Back to Exercise: Know the Fundamental Theorem of Algebra

Exercises: Know the Fundamental Theorem of Algebra

For quadratic problems, show the discriminant. For root-counting, include multiplicity.

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

Printable layout

Warm-Up: Review What You Know

The Fundamental Theorem of Algebra states that every non-constant polynomial of degree $n$ has exactly how many roots in the complex numbers?

At least $n$ roots (could be more)

Exactly $n$ distinct roots, all different

Exactly $n$ roots, counting multiplicity

At most $n$ roots

The polynomial $p(x) = x^2 + 1$ has no real roots. Does the Fundamental Theorem of Algebra apply to $p(x)$ ?

No — the FTA only applies to polynomials with complex coefficients, not real-coefficient polynomials like $p(x)$ .

Yes — the FTA applies, but only if $p(x)$ has at least one real root.

Yes — the FTA guarantees $p(x)$ has exactly $2$ roots in the complex numbers. They are $i$ and $-i$ .

No — $p(x)$ has degree $2$ but no real roots, so the FTA fails here.

The polynomial $p(x) = (x - 2)^2(x + 5)$ has degree $3$ .

According to the FTA, how many roots does $p(x)$ have in the complex numbers, counting multiplicity?

Fluency Practice

Classify the discriminant case for $x^2 - 5x + 6 = 0$ and identify the number of roots.

Discriminant $< 0$ : two complex roots

Discriminant $= 0$ : one real root (multiplicity 2)

Discriminant $> 0$ : two distinct real roots

Discriminant $< 0$ : no roots

Classify the discriminant case for $x^2 - 4x + 4 = 0$ and identify the roots.

Discriminant $= 0$ : one root $x = 2$ (multiplicity 2); total root count is $2$

Discriminant $= 0$ : one root $x = 2$ (multiplicity 1); total root count is $1$

Discriminant $> 0$ : two distinct roots

Discriminant $< 0$ : two complex roots

Classify the discriminant case for $x^2 - 4x + 5 = 0$ and identify the roots.

Discriminant $= $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ (enter an integer). The roots are $x = 2 \pm$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $i$ .

The polynomial $p(x) = (x - 1)(x + 3)(x^2 + 4)$ has degree $4$ .

How many total roots does $p(x)$ have over the complex numbers, counting multiplicity?

The polynomial $p(x) = x^4 + 1$ has degree $4$ and real coefficients. How many real roots does it have?

$4$ real roots (since degree $= 4$ )

$2$ real roots

$0$ real roots (all $4$ roots are complex non-real)

$1$ real root

You're viewing 2 of 6 sections.

Create a free account to continue the full exercise set and save your progress.

Create free account