Question

Factorize and find the roots of the following equation:
${x^3} - 3{x^2} - x + 3 = 0$.

Answer 1

Hint: The given equation is having a cubic polynomial. To factorize it, separate the terms in pairs and take the common terms outside. Further, equate each factor to 0 to find the roots of the equation.

Complete step by step answer:
According to the question, the given equation is ${x^3} - 3{x^2} - x + 3 = 0$.
For factorizing it, we can take ${x^2}$ common from the first two terms and -1 from the other two terms. Doing this and simplifying it further, we’ll get:
\[
   \Rightarrow {x^3} - 3{x^2} - x + 3 = 0 \\
   \Rightarrow {x^2}\left( {x - 3} \right) - 1\left( {x - 3} \right) = 0 \\
   \Rightarrow \left( {{x^2} - 1} \right)\left( {x - 3} \right) = 0 \\
\]
And we know that ${x^2} - {a^2} = \left( {x + a} \right)\left( {x - a} \right)$. Using this for above equation, we’ll get:
\[
   \Rightarrow \left( {{x^2} - 1} \right)\left( {x - 3} \right) = 0 \\
   \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right) = 0 \\
\]
So we are getting three different factors of the polynomial. For finding the roots, we will equate each factor to 0 separately. We’ll get:
\[
   \Rightarrow \left( {x + 1} \right) = 0{\text{ or }}\left( {x - 1} \right) = 0{\text{ or }}\left( {x - 3} \right) = 0 \\
   \Rightarrow x = - 1{\text{ or }}x = 1{\text{ or }}x = 3 \\
\]

Thus, the factorized form of the equation is \[\left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right) = 0\], and its roots are -1, 1 and 3.

Note: If we are facing any difficulty factorizing any cubic polynomial by separating terms then we try to find one of the roots of the polynomial by hit and trial and convert the cubic polynomial into a product of a linear and a quadratic polynomial. And then we solve quadratic expressions separately to get all the roots.