Question

Evaluate ${{\left( 3a+5b+c \right)}^{2}}$ .
(a) $9{{a}^{2}}+25{{b}^{2}}+{{c}^{2}}+30ab+10bc+6ac$
(b) $9{{a}^{2}}+25{{b}^{2}}+{{c}^{2}}+90ab+10bc+6ac$
(c) ${{a}^{2}}+25{{b}^{2}}+{{c}^{2}}30ab+10bc+6ac$
(d) ${{a}^{2}}+25{{b}^{2}}+{{c}^{2}}30ab+10bc+ac$

Answer 1

Hint: To evaluate ${{\left( 3a+5b+c \right)}^{2}}$ , we have to use the algebraic identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$ to expand the LHS. We have to substitute a as 3a and b as 5b in this identity and simplify.

Complete step by step solution:
We have to evaluate ${{\left( 3a+5b+c \right)}^{2}}$ . We have to use algebraic identity to expand this expression. We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$ . When we compare the LHS of this identity with ${{\left( 3a+5b+c \right)}^{2}}$ , we have to change a as 3a and b as 5b in the identity.
$\Rightarrow {{\left( 3a+5b+c \right)}^{2}}={{\left( 3a \right)}^{2}}+{{\left( 5b \right)}^{2}}+{{c}^{2}}+\left( 2\times 3a\times 5b \right)+\left( 2\times 5b\times c \right)+\left( 2\times 3a\times c \right)$
Let us solve the terms inside the bracket.
\[\Rightarrow {{\left( 3a+5b+c \right)}^{2}}=9{{a}^{2}}+25{{b}^{2}}+{{c}^{2}}+30ab+10bc+6ac\]

So, the correct answer is “Option a”.

Note: Students must be thorough with the algebraic identities to expand the terms.Students have a chance of making mistake by writing the expansion of ${{\left( a+b+c \right)}^{2}}$ as ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab+2bc-2ac$ . This formula is for ${{\left( a-b-c \right)}^{2}}$ . If the students forget this identity, they can use the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to evaluate ${{\left( 3a+5b+c \right)}^{2}}$ . For this, we have to combine the first two terms inside the bracket.
$\Rightarrow {{\left( 3a+5b+c \right)}^{2}}={{\left[ \left( 3a+5b \right)+c \right]}^{2}}$
Now, we have to substitute for a as $3a+5b$ and b as c in the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
$\Rightarrow {{\left[ \left( 3a+5b \right)+c \right]}^{2}}={{\left( 3a+5b \right)}^{2}}+2\left( 3a+5b \right)c+{{c}^{2}}$
Let us simplify the second term of RHS by applying distributive property. Then, we can write the above equation as
$\Rightarrow {{\left[ \left( 3a+5b \right)+c \right]}^{2}}={{\left( 3a+5b \right)}^{2}}+6ac+10bc+{{c}^{2}}$
Now, we have to expand ${{\left( 3a+5b \right)}^{2}}$ in the RHS using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
$\Rightarrow {{\left[ \left( 3a+5b \right)+c \right]}^{2}}={{\left( 3a \right)}^{2}}+\left( 2\times 3a\times 5b \right)+{{\left( 5b \right)}^{2}}+6ac+10bc+{{c}^{2}}$
Let us simplify the terms inside the bracket.
$\Rightarrow {{\left[ \left( 3a+5b \right)+c \right]}^{2}}=9{{a}^{2}}+30ab+25{{b}^{2}}+6ac+10bc+{{c}^{2}}$
Let us rearrange the terms. We can write the above equation as
$\Rightarrow {{\left( 3a+5b+c \right)}^{2}}=9{{a}^{2}}+25{{b}^{2}}+{{c}^{2}}+30ab+10bc+6ac$
Similarly, we can also combine ${{\left( 3a+5b+c \right)}^{2}}$ as ${{\left[ 3a+\left( 5b+c \right) \right]}^{2}}$ and simplify.