Factorise \[{x^4} - 5{x^2} + 4 < 0\]

Hint: Firstly try to simplify the given higher degree equation into a smaller degree equation. Then depending upon the degree of the equation, try to factorize it into as many factors as possible. Based on the factors obtained, deduce the interval in which the value of the given variable lies.

Complete step-by-step answer:
We are given with the inequality \[{x^4} - 5{x^2} + 4 < 0\]
This can be rewritten as \[{\left( {{x^2}} \right)^2} - 5{x^2} + 4 < 0\]
This becomes a quadratic equation in terms of \[{x^2}\].
We know that any quadratic equation in the variable \[x\] is of the form \[a{x^2} + bx + c = 0\] .
Therefore using the quadratic formula to find the values of roots we get
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here in this question we have the inequality \[{x^4} - 5{x^2} + 4 < 0\] which is a quadratic in terms of \[{x^2}\].
Therefore using the quadratic formula to find the values of roots we get
\[{x^2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here \[a = 1\,,\,b = 5\,,\,c = 4\]
Hence we get \[{x^2} = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4(1)(4)} }}{{2(1)}}\]
On doing the calculations in the root part we get
\[{x^2} = \dfrac{{5 \pm \sqrt {25 - 16} }}{2}\]
\[{x^2} = \dfrac{{5 \pm 3}}{2}\]
Therefore we get \[{x^2} = 4,1\]
Hence the inequality \[{x^4} - 5{x^2} + 4 < 0\] can be written as \[\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right) < 0\]
Using the identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] we get
So, the correct answer is “\[\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right) < 0\]”.

Note: Factorization or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together gives the original number. In the factorization method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets.