Question

How do you solve the polynomial ${x^5} - 5{x^3} + 4x = 0$?

Answer 1

Hint: Given a polynomial of degree five. We have to find the solution of the polynomial equation. First, we will take out the common term from the left-hand side of the polynomial. Then, we will factorize the equation. Then, set each factor equal to zero to solve the equation. Then, we will apply the algebraic identity to the factor. Then, we will find all the values that will make the polynomial true.
Formula used:
The algebraic identity for $\left( {{a^2} - {b^2}} \right)$ is given by:
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$

Complete step by step solution:
We are given the polynomial. First, we will factor out the common term, $x$ on the left hand side of the expression.
$ \Rightarrow x\left( {{x^4} - 5{x^2} + 4} \right) = 0$
Rewrite the expression by representing ${x^4}$ as ${\left( {{x^2}} \right)^2}$
$ \Rightarrow x\left[ {{{\left( {{x^2}} \right)}^2} - 5{x^2} + 4} \right] = 0$
Now, we will substitute ${x^2} = t$ into the expression.
$ \Rightarrow x\left( {{t^2} - 5t + 4} \right) = 0$
Now, factorize the equation by splitting the middle term whose sum is $ - 5$ and product is $4$
$ \Rightarrow x\left( {{t^2} - 4t - t + 4} \right) = 0$
On simplifying the expression further, we get:
$ \Rightarrow x\left[ {t\left( {t - 4} \right) - 1\left( {t - 4} \right)} \right] = 0$
$ \Rightarrow x\left[ {\left( {t - 1} \right)\left( {t - 4} \right)} \right] = 0$
Now, we will replace $t$ by ${x^2}$.
$ \Rightarrow x\left[ {\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)} \right] = 0$
Now, we will apply the algebraic identity $\left( {{a^2} - {b^2}} \right)$ to the expression.
$ \Rightarrow x\left[ {\left( {{x^2} - {1^2}} \right)\left( {{x^2} - {2^2}} \right)} \right] = 0$
$ \Rightarrow x\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0$
Set each factor equal to zero.
$ \Rightarrow x = 0{\text{ or }}\left( {x - 1} \right) = 0{\text{ or }}\left( {x + 1} \right) = 0{\text{ or }}\left( {x - 2} \right) = 0{\text{ or }}\left( {x + 2} \right) = 0$
Now, solve the first factor for $x$.
$ \Rightarrow x = 0$
Now, solve the second factor for $x$ by adding one to both sides of the expression $x - 1 = 0$
$ \Rightarrow x = 1$
Now, solve the third factor for $x$ by subtracting one from both sides of the expression $x + 1 = 0$
$ \Rightarrow x = - 1$
Now, solve the next factor for $x$ by adding two to both sides of the expression $x - 2 = 0$
$ \Rightarrow x = 2$
Now, solve the next factor for $x$ by subtracting two from both sides of the expression $x + 2 = 0$
$ \Rightarrow x = - 2$
Hence, the solution of the equation is $x = 0,1, - 1,2, - 2$

Note: To determine the required roots/zeros of the quadratic expression it is necessary that we have to rearrange the terms in the given expression.
Students must also remember that on solving a polynomial with five degrees at most five zeroes/roots can be obtained which will satisfy the expression means on substituting the roots/zeroes will make the expression 0.