
How do you find the product of \[\left( 10x \right)\left( 4{{x}^{7}} \right)\]?
Answer
561.6k+ views
Hint: Take the product of the constant terms, i.e., 10 and 4 separately and the product of the variables x and \[{{x}^{7}}\] separately. Use the formula of exponents and powers given as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] to simplify the product of variables. Now, multiply the resultant of the two products considered at the initial step to get the answer.
Complete step by step answer:
Here, we have been provided with a pair of monomials: 10x and \[4{{x}^{7}}\]. We have been asked to find their product. But first let us understand the meaning of the term ‘monomial’.
Now, monomial is an expression that contains only one term. For example: - a, 6n, 5x, \[9{{y}^{2}}\], \[10{{a}^{3}}{{b}^{2}}{{c}^{5}}\] etc. These are all examples of monomials as they contain only one term. A monomial converts into a binomial when the variables are separated with a (+) or (-) sign.
Now, let us come to the question. We have two monomials 10x and \[4{{x}^{7}}\]. So, taking their product, we get,
\[\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=\left( 10\times 4 \right)\left( x\times {{x}^{7}} \right)\]
Here, we have grouped the constants together and the variables together and considering their product, we get,
\[\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40\left( x\times {{x}^{7}} \right)\]
Using the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get,
\[\begin{align}
& \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{1+7}} \\
& \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{8}} \\
\end{align}\]
Hence, the required product is \[40{{x}^{8}}\].
Note: One may check the answer by substituting some small value to the variable x. For example: - let us substitute x = 1, then in the L.H.S. we will have \[\left( 10\times 1 \right)\left( 4\times {{1}^{7}} \right)=10\times 4=40\]. Now, in the R.H.S. we will have \[4\times {{1}^{8}}=40\]. Do not substitute any larger value of x because the exponent is 8 which is a large number. You must remember the formulas of exponents and powers like: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] etc.
Complete step by step answer:
Here, we have been provided with a pair of monomials: 10x and \[4{{x}^{7}}\]. We have been asked to find their product. But first let us understand the meaning of the term ‘monomial’.
Now, monomial is an expression that contains only one term. For example: - a, 6n, 5x, \[9{{y}^{2}}\], \[10{{a}^{3}}{{b}^{2}}{{c}^{5}}\] etc. These are all examples of monomials as they contain only one term. A monomial converts into a binomial when the variables are separated with a (+) or (-) sign.
Now, let us come to the question. We have two monomials 10x and \[4{{x}^{7}}\]. So, taking their product, we get,
\[\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=\left( 10\times 4 \right)\left( x\times {{x}^{7}} \right)\]
Here, we have grouped the constants together and the variables together and considering their product, we get,
\[\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40\left( x\times {{x}^{7}} \right)\]
Using the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get,
\[\begin{align}
& \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{1+7}} \\
& \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{8}} \\
\end{align}\]
Hence, the required product is \[40{{x}^{8}}\].
Note: One may check the answer by substituting some small value to the variable x. For example: - let us substitute x = 1, then in the L.H.S. we will have \[\left( 10\times 1 \right)\left( 4\times {{1}^{7}} \right)=10\times 4=40\]. Now, in the R.H.S. we will have \[4\times {{1}^{8}}=40\]. Do not substitute any larger value of x because the exponent is 8 which is a large number. You must remember the formulas of exponents and powers like: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] etc.
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