Question

(2a-5b-3c)whole square expand
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Answer 1

$\longrightarrow{\tt{ {(2a - 5b - 3c)}^{2}}}$

The expansion of identity used :

$\longrightarrow{\tt{{(a - b - c)}^{2}}}$

$\tt{\implies {(a)}^{2}+{(b)}^{2}+ {(c)}^{2} + 2ab + 2bc - 2ca}$

Following the same expansion we can expand the given problem.

$\longrightarrow{\tt{{(2a - 5b - 3c)}^{2}}}$

$\tt{ a \implies 2a}$

$\tt{b \implies 5b}$

$\tt{c \implies 3c}$

Substituting the values ; we get :

$\longrightarrow{\tt{{(2a - 5b - 3c)}^{2}}}$

$\longrightarrow{\tt{{(2a)}^{2} + {(-5b)}^{2} + {(-3c)}^{2} + 2 \times 2a \times (-5b) + 2 \times (-5b) \times (-3c) + 2 \times 2a \times (-3c)}}$

Further simplification gives :

$\longrightarrow{\tt{{(2a - 5b - 3c)}^{2}}}$

$\longrightarrow{\tt{{(4a)}^{2} + {(25b)}^{2} + {(9c)}^{2} - 20ab + 30bc - 12ac}}$

So , the expansion of :

$\tt{{(2a - 5b - 3c)}^{2}}$ is :

$\longrightarrow{\boxed{\tt\red{{(4a)}^{2} + {(25b)}^{2} + {(9c)}^{2} - 20ab + 30bc - 12ac}}}$

Answer 2

Answer:

$\huge\bold\star\red{Answer}\star$

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Given:-

• An expression (2a-5b-3c)

To find:-

• (2a-5b-3c)^2

Solution:-

(2a-5b-3c)^2

={2a+(-5b)+(-3c)}^2

We will use the identity:-

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

let,a=2a

b=(-5b)

c=(-3c)

putting the value of a,b,c in the expression

${(2a)}^{2} + {( - 5b)}^{2} + {( - 5c)}^{2} + 2.2a.( - 5b) + 2.( - 5b)( - 3c) + 2.( - 3c).2a$

$= 4 {a}^{2} + 25 {b}^{2} + 9 {c}^{2} - 20ab + 30bc - 12ca$

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$\huge\green{Hope\:this\:help\:you}$