Answer :
We wish to factorize the polynomial
[tex]$$
P(x)=x^4-2x^3-21x^2+x+20.
$$[/tex]
A natural first step is to look for possible factors by testing the candidates predicted by the Rational Root Theorem. According to the theorem, any rational root must be of the form
[tex]$$
\pm \frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] divides the constant term (20) and [tex]$q$[/tex] divides the leading coefficient (1). Thus the only possibilities are
[tex]$$
\pm 1,\, \pm 2,\, \pm 4,\, \pm 5,\, \pm 10,\, \pm 20.
$$[/tex]
A brief check of these potential roots shows that none of them makes [tex]$P(x)=0$[/tex]. (For example, substituting [tex]$x=1$[/tex] gives
[tex]$$
1-2-21+1+20=-1,
$$[/tex]
which is not zero.) Consequently, there are no linear factors with rational (or integer) coefficients.
One might then try to factor [tex]$P(x)$[/tex] as a product of two quadratic polynomials
[tex]$$
P(x)=(x^2+ax+b)(x^2+cx+d).
$$[/tex]
Expanding this product gives
[tex]$$
(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd.
$$[/tex]
Matching coefficients with [tex]$P(x)$[/tex] leads to the system
[tex]\[
\begin{cases}
a+c=-2,\\[1mm]
ac+b+d=-21,\\[1mm]
ad+bc=1,\\[1mm]
bd=20.
\end{cases}
\][/tex]
A careful inspection of the possible factorizations of 20 (the value of [tex]$bd$[/tex]) and the corresponding values for [tex]$a$[/tex], [tex]$b$[/tex], [tex]$c$[/tex], and [tex]$d$[/tex] shows that no choice of integers (or even nice rational numbers) satisfies all these conditions simultaneously.
Because we cannot factor [tex]$P(x)$[/tex] nontrivially into factors with rational coefficients, the factorization of [tex]$P(x)$[/tex] remains in its original form. In other words, the polynomial is irreducible (with respect to factorization over the rational numbers).
Thus, the factorized form of
[tex]$$
x^4-2x^3-21x^2+x+20
$$[/tex]
is simply
[tex]$$
x^4-2x^3-21x^2+x+20.
$$[/tex]
[tex]$$
P(x)=x^4-2x^3-21x^2+x+20.
$$[/tex]
A natural first step is to look for possible factors by testing the candidates predicted by the Rational Root Theorem. According to the theorem, any rational root must be of the form
[tex]$$
\pm \frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] divides the constant term (20) and [tex]$q$[/tex] divides the leading coefficient (1). Thus the only possibilities are
[tex]$$
\pm 1,\, \pm 2,\, \pm 4,\, \pm 5,\, \pm 10,\, \pm 20.
$$[/tex]
A brief check of these potential roots shows that none of them makes [tex]$P(x)=0$[/tex]. (For example, substituting [tex]$x=1$[/tex] gives
[tex]$$
1-2-21+1+20=-1,
$$[/tex]
which is not zero.) Consequently, there are no linear factors with rational (or integer) coefficients.
One might then try to factor [tex]$P(x)$[/tex] as a product of two quadratic polynomials
[tex]$$
P(x)=(x^2+ax+b)(x^2+cx+d).
$$[/tex]
Expanding this product gives
[tex]$$
(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd.
$$[/tex]
Matching coefficients with [tex]$P(x)$[/tex] leads to the system
[tex]\[
\begin{cases}
a+c=-2,\\[1mm]
ac+b+d=-21,\\[1mm]
ad+bc=1,\\[1mm]
bd=20.
\end{cases}
\][/tex]
A careful inspection of the possible factorizations of 20 (the value of [tex]$bd$[/tex]) and the corresponding values for [tex]$a$[/tex], [tex]$b$[/tex], [tex]$c$[/tex], and [tex]$d$[/tex] shows that no choice of integers (or even nice rational numbers) satisfies all these conditions simultaneously.
Because we cannot factor [tex]$P(x)$[/tex] nontrivially into factors with rational coefficients, the factorization of [tex]$P(x)$[/tex] remains in its original form. In other words, the polynomial is irreducible (with respect to factorization over the rational numbers).
Thus, the factorized form of
[tex]$$
x^4-2x^3-21x^2+x+20
$$[/tex]
is simply
[tex]$$
x^4-2x^3-21x^2+x+20.
$$[/tex]
