Factoring Polynomials of Higher Degree

The process of factoring polynomials of higher degree is generally by using the rational root theorem, although there are special cases that can be done otherwise.

If a given cubic polynomial has rational coefficients and a rational root, it can be found using the rational root theorem.

Factor the polynomial \(3x^3 + 4x^2+6x-35\) over the real numbers.

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Any rational root of the polynomial has numerator dividing \( 35\) and denominator dividing \( 3.\) The possibilities are \[ \pm 1, \pm 5, \pm 7, \pm 35, \pm \frac13, \pm \frac53, \pm \frac73, \pm \frac{35}3. \] Inspection of these sixteen possibilities yields the unique root \( x=\frac53.\) By the remainder factor theorem, \( x-\frac53 \) (or \( 3x-5)\) can be factored out of the original polynomial.

Polynomial division of \( 3x^3+4x^2+6x-35 \) by \( 3x-5 \) yields \( x^2+3x+7,\) so the polynomial factors as \[ 3x^3+4x^2+6x+35 = (3x-5)(x^2+3x+7). \] Note that the quadratic factor is irreducible over the real numbers, since its discriminant is \( 3^2-4\cdot 7 = - 19 <0.\) \(_\square\)

Factorize the cubic polynomial \( f(x) = 2x^3 + 7x^2 + 5x + 1 \) over the rational numbers.

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By the rational root theorem, any rational root of \(f(x)\) has the form \(r= \frac{a}{b},\) where \( a \vert 1\) and \( b \vert 2\). Thus, we only need to try numbers \( \pm \frac {1}{1}, \pm \frac {1}{2}\). Substituting all the possible values,

\[\begin{aligned} f(1) &> 0 \\ f(-1) &= -2 + 7 - 5 + 1 = 1 \neq 0 \\ f\bigg (\frac {1}{2}\bigg ) &> 0 \\ f\bigg (-\frac {1}{2}\bigg ) &= -\frac {2}{8} + \frac {7}{4} - \frac {5}{2} + 1 = 0. \end{aligned}\]

By the remainder-factor theorem, \( (2x+1)\) is a factor of \(f(x)\), implying \( f(x) = (2x+1) (x^2 + 3x + 1)\). We can then use the quadratic formula to factorize the quadratic if irrational roots are desired. \(_\square\)