Question

(x+3y)^2 + (3x+y)^2​

Answer 1

Answer:

ln these type of questions we have to used the identity that is

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

now using this identity we can simplify it as

${(x + 3y)}^{2} + {(3x + y)}^{2} \\ = { {x}^{2}} + {(3y)}^{2} + 2 \times x \times 3y \\ + {(3x)}^{2} + {y}^{2} + 2 \times 3 x\times y \\ = {x}^{2} + 9 {y}^{2} + 6xy \\ + 9 {x}^{2} + {y}^{2} + 6xy \\ = 10 {x}^{2} + 10 {y}^{2} + 12xy$

$\: Now \: divide \: whole \: polynomial \: by \: 2 \: to \: get \: it \: in \: simplest \: form \: </p><p>$

$\implies \: 10 {x}^{2} + 10 {y}^{2} + 12xy = 5 {x}^{2} + 5 {y}^{2} + 6xy$

this is the required final polynomial in the term of x and y.

Answer 2

Solution :

∴ (x + 3y)² + (3x + y)²

= (x² + 6xy + 9y²) + (9x² + 6xy + y²)

= x² + 6xy + 9y² + 9x² + 6xy + y²

= 10x² + 12xy + 10y²