How do you factor\[2{{x}^{3}}+54\] ?
Answer
619.5k+ views
Hint:In the given question, we have been asked to factorize the cubic equation. In order to solve the question first we need to take a common factor from the whole equation and we obtain a quadratic equation. Then once you get a quadratic equation use the standard factorization method to factorize the quadratic equation.
Formula used:
A sum of two perfect cubes can be factorize in that way,
\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
Complete step by step answer:
We have the given equation:
\[2{{x}^{3}}+54\]
Taking 2 as a common factor, we get
\[2\left( {{x}^{3}}+27 \right)\]
Factorize \[\left( {{x}^{3}}+27 \right)\] as a sum of cubes.As we know that, \[{{x}^{3}}\] is a cube of \[x\] and 27 is the cube of 3.Therefore,
\[2\left( {{x}^{3}}+{{3}^{3}} \right)\]
Factorize\[\left( {{x}^{3}}+{{3}^{3}} \right)\], by using the property of\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\], we get
\[2\left( \left( x+3 \right)\times \left( {{x}^{2}}-3x+9 \right) \right)\]
Factorize the quadratic equation i.e. \[\left( {{x}^{2}}-3x+9 \right)\] by splitting the middle term.As we saw there will be no such factors can be found whose sum equals to the coefficient of the middle term, which is -3. Thus, \[\left( {{x}^{2}}-3x+9 \right)\] cannot be factored
Therefore, \[2\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)\] is the required solution.
Note:Remember that the number of roots of any given polynomial is always equal to the highest power that is given in the question. The number of roots of a cubic equation is 3 because the highest power in the cubic polynomial is 3. Similarly, the number of roots of any quadratic equation will be always 2 because the highest power of any quadratic equation is 2. Also try of solve the cubic equation by using the sum or difference of two perfect cubes property i.e. \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\]. Always examine carefully when you write down any equation or any property to get the correct solution.
Formula used:
A sum of two perfect cubes can be factorize in that way,
\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
Complete step by step answer:
We have the given equation:
\[2{{x}^{3}}+54\]
Taking 2 as a common factor, we get
\[2\left( {{x}^{3}}+27 \right)\]
Factorize \[\left( {{x}^{3}}+27 \right)\] as a sum of cubes.As we know that, \[{{x}^{3}}\] is a cube of \[x\] and 27 is the cube of 3.Therefore,
\[2\left( {{x}^{3}}+{{3}^{3}} \right)\]
Factorize\[\left( {{x}^{3}}+{{3}^{3}} \right)\], by using the property of\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\], we get
\[2\left( \left( x+3 \right)\times \left( {{x}^{2}}-3x+9 \right) \right)\]
Factorize the quadratic equation i.e. \[\left( {{x}^{2}}-3x+9 \right)\] by splitting the middle term.As we saw there will be no such factors can be found whose sum equals to the coefficient of the middle term, which is -3. Thus, \[\left( {{x}^{2}}-3x+9 \right)\] cannot be factored
Therefore, \[2\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)\] is the required solution.
Note:Remember that the number of roots of any given polynomial is always equal to the highest power that is given in the question. The number of roots of a cubic equation is 3 because the highest power in the cubic polynomial is 3. Similarly, the number of roots of any quadratic equation will be always 2 because the highest power of any quadratic equation is 2. Also try of solve the cubic equation by using the sum or difference of two perfect cubes property i.e. \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\times \left( {{a}^{2}}-ab+{{b}^{2}} \right)\]. Always examine carefully when you write down any equation or any property to get the correct solution.
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