Question

Factorize the following expression:
\[{{x}^{3}}-3{{x}^{2}}-9x-5\]

Answer 1

Hint: Factoring means creating factors. By hit and trial method, we can find any one factor. If the degree of the polynomial is greater than 2. With this method, we start substituting the value of \[x=0,\pm 1,\pm 2\]… and collect the one which makes the whole polynomial equal to 0.

Complete step-by-step solution -

Consider the polynomial \[{{x}^{3}}-3{{x}^{2}}-9x-5....\left( i \right)\]
Since the degree of the polynomial is equal to 3, this means that there will be three factors of the polynomial. By hit and trial method, we start substituting the value of \[x=0,\pm 1,\pm 2\]…. Until we get the polynomial equals to 0. By this, we will get one factor and with the help of that factor, we will factorize the polynomial.
Let’s start by substituting x = 0 this results into
\[{{x}^{3}}-3{{x}^{2}}-9x-5={{\left( 0 \right)}^{3}}-3{{\left( 0 \right)}^{2}}-9\left( 0 \right)-5\]
\[{{x}^{3}}-3{{x}^{2}}-9x-5=-5\]
Since – 5 is not equal to 0, therefore, x = 0 cannot be a factor. Now, substitute x = – 1 in equation (i). This results as follows:
\[{{x}^{3}}-3{{x}^{2}}-9x-5={{\left( -1 \right)}^{3}}-3{{\left( -1 \right)}^{2}}-9\left( -1 \right)-5\]
\[{{x}^{3}}-3{{x}^{2}}-9x-5=-1-3+9-5\]
\[{{x}^{3}}-3{{x}^{2}}-9x-5=-9+9\]
By canceling – 9 and 9, we have,
\[{{x}^{3}}-3{{x}^{2}}-9x-5=0\]
Therefore, x = – 1 is a factor of the polynomial or we can say that (x + 1) is one of the three factors of the equation (i).
Now, we will try to split the equation (i) such that it converts into the terms of (x + 1). Thus we get,
\[{{x}^{3}}+\left( {{x}^{2}}-4{{x}^{2}} \right)+\left( -4x-5x \right)-5=0\]
\[\Rightarrow {{x}^{3}}+{{x}^{2}}-4{{x}^{2}}-4x-5x-5=0\]
Now, we will take \[{{x}^{2}}\] as a common term from \[{{x}^{3}}+{{x}^{2}}\] , – 4x from \[-4{{x}^{2}}-4x\] and 5 from – 5x – 5. Therefore, we get,
\[{{x}^{3}}-3{{x}^{2}}-9x-5={{x}^{2}}\left( x+1 \right)-4x\left( x+1 \right)-5\left( x+1 \right)\]
Now we will take (x + 1) common.
\[{{x}^{3}}-3{{x}^{2}}-9x-5=\left( x+1 \right)\left( {{x}^{2}}-4x-5 \right)\]
Now, we will substitute – 4x as – 5x + x. Thus, we get,
\[{{x}^{3}}-3{{x}^{2}}-9x-5=\left( x+1 \right)\left( {{x}^{2}}-5x+x-5 \right)\]
Now, we will take x as a common term from \[{{x}^{2}}-5x\] and 1 from x – 5. Therefore, we have
\[{{x}^{3}}-3{{x}^{2}}-9x-5=\left( x+1 \right)\left( x\left( x-5 \right)+1\left( x-5 \right) \right)\]
Now, we will take out the common term x – 5 to get the following equation.
\[{{x}^{3}}-3{{x}^{2}}-9x-5=\left( x+1 \right)\left( x-5 \right)\left( x+1 \right)\]
Since these are the three factors of the polynomial and the polynomial itself has degree 3. Therefore, these are the total number of factors by factorization.
Hence, the factors are (x + 1) (x – 5) (x + 1)

Note: Alternatively, we could have also solved the expression \[{{x}^{3}}-3{{x}^{2}}-9x-5\] by using the hit and trial method followed by a square root formula instead of splitting this equation. The formula for square root is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where \[a{{x}^{2}}+bx+c=0\] is the equation.