Simplify (4+i)-(3-5i)(-2+5i)

$(4 + i) - (3 - 5 i) (- 2 + 5 i)$

Step 1

Simplify each term.

Apply the distributive property.

$4 + i + (- 1 \cdot 3 - (- 5 i)) (- 2 + 5 i)$

Multiply $- 1$ by $3$ .

$4 + i + (- 3 - (- 5 i)) (- 2 + 5 i)$

Multiply $- 5$ by $- 1$ .

$4 + i + (- 3 + 5 i) (- 2 + 5 i)$

Expand $(- 3 + 5 i) (- 2 + 5 i)$ using the FOIL Method.

Apply the distributive property.

$4 + i - 3 (- 2 + 5 i) + 5 i (- 2 + 5 i)$

Apply the distributive property.

$4 + i - 3 \cdot - 2 - 3 (5 i) + 5 i (- 2 + 5 i)$

Apply the distributive property.

$4 + i - 3 \cdot - 2 - 3 (5 i) + 5 i \cdot - 2 + 5 i (5 i)$

Simplify and combine like terms.

Simplify each term.

Multiply $- 3$ by $- 2$ .

$4 + i + 6 - 3 (5 i) + 5 i \cdot - 2 + 5 i (5 i)$

Multiply $5$ by $- 3$ .

$4 + i + 6 - 15 i + 5 i \cdot - 2 + 5 i (5 i)$

Multiply $- 2$ by $5$ .

$4 + i + 6 - 15 i - 10 i + 5 i (5 i)$

Multiply $5 i (5 i)$ .

Multiply $5$ by $5$ .

$4 + i + 6 - 15 i - 10 i + 25 i i$

Raise $i$ to the power of $1$ .

$4 + i + 6 - 15 i - 10 i + 25 (i^{1} i)$

Raise $i$ to the power of $1$ .

$4 + i + 6 - 15 i - 10 i + 25 (i^{1} i^{1})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 + i + 6 - 15 i - 10 i + 25 i^{1 + 1}$

Add $1$ and $1$ .

$4 + i + 6 - 15 i - 10 i + 25 i^{2}$

Rewrite $i^{2}$ as $- 1$ .

$4 + i + 6 - 15 i - 10 i + 25 \cdot - 1$

Multiply $25$ by $- 1$ .

$4 + i + 6 - 15 i - 10 i - 25$

Subtract $25$ from $6$ .

$4 + i - 19 - 15 i - 10 i$

Subtract $10 i$ from $- 15 i$ .

$4 + i - 19 - 25 i$

Step 2

Simplify by adding terms.

Subtract $19$ from $4$ .

$- 15 + i - 25 i$

Subtract $25 i$ from $i$ .

$- 15 - 24 i$

Simplify (4+i)-(3-5i)(-2+5i)

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