Exercises: Extend Polynomial Identities to Complex Numbers

Factor completely over the complex numbers. Verify by expanding when asked.

Grade 9·20 problems·~40 min·Common Core Math - HS Number and Quantity·standard·hsn-cn-c-8

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Warm-Up: Review What You Know

Factor $x^2 - 9$ over the real numbers.

$(x + 3i)(x - 3i)$

$(x + 3)(x - 3)$

$(x - 3)^2$

Prime (does not factor)

Compute $(x + 2i)(x - 2i)$ . Simplify using $i^2 = -1$ .

$x^2 - 4$

$x^2 + 4ix - 4$

$x^2 + 4$

$x^2 - 4i^2$

The quadratic $x^2 - 4x + 5 = 0$ has roots $2 + i$ and $2 - i$ (from CN.C.7).

By the root-to-factor connection: $x^2 - 4x + 5 = (x - r_1)(x - r_2)$ .

Verify: expand $(x - (2+i))(x - (2-i))$ and find the constant term. Enter the constant term.

Factor $x^2 + 9$ over the complex numbers.

$(x + 3)(x - 3)$

$(x + 3i)(x - 3i)$

$(x + 3i)^2$

Prime over complex numbers

The canonical example: factor $x^2 + 4$ over the complex numbers.

$(x + 2)(x - 2)$

$(x + 2i)^2$

$(x + 2i)(x - 2i)$

$(x + 4i)(x - i)$

Factor $x^2 + 5$ over the complex numbers.

Cannot be factored because $5$ is not a perfect square

$(x + \sqrt{5}\,i)(x - \sqrt{5}\,i)$

$(x + \sqrt{5})(x - \sqrt{5})$

$(x + 5i)(x - i)$

The quadratic $x^2 + 2x + 5 = 0$ has roots $-1 + 2i$ and $-1 - 2i$ .

Write the factored form: $x^2 + 2x + 5 = (x - $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $)(x - $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ )$.

Enter the two roots (the numbers being subtracted), separated.

Factor $4x^2 + 25$ over the complex numbers.

$(2x + 5)(2x - 5)$

$(2x + 5i)(2x - 5i)$

$(4x + 25i)(x - i)$

Cannot be factored

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