Factorize the solution: $8{{x}^{3}}-16x-85$.
Answer
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Hint: Mathematics includes the study of topics which are related to quantity, structure, space and change. It has no generally accepted definition. It involves various operations which are performed using various operators such as addition multiplication division and subtraction. For our problem, we are required to factorize the given equation.
Complete step-by-step answer:
In mathematics, the number system is the branch that deals with various types of numbers possible to form and easy to operate with different operators such as addition, multiplication and so on. Another system for variables in mathematics is an algebra system in which we have equations corresponding to some variables through which we can evaluate values of those variables.
Here, we have one equation to factorize which is involving variable x. Factors are those numbers which completely divides the given number without leaving any remainder. Similarly factors of an equation will completely divide the equation without leaving any remainder.
Now, we are given the equation: $8{{x}^{3}}-16x-85$.
It is a cubic equation so by hit and trial we find the factor of the equation. For x = 1, $f(x)\ne 0$ so it is not a factor. For x = 1.5, $f(x)\ne 0$ so it is not a factor. For x = 2, $f(x)\ne 0$ so it is not a factor. For x = 2.5, $f(x)=0$, so one possible factor is \[x=\dfrac{5}{2}\].
Now using this factor, we can easily factorize the equation.
$8{{x}^{3}}-16x-85=\_\_(2x-5)+\_\_(2x-5)+\_\_(2x-5)$
We try to fill the blanks in such a way that L.H.S = R.H.S.
$\begin{align}
& \therefore 4{{x}^{2}}(2x-5)+10x(2x-5)+17(2x-5) \\
& \Rightarrow (2x-5)(4{{x}^{2}}+10x+17) \\
\end{align}$
This implies that the required factors are $(2x-5)(4{{x}^{2}}+10x+17)$.
Note: The key step for solving this problem is the knowledge of the algebraic system of equations. To solve any particular algebraic equation, we require the same number of equations as there are a number of variables present. Since, we have one equation and a single variable so we are able to factorize the problem.
Complete step-by-step answer:
In mathematics, the number system is the branch that deals with various types of numbers possible to form and easy to operate with different operators such as addition, multiplication and so on. Another system for variables in mathematics is an algebra system in which we have equations corresponding to some variables through which we can evaluate values of those variables.
Here, we have one equation to factorize which is involving variable x. Factors are those numbers which completely divides the given number without leaving any remainder. Similarly factors of an equation will completely divide the equation without leaving any remainder.
Now, we are given the equation: $8{{x}^{3}}-16x-85$.
It is a cubic equation so by hit and trial we find the factor of the equation. For x = 1, $f(x)\ne 0$ so it is not a factor. For x = 1.5, $f(x)\ne 0$ so it is not a factor. For x = 2, $f(x)\ne 0$ so it is not a factor. For x = 2.5, $f(x)=0$, so one possible factor is \[x=\dfrac{5}{2}\].
Now using this factor, we can easily factorize the equation.
$8{{x}^{3}}-16x-85=\_\_(2x-5)+\_\_(2x-5)+\_\_(2x-5)$
We try to fill the blanks in such a way that L.H.S = R.H.S.
$\begin{align}
& \therefore 4{{x}^{2}}(2x-5)+10x(2x-5)+17(2x-5) \\
& \Rightarrow (2x-5)(4{{x}^{2}}+10x+17) \\
\end{align}$
This implies that the required factors are $(2x-5)(4{{x}^{2}}+10x+17)$.
Note: The key step for solving this problem is the knowledge of the algebraic system of equations. To solve any particular algebraic equation, we require the same number of equations as there are a number of variables present. Since, we have one equation and a single variable so we are able to factorize the problem.
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