Question

find all solutions to the equation \\(x^3 - 4x^2 + 9x - 10 = 0\\).

\\(\bigcirc\\) \\(-10, 1 + 2i, 1 - 2i\\)
\\(\bigcirc\\) \\(-2, -1 + 2i, -1 - 2i\\)
\\(\bigcirc\\) \\(2, 1 + 2i, 1 - 2i\\)
\\(\bigcirc\\) \\(2, -1 + 2i, -1 - 2i\\)

Explanation:

Find a rational root

Test integer factors of the constant term \(-10\).
Let \(P(x) = x^3 - 4x^2 + 9x - 10\).
Evaluate \(P(2)\):

$$ P(2) = 2^3 - 4(2)^2 + 9(2) - 10 = 8 - 16 + 18 - 10 = 0 $$

Thus, \(x = 2\) is a root.

Perform polynomial division

Divide \(P(x)\) by \((x - 2)\) using synthetic division:

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The quotient is \(x^2 - 2x + 5\).

Solve the quadratic equation

Set the quotient to zero:

$$ x^2 - 2x + 5 = 0 $$

Apply the quadratic formula:

$$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} $$

Simplify the expression:

$$ x = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i $$

Combine all solutions

The complete set of solutions is:

$$ x = 2, \quad x = 1 + 2i, \quad x = 1 - 2i $$

Answer:

• (A) \(-10, 1 + 2i, 1 - 2i\)
• (B) \(-2, -1 + 2i, -1 - 2i\)
• (C) \(2, 1 + 2i, 1 - 2i\) (Correct answer)
• (D) \(2, -1 + 2i, -1 - 2i\)