Question

Use factor theorem to factorize the polynomial $x^3 - 13x -12$.

Answer 1

Hint: According to Factor Theorem, zeroes and factors of polynomial are linked to each other.
Cubic polynomial is factorized by finding one factor by hit-and trial method and finding remaining two factors by factorizing the remaining quadratic polynomial.

Complete step by step solution:
To find the factors or zeroes of a cubic polynomial, firstly a factor is determined by determining the factors of constant number in the given cubic polynomial. To find factors of given cubic polynomial, factors of constant number, -12 have to be determined.
Factors of number $-12$ are: $1,-1,2,-2,3,-3,4,-4,6,-6,12,-12$.
By putting values of different factors of $-12$ into cubic polynomials, the factor of polynomial p(x) = ${{x}^{3}}-13x-12$ has to be determined.
At x = $4$, the value of polynomial p(x) is equal to zero. This indicates that x = $4$ is one of the zeroes of polynomials.
Now p(x) is to be divided by $4$ to determine other factors.
\[\begin{matrix}
  x-4\overline{\left){\begin{align}
  & {{x}^{3}}-13x-12\text{ } \\
 & \underline{-{{x}^{3}}+4{{x}^{2}}} \\
\end{align}}\right.}({{x}^{2}}+4x+3) \\
  0+4{{x}^{2}}-13x-12 \\
  \underline{-4{{x}^{2}}+16x} \\
  3x-12 \\
  \underline{3x-12} \\
  \underline{0} \\
\end{matrix}\]
Further, the polynomial $({{x}^{2}}+4x+3)$ has to be factorized to get the factors.
$\begin{align}
  & {{x}^{2}}+4x+3={{x}^{2}}+3x+x+3 \\
 & =x(x+3)+1(x+3) \\
 & =(x+3)(x+1)
\end{align}$
When (x + $3$) (x + $1$) is set equal to zero, the zeroes obtained are:
x = $-3$, x = $-1$
This indicates that the factorization of cubic polynomial, p(x) is denoted as:
$p(x)=(x-4)(x+1)(x+3)$
Therefore, (x + $3$)(x + $1$) (x – $4$) are the factors of the given polynomial.

Note:
> Factors or zeroes of a polynomial are those numbers which when put in the polynomial make the value of the polynomial equal to zero.
> Factor theorem is used to find factors of different kinds of polynomials.