Answer :
To solve for [tex]\((11 + 4i)(2 - 5i)\)[/tex], let's follow these steps:
1. Distribute each term in the first binomial by each term in the second binomial:
[tex]\[ (11 + 4i)(2 - 5i) = 11 \cdot 2 + 11 \cdot (-5i) + 4i \cdot 2 + 4i \cdot (-5i) \][/tex]
2. Compute each of these products:
[tex]\[ 11 \cdot 2 = 22 \][/tex]
[tex]\[ 11 \cdot (-5i) = -55i \][/tex]
[tex]\[ 4i \cdot 2 = 8i \][/tex]
[tex]\[ 4i \cdot (-5i) = -20i^2 \][/tex]
3. Simplify the product of imaginary units [tex]\(i^2\)[/tex]:
Since [tex]\(i^2 = -1\)[/tex], we have:
[tex]\[ -20i^2 = -20(-1) = 20 \][/tex]
4. Add up all the terms, combining like terms:
[tex]\[ 22 + (-55i) + 8i + 20 \][/tex]
5. Combine the real and imaginary parts separately:
The real parts:
[tex]\[ 22 + 20 = 42 \][/tex]
The imaginary parts:
[tex]\[ -55i + 8i = -47i \][/tex]
Therefore, the expression [tex]\((11 + 4i)(2 - 5i)\)[/tex] simplifies to:
[tex]\[ 42 - 47i \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{42 - 47i} \][/tex]
From the given options, the correct match is:
[tex]\[ \boxed{E} \][/tex]
1. Distribute each term in the first binomial by each term in the second binomial:
[tex]\[ (11 + 4i)(2 - 5i) = 11 \cdot 2 + 11 \cdot (-5i) + 4i \cdot 2 + 4i \cdot (-5i) \][/tex]
2. Compute each of these products:
[tex]\[ 11 \cdot 2 = 22 \][/tex]
[tex]\[ 11 \cdot (-5i) = -55i \][/tex]
[tex]\[ 4i \cdot 2 = 8i \][/tex]
[tex]\[ 4i \cdot (-5i) = -20i^2 \][/tex]
3. Simplify the product of imaginary units [tex]\(i^2\)[/tex]:
Since [tex]\(i^2 = -1\)[/tex], we have:
[tex]\[ -20i^2 = -20(-1) = 20 \][/tex]
4. Add up all the terms, combining like terms:
[tex]\[ 22 + (-55i) + 8i + 20 \][/tex]
5. Combine the real and imaginary parts separately:
The real parts:
[tex]\[ 22 + 20 = 42 \][/tex]
The imaginary parts:
[tex]\[ -55i + 8i = -47i \][/tex]
Therefore, the expression [tex]\((11 + 4i)(2 - 5i)\)[/tex] simplifies to:
[tex]\[ 42 - 47i \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{42 - 47i} \][/tex]
From the given options, the correct match is:
[tex]\[ \boxed{E} \][/tex]
