How do you solve \[8{x^5} + 10{x^4} = 4{x^3} + 5{x^2}\] ?

Hint: Here we have a polynomial of degree 5 and it is called quintic polynomial. It is a straightforward equation and we can solve this by grouping the terms \[8{x^5}\] and \[10{x^4}\] . We group \[4{x^3}\] and \[5{x^2}\] . On further simplification we will have the desired result.

Complete step-by-step answer:
Given \[8{x^5} + 10{x^4} = 4{x^3} + 5{x^2}\]
That is \[8{x^5} + 10{x^4} - 4{x^3} - 5{x^2} = 0\]
 \[\left( {8{x^5} + 10{x^4}} \right) - \left( {4{x^3} + 5{x^2}} \right) = 0\]
Now take \[2{x^4}\] common in the first group and \[{x^2}\] in the reaming group,
 \[2{x^4}\left( {4x + 5} \right) - {x^2}\left( {4x + 5} \right) = 0\]
Now we can take \[\left( {4x + 5} \right)\] common,
 \[\left( {4x + 5} \right)\left( {2{x^4} - {x^2}} \right) = 0\]
 \[{x^2}\left( {4x + 5} \right)\left( {2{x^2} - 1} \right) = 0\]
Thus we have the factors, now by zero product principle we have,
 \[{x^2} = 0\] or \[\left( {4x + 5} \right) = 0\] or \[\left( {2{x^2} - 1} \right) = 0\]
 \[ \Rightarrow {x^2} = 0\] or \[4x = - 5\] or \[2{x^2} = 1\]
 \[ \Rightarrow x = 0\] or \[x = - \dfrac{5}{4}\] or \[{x^2} = \dfrac{1}{2}\]
 \[ \Rightarrow x = 0\] or \[x = - \dfrac{5}{4}\] or \[x = \pm \sqrt {\dfrac{1}{2}} \] .
Thus the solution of \[8{x^5} + 10{x^4} = 4{x^3} + 5{x^2}\] is \[ \Rightarrow x = 0\] , \[x = - \dfrac{5}{4}\] and \[x = \pm \sqrt {\dfrac{1}{2}} \]
So, the correct answer is “\[ \Rightarrow x = 0\] , \[x = - \dfrac{5}{4}\] and \[x = \pm \sqrt {\dfrac{1}{2}} \] ”.

Note: If we have a polynomial of degree n then we will have n roots. Here we have degree 5 and we will have 5 roots (zero is repeated twice). Usually we solve this type of problem by synthetic division that is by reducing quintic to quartic and then we reduce quartic to cubic and then we reduce it to quadratic and we can solve the quadratic equation using factorization method. In both the cases we will have the same answer.