
How do you factor \[21{{x}^{3}}-28x\]?
Answer
534.6k+ views
Hint: In this problem, we have to find the factor for the given expression. We can see that the expression contains a common factor 7x. We can take the common factor 7x outside the expression and write the remaining terms inside the brackets as a factor. We should also be clear that the factor which we have taken outside must be correct as we multiply it inside, we should get the given expression. We will get the required factor for the expression.
Complete step by step solution:
We know that the expression given is,
\[21{{x}^{3}}-28x\]
We can see that the expression contains a common factor 7x.
We can take the common factor 7x outside the expression and write the remaining terms inside the brackets as a factor.
We can split the term as,
\[\begin{align}
& \Rightarrow 7x\times 3{{x}^{2}}=21{{x}^{3}} \\
& \Rightarrow 7x\times 4x=28x \\
\end{align}\]
We can write the given expression as by taking the common term 7, first, we get
\[\Rightarrow 7\left( 3{{x}^{3}}-4x \right)\]
We can now take the common term x outside the expression and write the remaining terms inside the brackets as a factor.
\[\Rightarrow 7x\left( 3{{x}^{2}}-4 \right)\]
Therefore, the required factor is \[7x\left( 3{{x}^{2}}-4 \right)\]
Note: We should also be cleared that the factor which we have taken outside must be correct as we multiply it inside, we should get the given expression. We can also multiply the factor to check for the correct answer.
\[\begin{align}
& \Rightarrow 7x\times \left( 3{{x}^{2}}-4 \right) \\
& \Rightarrow 7x\times 3{{x}^{2}}-7x\times 4 \\
& \Rightarrow 21{{x}^{3}}-28x \\
\end{align}\]
Therefore, the factor that we got in the result is correct.
Complete step by step solution:
We know that the expression given is,
\[21{{x}^{3}}-28x\]
We can see that the expression contains a common factor 7x.
We can take the common factor 7x outside the expression and write the remaining terms inside the brackets as a factor.
We can split the term as,
\[\begin{align}
& \Rightarrow 7x\times 3{{x}^{2}}=21{{x}^{3}} \\
& \Rightarrow 7x\times 4x=28x \\
\end{align}\]
We can write the given expression as by taking the common term 7, first, we get
\[\Rightarrow 7\left( 3{{x}^{3}}-4x \right)\]
We can now take the common term x outside the expression and write the remaining terms inside the brackets as a factor.
\[\Rightarrow 7x\left( 3{{x}^{2}}-4 \right)\]
Therefore, the required factor is \[7x\left( 3{{x}^{2}}-4 \right)\]
Note: We should also be cleared that the factor which we have taken outside must be correct as we multiply it inside, we should get the given expression. We can also multiply the factor to check for the correct answer.
\[\begin{align}
& \Rightarrow 7x\times \left( 3{{x}^{2}}-4 \right) \\
& \Rightarrow 7x\times 3{{x}^{2}}-7x\times 4 \\
& \Rightarrow 21{{x}^{3}}-28x \\
\end{align}\]
Therefore, the factor that we got in the result is correct.
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