Find the Roots (Zeros) f(x)=(x+2)(x-1)(x-(4+3i))(x-(4-3i))

$f (x) = (x + 2) (x - 1) (x - (4 + 3 i)) (x - (4 - 3 i))$

Step 1

Set $(x + 2) (x - 1) (x - (4 + 3 i)) (x - (4 - 3 i))$ equal to $0$ .

$(x + 2) (x - 1) (x - (4 + 3 i)) (x - (4 - 3 i)) = 0$

Step 2

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$x - 1 = 0$

$x - (4 + 3 i) = 0$

$x - (4 - 3 i) = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - (4 + 3 i)$ equal to $0$ and solve for $x$ .

Set $x - (4 + 3 i)$ equal to $0$ .

$x - (4 + 3 i) = 0$

Solve $x - (4 + 3 i) = 0$ for $x$ .

Simplify each term.

Apply the distributive property.

$x - 1 \cdot 4 - (3 i) = 0$

Multiply $- 1$ by $4$ .

$x - 4 - (3 i) = 0$

Multiply $3$ by $- 1$ .

$x - 4 - 3 i = 0$

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$x - 3 i = 4$

Add $3 i$ to both sides of the equation.

$x = 4 + 3 i$

Set $x - (4 - 3 i)$ equal to $0$ and solve for $x$ .

Set $x - (4 - 3 i)$ equal to $0$ .

$x - (4 - 3 i) = 0$

Solve $x - (4 - 3 i) = 0$ for $x$ .

Simplify each term.

Apply the distributive property.

$x - 1 \cdot 4 - (- 3 i) = 0$

Multiply $- 1$ by $4$ .

$x - 4 - (- 3 i) = 0$

Multiply $- 3$ by $- 1$ .

$x - 4 + 3 i = 0$

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$x + 3 i = 4$

Subtract $3 i$ from both sides of the equation.

$x = 4 - 3 i$

The final solution is all the values that make $(x + 2) (x - 1) (x - (4 + 3 i)) (x - (4 - 3 i)) = 0$ true.

$x = - 2, 1, 4 + 3 i, 4 - 3 i$

Step 3

Find the Roots (Zeros) f(x)=(x+2)(x-1)(x-(4+3i))(x-(4-3i))

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