Find the product of (x + (3+5i))^2

x^2 +6x +10xi -16 +30i

Step-by-step explanation:

(x + (3+5i))^2

Rewriting as

(x + (3+5i))  (x + (3+5i))

Let 3+5i = m

(x + m)  (x + m)

FOILing

x^2 + xm+xm+m^2

x^2 +2xm + m^2

Replacing m with 3+5i

x^2 +2x (3+5i) +(3+5i)^2

Distribute

x^2 +6x +10xi +(3+5i)^2

Now FOIL (3+5i)^2

( 3+5i) ( 3+5i)

 3*3+ 3*5i + 3*5i+ 25i^2

= 9 +15i+15i +25(-1)

9+30i -25

-16 +30 i

Replacing that in the equation

x^2 +6x +10xi -16 +30i

x² + 6x + 10xi -16 + 30i

Step-by-step explanation:

(x + (3 + 5i) )²

use the binomial theorem:

(a + b)² = a² + 2ab + b² to expand (x + (3 + 5i) )²

substitute the values,

x² + (6 + 10i)x + (-16 + 30i)

therefore

x² + 6x + 10xi -16 + 30i

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to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk