Verified Answer :

\(i^2(2i^2-5) \rightarrow 7\)
\(i^2(3+i^2) \rightarrow -2\)
\(2i(2i-i^3) \rightarrow -6\)
\(i(4i^3-i) \rightarrow 5\)

Answer:

Distributive property says that:

\(a \cdot (b+c) =a\cdot b+ a\cdot c\)

We know that

\(i^2= -1\) where i is the imaginary

Given the complex numbers:

A.

\(i^2(2i^2-5)\)

⇒\(-1(2(-1)-5) = -1(-2-5)= -1(-7) = 7\)

B.

\(i^2(3+i^2)\)

⇒\(-1(3+(-1)) = -1(3-1)= -1(2) = -2\)

C.

\(2i(2i-i^3)\)

Using distributive property

⇒\(4i^2-2i^4\)

⇒\(4(-1)-2(i^2)^2 = -4-2(-1)^2 = -4 -2 = -6\)

D.

\(i(4i^3-i)\)

Using distributive property

⇒\(4i^4-i^2\)

⇒\(4(i^2)^2-i^2\)

⇒\(4(-1)^2-(-1) = 4+1 = 5\)

Therefore. matching defined as:

A. \(i^2(2i^2-5)\)  → 7

B. \(i^2(3+i^2)\)  → -2

C. \(2i(2i-i^3)\)  → -6

D. \(i(4i^3-i)\)    →  5