Question

How do you find the roots of ${x^3} + {x^2} - 5x + 3 = 0$?

Answer 1

Hint: First we have to factorize this cubic polynomial and then equate each factor to zero. To factorise a cubic polynomial \[p\left( x \right) = a{x^3} + b{x^2} + cx + d\], we first make the coefficient of ${x^3}$ equal to one if it is not one and then find the constant term.
Then, find all the factors of constant term. Then, check at which factor of constant term, $p\left( x \right)$ is zero by using trial method and get one factor of $p\left( x \right)$. Then, write $p\left( x \right)$ as the product of this factor and a quadratic polynomial.
Then, apply splitting the middle term method or factor theorem in quadratic polynomial to get the other two factors.
Then, we get all the three factors of a given cubic polynomial. Next, equate each factor to zero to find the roots of the given cubic equation.

Formula used: To factorise a cubic polynomial \[p\left( x \right) = a{x^3} + b{x^2} + cx + d\], we use the following steps:
First make the coefficient of ${x^3}$ equal to one if it is not one and then find the constant term.
Find all the factors of constant term.
Check at which factor of constant term, $p\left( x \right)$ is zero by using trial method and get one factor of $p\left( x \right)$.
Write $p\left( x \right)$ as the product of this factor and a quadratic polynomial.
Apply splitting the middle term method or factor theorem in quadratic polynomial to get the other two factors.
Thus, we get all the three factors of a given cubic polynomial.

Complete step-by-step solution:
Given cubic equation: ${x^3} + {x^2} - 5x + 3 = 0$
We have to find the roots of a given cubic equation.
For this we first have to factorize this cubic polynomial and then equate each factor to zero.
So, assuming a given polynomial to be$p\left( x \right)$.
Let \[p\left( x \right) = {x^3} + {x^2} - 5x + 3\]
Next step is to make the coefficient of ${x^3}$ equal to one if it is not one. Here, the given polynomial has already $1$ as the coefficient of ${x^3}$.
So, \[p\left( x \right) = {x^3} + {x^2} - 5x + 3\]
Next step is to find the constant term and all the factors of the constant term.
So, we have to find the constant term of $p\left( x \right)$ and all its factors.
Constant term is $3$ and it’s all factors are \[ \pm 1, \pm 3\].
Next step is to check at which factor of constant term, $p\left( x \right)$ is zero by using trial method and get one factor of $p\left( x \right)$.
So, put $x = 1$ in\[p\left( x \right) = {x^3} + {x^2} - 5x + 3\].
\[p\left( 1 \right) = {\left( 1 \right)^3} + {\left( 1 \right)^2} - 5\left( 1 \right) + 3\]
On simplify the term and we get,
\[ \Rightarrow 1 + 1 - 5 + 3\]
On adding the term and we get
\[ \Rightarrow 5 - 5\]
Let us subtract the term and we get,
\[ \Rightarrow 0\]
So, $\left( {x - 1} \right)$ is a factor of $p\left( x \right)$.
Next step is to write $p\left( x \right)$ as the product of this factor and a quadratic polynomial.
So, divide $p\left( x \right)$ by $\left( {x - 1} \right)$.
\[\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x - 3} \\
  {x - 1)\overline {{x^3} + {x^2} - 5x + 3} } \\
  {\,\,\,\underline {{x^3} - {x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{x^2} - 5x + 3} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,2{x^2} - 2x\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3x + 3} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline { - 3x + 3} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0}
\end{array}\]
 $p\left( x \right) = \left( {x - 1} \right)\left( {{x^2} + 2x - 3} \right)$
Next step is to apply the splitting of the middle term method or factor theorem in quadratic polynomials to get the other two factors.
So, factorise ${x^2} + 2x - 3$.
$ \Rightarrow {x^2} + 2x - 3 = {x^2} - x + 3x - 3$
$ \Rightarrow x\left( {x - 1} \right) + 3\left( {x - 1} \right)$
$ \Rightarrow \left( {x - 1} \right)\left( {x + 3} \right)$
Next, put the value of ${x^2} + 2x - 3$ in $p\left( x \right) = \left( {x - 1} \right)\left( {{x^2} + 2x - 3} \right)$.
$p\left( x \right) = \left( {x - 1} \right)\left( {{x^2} + 2x - 3} \right)$
$ \Rightarrow {\left( {x - 1} \right)^2}\left( {x + 3} \right)$
Next, equate each factor to zero to find the roots of the given cubic equation.
$ \Rightarrow {\left( {x - 1} \right)^2} = 0$
$ \Rightarrow x - 1 = 0$
$ \Rightarrow x = 1$
Also,
$ \Rightarrow x + 3 = 0$
$ \Rightarrow x = - 3$

Therefore, $x = 1$(double root) and $x = - 3$ are roots of ${x^3} + {x^2} - 5x + 3 = 0$.

Note: In above question, it should be noted that $x = 1$ is a double root of the equation ${x^3} + {x^2} - 5x + 3 = 0$. So, $x = 1$ is a root of the algebraic equation that appears twice in the solution.
It can also be said like $x = 1$ is a root of the algebraic equation with multiplicity$2$.
Therefore, $x = 1$ and $x = - 3$ are roots of ${x^3} + {x^2} - 5x + 3 = 0$ with multiplicity $2$ and $1$ respectively.