Polynomial Roots Practice Test

Find polynomial zeros, analyze multiplicity, use root relationships, and connect factors with graph behavior.

Polynomial Roots Practice Test

20 polynomial-root questions with detailed solutions and distractor analysis.

Instant feedback · Worked explanations

Polynomial root constellation lab

A root is not just a number. It links factors, graphs, and structure.

Polynomial roots connect algebraic factorization to graph behavior. A zero identifies a linear factor, a repeated zero carries multiplicity, and the same information can be read through the Factor Theorem, Vieta relationships, rational-root candidates, or a constructed factored form. The goal is not only to find roots, but to interpret what each root tells you.

Anchor 01A root

r

corresponds to factor

x - r

Anchor 02Multiplicity controls whether the graph touches or crosses.

Anchor 03Vieta links coefficients to root sums and products.

Anchor 04Rational Root Theorem generates candidates, not guaranteed roots.

1. Move between root, factor, equation, and graph

The fastest root problems are often solved by translating the same information between equivalent forms.

x = r \Leftrightarrow (x - r) = 0

Root

r

→

Linear factor

x - r

→

Zero equation

x - r = 0

→

horizontal-axis intercept

(r, 0)

Sign reversal

Factor sign is opposite root sign

Factor $x + 5$ gives root $x = - 5$ , not positive $5$ .

Zero-product property

Set each factor equal to zero

If a factored polynomial equals zero, each factor can independently create a root.

Zero itself

Do not lose a factored-out variable

If $x$ is a factor, then $x = 0$ is a root.

2. Multiplicity changes how the graph meets the horizontal axis

Repeated roots are not merely duplicates. Their multiplicity affects local graph behavior at the intercept.

Even multiplicity: touch and turn

A factor such as ${(x - 2)}^{2}$ gives root $x = 2$ with even multiplicity. The graph usually touches the axis and turns around.

Odd multiplicity: cross through

A factor such as ${(x + 1)}^{3}$ gives root $x = - 1$ with odd multiplicity. The graph usually crosses the axis.

3. Factor and Remainder Theorems test candidate roots quickly

A number is a root exactly when substituting it into the polynomial produces zero. The same evaluation is also the remainder when dividing by the corresponding linear factor.

f (r) = 0

Root testIf

f (r) = 0

, then

r

is a root.

Factor testIf

f (r) = 0

, then

x - r

is a factor.

Remainder interpretationIf

f (r) \neq 0

, that nonzero value is the remainder on division by

x - r

4. Vieta reads root relationships directly from coefficients

For a quadratic, you do not always need to solve for each root individually. The coefficient pattern already tells you the sum and product of the two roots.

a x^{2} + b x + c

Sum of roots

r_{1} + r_{2} = - \frac{b}{a}

The middle coefficient controls the root sum, including its sign.

Product of roots

r_{1} \cdot r_{2} = \frac{c}{a}

The constant term relative to the leading coefficient controls the root product.

Illustrative example

Read without solving

For $2 x^{2} - 7 x + 3$ , the root sum is $\frac{7}{2}$ and the root product is $\frac{3}{2}$ .

Use case

Check candidate roots

If two proposed roots have the wrong sum or product, they cannot both be the roots of the quadratic.

5. Rational Root Theorem narrows the search

For integer coefficients, possible rational roots are built from factors of the constant term divided by factors of the leading coefficient. The theorem creates a candidate list; each candidate still has to be tested.

Candidate generator

possible root = \pm \frac{factor of constant}{factor of leading coefficient}

Illustrative example for $2 x^{3} - 3 x^{2} - 8 x + 12$ :

\pm 1

\pm 2

\pm 3

\pm 4

\pm 6

\pm 12

\pm \frac{1}{2}

\pm \frac{3}{2}

Numerator sourceUse factors of the constant term.

Denominator sourceUse factors of the leading coefficient.

Include both signsPositive and negative candidates must both be considered unless other information rules them out.

