Special trinomial

In algebra, specifically in the factorization of polynomials, a special trinomial is a polynomial that can be expressed in the form:

$a{x}^2+bx+c(\text{with }a,b,c\ne 0)$

and for which there is a known method for factoring it as the product of two first-degree binomials.

Factoring Method

We can distinguish between the cases in which the coefficient of the quadratic term is equal to or different from $1$.

a = 1

If the coefficient of the quadratic term is equal to $1$, the trinomial takes the form:

$x^2+bx+c$

in this case, it can be factored into the product of two first-degree binomials, in the form:

$(x+u)(x+v)$,

where $u$ and $v$ are two terms with the following two properties:

• $u+v=b$
• $u\cdot v=c$.

In fact, performing the calculations yields:

$\begin{array}{rl}(x+u)(x+v)& =x^2+vx+ux+uv\\ & =x^2+x(v+u)+uv\\ & =x^2+bx+c\\ \end{array}$

A practical method to find $u$ and $v$ is to solve for the two roots of the polynomial. In fact, if

$x^2+bx+c=0$,

then:

$(x-u)(x-v)=0\Rightarrow \{\begin{array}{l}{x}_1=u\\ x_2=v\end{array}$

To find the roots of the quadratic trinomial, simply use the quadratic formula:

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

a ≠ 1

If the coefficient of the quadratic term is not $1$, the trinomial can be factored as follows:

$a{x}^2+bx+c=a(x+u)(x+v)$,

where $u$ and $v$ have the following properties:

• $u+v=\frac{b}{a}$
• $u\cdot v=\frac{c}{a}$.

In this case as well, the factorization can be demonstrated as follows:

$\begin{array}{rl}a(x+u)(x+v)& =a{x}^2+avx+aux+auv\\ & =a{x}^2+ax(v+u)+auv\\ & =a{x}^2+bx+c\\ \end{array}$

As in the previous case, $u$ and $v$ can be found by solving for the roots of the polynomial using the quadratic formula.

Trinomials of degree greater than 2

Factoring by substitution

More generally, if we consider the trinomial:

$a{x}^{2n}+b{x}^n+c$

This can be factored by making the substitution $x^n=t$, which gives the trinomial:

$a{t}^2+bt+c$,

that can be factored using the methods described above and then by reapplying the substitution in reverse.

Direct factoring

Considering the previous properties:

• $u+v=\frac{b}{a}$
• $u\cdot v=\frac{c}{a}$.

The trinomial can also be directly factored as follows:

$a{x}^{2n}+b{x}^n+c=a(x^n+u)(x^n+v)$

The factorization can be demonstrated as follows:

$\begin{array}{rl}a(x^n+u)(x^n+v)& =a{x}^{2n}+a{x}^{n}u+a{x}^{n}v+auv\\ & =a{x}^{2n}+a{x}^n(u+v)+auv\\ & =a{x}^{2n}+a{x}^n\frac{b}{a}+a\frac{c}{a}\\ & =a{x}^{2n}+b{x}^n+c\end{array}$

Also in this case $u$ and $v$ can be found by solving for the roots of the polynomial using the quadratic formula.

Bibliography

• Marzia Re Fraschini, Gabriella Grazzi (2012). I principi della matematica (Volume 3). Atlas. ISBN 978-88-268-1711-8. 
• Massimo Bergamini, Graziella Barozzi, Anna Trifone. Matematica.blu (seconda edizione) Vol.1. Zanichelli - Bologna, 2018. ISBN 978-88-08-22085-1.