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Question

Question: How do you expand \[{{\left( b+2 \right)}^{2}}\] ?...

How do you expand (b+2)2{{\left( b+2 \right)}^{2}} ?

Explanation

Solution

From the given question we have to expand the binomial(b+2)2{{\left( b+2 \right)}^{2}}. To expand this, we have to use binomial theorem i.e., the expansion of (a+b)n=k=0nnCk.(ankbk){{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}. Here we have to substitute b in place of a and 22 in place of b. by this we can expand the above binomial (b+2)2{{\left( b+2 \right)}^{2}}.

Complete step by step solution:
From the given question we have to expand the binomial (b+2)2{{\left( b+2 \right)}^{2}}
As we know that we have to expand this by using binomial theorem. Binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial (a+b)n{{\left( a+b \right)}^{n}} into a sum involving terms of the form caxbyc{{a}^{x}}{{b}^{y}}, where the exponents x and y are nonnegative integers with x+y=nx+y=n, and the coefficient c of each term is a specific positive integer depending on n and x. the coefficient c in the term of caxbyc{{a}^{x}}{{b}^{y}} is known as the binomial coefficient.
Now, by using binomial theorem we have to expand the binomial (b+2)2{{\left( b+2 \right)}^{2}}.
(b+2)2=k=022!(2k)!k!.(b2k).(2)k\Rightarrow {{\left( b+2 \right)}^{2}}=\sum\limits_{k=0}^{2}{\dfrac{2!}{\left( 2-k \right)!k!}.\left( {{b}^{2-k}} \right)}.{{\left( 2 \right)}^{k}}
Now we have to expand the summation.
(b+2)2=2!(20)!0!.(b20).20+2!(21)!1!.(b21).(2)1+2!(22)!2!.(b22).(2)2\Rightarrow {{\left( b+2 \right)}^{2}}=\dfrac{2!}{\left( 2-0 \right)!0!}.\left( {{b}^{2-0}} \right){{.2}^{0}}+\dfrac{2!}{\left( 2-1 \right)!1!}.\left( {{b}^{2-1}} \right).{{\left( 2 \right)}^{1}}+\dfrac{2!}{\left( 2-2 \right)!2!}.\left( {{b}^{2-2}} \right).{{\left( 2 \right)}^{2}}
Now, we have to simplify the above form.
(b+2)2=(1.(2)0.b2)+(2.(2)1.b1)+(1.(2)2.b0)\Rightarrow {{\left( b+2 \right)}^{2}}=\left( 1.{{\left( 2 \right)}^{0}}.{{b}^{2}} \right)+\left( 2.{{\left( 2 \right)}^{1}}.{{b}^{1}} \right)+\left( 1.{{\left( 2 \right)}^{2}}.{{b}^{0}} \right)
After the simplification the above binomial expression is
(b+2)2=b2+4b+4\Rightarrow {{\left( b+2 \right)}^{2}}={{b}^{2}}+4b+4
Therefore, this is the required binomial expansion for the given binomial (b+2)2{{\left( b+2 \right)}^{2}}.

Note: Students should know the expansions and binomial theorem. Student should be careful with signs and calculation. Student can do this problem by simply formula as we know that (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}. By substituting in this formula also we can expand the (b+2)2{{\left( b+2 \right)}^{2}}.