ax^2 + bx +c =0$$ is $$ x_1+x_2 = - \frac {b}{a}.$$

For the quadratic $$(1+m^2)x^2 + 2m^2x + m^2 - 1 = 0$$ we have $$ x_1+x_2 = - \frac {b}{a}=\frac {-2m^2}{1+m^2}$$

With $$x_1=-1$$

We get $$x_2= \frac {-2m^2}{1+m^2}+1 =\frac {1-m^2}{1+m^2}$$