Test candidatesUse substitution or synthetic division; being on the list does not make a candidate a root.

6. Complex roots complete the solution set

A polynomial can have roots that are not real. For polynomials with real coefficients, nonreal complex roots occur in conjugate pairs.

Conjugate pair

a + b i \leftrightarrow a - b i

If one nonreal root appears, its conjugate must also appear for real coefficients.

Quadratic example

$x^{2} + 4 = 0$ gives $x = \pm 2 i$ .

Graph meaning

Nonreal roots do not create horizontal-axis intercepts on the real coordinate plane, even though they remain algebraic roots.

7. Build a polynomial from its roots

To construct a polynomial, convert each root into a linear factor, repeat factors according to multiplicity, multiply, and optionally scale by a nonzero leading constant.

Root-to-factor construction

Roots

2, - 3

→

Factors

(x - 2) (x + 3)

→

Polynomial

x^{2} + x - 6

Repeated roots become repeated factors

If root $x = 4$ has multiplicity $3$ , then the polynomial contains factor ${(x - 4)}^{3}$ .

f (x) = a {(x - 4)}^{3} (x + 1)

Any nonzero choice of $a$ preserves the same roots and multiplicities.

8. Distinct roots and repeated roots answer different questions

A repeated root can count several times toward degree while still representing only one distinct solution value.

Multiplicity count

Count repetitions

${(x - 1)}^{3} (x + 2)$ has four roots counted with multiplicity.

Distinct count

Count unique values

The same polynomial has only two distinct roots: $1$ and $- 2$ .

Degree link

Total multiplicity matches degree

Over the complex numbers, a degree- $4$ polynomial has four roots counted with multiplicity.

9. Error analysis: common root misconceptions

Most distractors come from reversing factor signs incorrectly, dropping a zero root, or confusing multiplicity with the number of distinct solutions.

Factor sign copied directly

Factor $x + 6$ gives root $- 6$ , not positive $6$ .

Zero root omitted

If $x$ is factored out, $x = 0$ must be included as a root.

Repeated root counted as several distinct roots

Multiplicity and distinct-solution count are different quantities.

Rational candidate assumed to be a root

Rational Root Theorem generates possibilities; substitution or division must confirm them.

Even multiplicity shown crossing

Even multiplicity typically touches and turns rather than crossing the horizontal axis.

Complex root discarded

Nonreal solutions are still polynomial roots even though they do not appear as real horizontal-axis intercepts.

10. Worked mini-set: one root idea at a time

These examples are illustrative teaching examples, not questions copied from the test.

Example A

Factor to roots

$(x - 5) (x + 2) = 0$ gives roots $5$ and $- 2$ .

Example B

Repeated root

${(x - 3)}^{2}$ gives root $3$ with multiplicity $2$ .

Example C

Factor Theorem

If $f (4) = 0$ , then $x - 4$ is a factor.

Example D

Vieta

For $x^{2} - 5 x + 6$ , the root sum is $5$ and the product is $6$ .

Example E

Complex pair

If $1 + 2 i$ is a root of a real-coefficient polynomial, then $1 - 2 i$ is also a root.

Example F

Construct polynomial

Roots $1$ and $- 4$ give factor product $(x - 1) (x + 4)$ .

Final polynomial-roots checklist

Before accepting a root set, check the factor signs, multiplicities, theorem conditions, and whether the question asks for distinct roots or roots counted with repetition.

Did I convert factor signs correctly?Factor

x - r

corresponds to root

r

Did I include zero if $x$ is a factor?Factoring out the variable creates root

0

Did I preserve multiplicity?Repeated factors create repeated roots and affect graph behavior.

Did I test rational candidates?Candidate status alone does not prove a root.

Do Vieta relationships match?For quadratics, root sum and product provide a fast consistency check.

Are complex conjugates accounted for?Real-coefficient polynomials pair nonreal roots with their conjugates.