Question

(4p+3q+2r)square find the product ​

Answer 1

Answer:

We know,

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

Therefore, we will use this algebraic identity to find the product of (4p+3q+2r)² where,

• 4p = a
• 3q = b
• 2r = c

Hence,

$\tt{ {(4p + 3q + 2r)}^{2} }$

-:- Here, we have formed an equation to find the final product of the algebraic expression -:-

$\tt{ ={(4p)}^{2} + {(3q)}^{2} + {(2r)}^{2} + 2 (4p)(3q)+ 2(3q)(2r) + 2(2r)(4p)}$

-:- We have written the product of

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca -:-

$\tt\orange{=16 {p}^{2} + 9 {q}^{2} + 4 {r}^{2} + 24pq + 12qr + 16pr }$

$\text{-:-ANSWER-:-}$

REQUIRED ANSWER:-

$\tt{\therefore Product \ of \ {(4p + 3q + 2r)}^{2} \ is }$

$\tt{ 16 {p}^{2} + 9 {q}^{2} + 4 {r}^{2} + 24pq + 12qr + 16pr .}$