
How do you find the factors of \[4{x^3} - 108\] ?
Answer
549k+ views
Hint: In this question, initially we should take the coefficient of the variable common out of the whole equation and then to find nth root of the constant part, where n is the power to which the variable is raised and then equate the equation to zero.
Complete step by step answer:
Since 4 is the coefficient of the variable term, we will take 4 common, which will give us with,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]
Now we will equate above equation to zero, which will give us with,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]= \[0\]
We know that 27 is cube of 3, therefore,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]= \[0\]
Now here we will use the identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\], where \[x\]=a and 3=b,
\[ \Rightarrow \]4\[\left( {x - 3} \right)\left( {{x^2} + 3x + {3^2}} \right)\]=\[0\]
Now either,
\[\left( {x - 3} \right)\]=\[0\] OR \[\left( {{x^2} + 3x + 9} \right)\]=\[0\]
So, \[x\]=3 and now we will use the formula for finding roots quadratic equation in other equation,
FORMULA : \[x\]=\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
In above equation, a=1, b=3, c=9, after substituting the values, we get:-
\[x\]=\[\dfrac{{ - 3 \pm \sqrt {{3^2} - 4 \times 1 \times 9} }}{{2 \times 1}}\],
\[x\]=\[\dfrac{{ - 3 \pm \sqrt { - 27} }}{2}\],
\[x\]=\[\dfrac{{ - 3 \pm 3\sqrt 3 i}}{2}\] , where \[i = \sqrt { - 1} \].
Therefore, the equation has three factors out of which one factor is real and two factors are imaginary.
The real factor of the equation is \[x\]=3 and the imaginary factors of the equation are \[x\]=\[\dfrac{{ - 3 \pm 3\sqrt 3 i}}{2}\].
Note: While solving the equation \[4\left( {{x^3} - 27} \right)\]=\[0\] , we generally forget to consider the complex factors i.e. the imaginary factors of the equation .Also while substituting the values in the formula. We should not at all forget to substitute the \[ \pm \]sign in \[x\]=\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. A cubic function generally has one or three real roots which are not necessarily distinct. The graph of a cubic function always has one inflection point.
Complete step by step answer:
Since 4 is the coefficient of the variable term, we will take 4 common, which will give us with,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]
Now we will equate above equation to zero, which will give us with,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]= \[0\]
We know that 27 is cube of 3, therefore,
\[ \Rightarrow \]\[4\left( {{x^3} - 27} \right)\]= \[0\]
Now here we will use the identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\], where \[x\]=a and 3=b,
\[ \Rightarrow \]4\[\left( {x - 3} \right)\left( {{x^2} + 3x + {3^2}} \right)\]=\[0\]
Now either,
\[\left( {x - 3} \right)\]=\[0\] OR \[\left( {{x^2} + 3x + 9} \right)\]=\[0\]
So, \[x\]=3 and now we will use the formula for finding roots quadratic equation in other equation,
FORMULA : \[x\]=\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
In above equation, a=1, b=3, c=9, after substituting the values, we get:-
\[x\]=\[\dfrac{{ - 3 \pm \sqrt {{3^2} - 4 \times 1 \times 9} }}{{2 \times 1}}\],
\[x\]=\[\dfrac{{ - 3 \pm \sqrt { - 27} }}{2}\],
\[x\]=\[\dfrac{{ - 3 \pm 3\sqrt 3 i}}{2}\] , where \[i = \sqrt { - 1} \].
Therefore, the equation has three factors out of which one factor is real and two factors are imaginary.
The real factor of the equation is \[x\]=3 and the imaginary factors of the equation are \[x\]=\[\dfrac{{ - 3 \pm 3\sqrt 3 i}}{2}\].
Note: While solving the equation \[4\left( {{x^3} - 27} \right)\]=\[0\] , we generally forget to consider the complex factors i.e. the imaginary factors of the equation .Also while substituting the values in the formula. We should not at all forget to substitute the \[ \pm \]sign in \[x\]=\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. A cubic function generally has one or three real roots which are not necessarily distinct. The graph of a cubic function always has one inflection point.
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