How do you multiply \[{{\left( 2x+5 \right)}^{3}}\]?
Answer
632.7k+ views
Hint: A polynomial is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). To multiply any polynomial repeated times, we use distributive law to separate each polynomial. To simplify it, we use the FOIL rule which is commonly used in factoring, meaning to start by multiplying the two first variables first, then outer, then inner, then last.
Complete step by step answer:
As per the given question, we have to multiply the given expression. And the given expression is \[{{\left( 2x+5 \right)}^{3}}\].
If we have multiplication of two binomials \[(a+b)(c+d)\], then we have to use distributive property to divide the expression as the sum of \[a(c+d)\] and \[b(c+d)\]. This can be further divided as the sum of four monomials \[a.c\], \[a.d\], \[b.c\] and \[b.d\]. That is, we can express this as:
\[\Rightarrow (a+b)(c+d)=a(c+d)+b(c+d)=a.c+a.d+b.c+b.d\] ---(1)
Using distributive property for \[{{\left( 2x+5 \right)}^{3}}\], we can write it as
\[\Rightarrow {{(2x+5)}^{3}}=(2x+5).(2x+5).(2x+5)\]
Now, we use the FOIL rule to multiply the first two binomials. While multiplying the first two we use distributive law to expand the same as in equation (1). So, we can rewrite the expression as
\[\begin{align}
& \Rightarrow (2x+5).(2x+5).(2x+5)=[(2x+5).(2x+5)].(2x+5) \\
& \Rightarrow \left[ 2x(2x+5)+5(2x+5) \right].(2x+5)=[2x(2x)+2x(5)+5(2x)+5(5)].(2x+5) \\
\end{align}\]
\[\begin{align}
& \Rightarrow [4{{x}^{2}}+10x+10x+25].(2x+5) \\
& \Rightarrow \left[ 4{{x}^{2}}+20x+25 \right].(2x+5) \\
\end{align}\] ------(2)
If we have to multiply a trinomial with binomial \[(a+b+c)(d+e)\], then we get the simplified form of this as \[(a+b+c)(d+e)=ad+ae+bd+be+cd+ce\]. --(3)
From equation (3), we can rewrite the equation (2) as \[\begin{align}
& \Rightarrow \left[ 4{{x}^{2}}+20x+25 \right].(2x+5)=\left[ 4{{x}^{2}}(2x+5)+20x(2x+5)+25(2x+5) \right] \\
& \Rightarrow \left[ 4{{x}^{2}}(2x)+4{{x}^{2}}(5)+20x(2x)+20x(5)+25(2x)+25(5) \right] \\
& \Rightarrow \left[ 8{{x}^{3}}+20{{x}^{2}}+40{{x}^{2}}+100x+50x+125 \right] \\
& \Rightarrow 8{{x}^{3}}+60{{x}^{2}}+150x+15 \\
\end{align}\]
\[\therefore 8{{x}^{3}}+60{{x}^{2}}+150x+15\] is the required answer.
Note: In order to solve these types of questions, we should know the distributive property and simplify. While multiplying two monomials with different powers, we should remember the fact that we need to add up the powers of like variables. We should avoid calculation mistakes to get the correct answer.
Complete step by step answer:
As per the given question, we have to multiply the given expression. And the given expression is \[{{\left( 2x+5 \right)}^{3}}\].
If we have multiplication of two binomials \[(a+b)(c+d)\], then we have to use distributive property to divide the expression as the sum of \[a(c+d)\] and \[b(c+d)\]. This can be further divided as the sum of four monomials \[a.c\], \[a.d\], \[b.c\] and \[b.d\]. That is, we can express this as:
\[\Rightarrow (a+b)(c+d)=a(c+d)+b(c+d)=a.c+a.d+b.c+b.d\] ---(1)
Using distributive property for \[{{\left( 2x+5 \right)}^{3}}\], we can write it as
\[\Rightarrow {{(2x+5)}^{3}}=(2x+5).(2x+5).(2x+5)\]
Now, we use the FOIL rule to multiply the first two binomials. While multiplying the first two we use distributive law to expand the same as in equation (1). So, we can rewrite the expression as
\[\begin{align}
& \Rightarrow (2x+5).(2x+5).(2x+5)=[(2x+5).(2x+5)].(2x+5) \\
& \Rightarrow \left[ 2x(2x+5)+5(2x+5) \right].(2x+5)=[2x(2x)+2x(5)+5(2x)+5(5)].(2x+5) \\
\end{align}\]
\[\begin{align}
& \Rightarrow [4{{x}^{2}}+10x+10x+25].(2x+5) \\
& \Rightarrow \left[ 4{{x}^{2}}+20x+25 \right].(2x+5) \\
\end{align}\] ------(2)
If we have to multiply a trinomial with binomial \[(a+b+c)(d+e)\], then we get the simplified form of this as \[(a+b+c)(d+e)=ad+ae+bd+be+cd+ce\]. --(3)
From equation (3), we can rewrite the equation (2) as \[\begin{align}
& \Rightarrow \left[ 4{{x}^{2}}+20x+25 \right].(2x+5)=\left[ 4{{x}^{2}}(2x+5)+20x(2x+5)+25(2x+5) \right] \\
& \Rightarrow \left[ 4{{x}^{2}}(2x)+4{{x}^{2}}(5)+20x(2x)+20x(5)+25(2x)+25(5) \right] \\
& \Rightarrow \left[ 8{{x}^{3}}+20{{x}^{2}}+40{{x}^{2}}+100x+50x+125 \right] \\
& \Rightarrow 8{{x}^{3}}+60{{x}^{2}}+150x+15 \\
\end{align}\]
\[\therefore 8{{x}^{3}}+60{{x}^{2}}+150x+15\] is the required answer.
Note: In order to solve these types of questions, we should know the distributive property and simplify. While multiplying two monomials with different powers, we should remember the fact that we need to add up the powers of like variables. We should avoid calculation mistakes to get the correct answer.
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