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Complex and Irrational Roots: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two evaluations

A polynomial $p$ has real coefficients and $p(5+i)=3-2i$ . Find $p(5-i)$ .
1. Hint 1
  Conjugation passes through a polynomial whose coefficients are real.
2. Hint 2
  Conjugate the known output along with the input.
Answer

$3+2i$ .

Full solution

Real coefficients are unchanged by conjugation, so
$p(\overline z)=\overline{p(z)}$

The input $5-i$ is the conjugate of $5+i$ .
Its output is therefore the conjugate of $3-2i$ , namely $3+2i$ .

Answer

$3+2i$ .

Key idea

A polynomial with real coefficients takes conjugate inputs to conjugate outputs.
Problem 1: Mark complete
Problem 2 Testing a substitution rule

For $q(x)=(1+i)x+2$ , compare $q(-i)$ with $\overline{q(i)}$ .
1. Hint 1
  Evaluate the polynomial before conjugating its output.
2. Hint 2
  The coefficient involving $i$ changes under conjugation.
Answer

$q(-i)=3-i$ ; $\overline{q(i)}=1-i$ ; they are unequal.

Full solution

Direct substitution gives
$q(i)=(1+i)i+2$
This simplifies to $1+i$ .
Similarly,
$q(-i)=(1+i)(-i)+2$
This simplifies to $3-i$ .

The conjugate of $1+i$ is $1-i$ , so $\overline{q(i)}=1-i$ , while $q(-i)=3-i$ , and these are unequal.
The coefficient $1+i$ is not real, so the real-coefficient substitution rule does not apply here.
In particular, the root of $q$ , which is $-1+i$ , has no conjugate partner:
$q(-1-i)=2-2i\neq0.$

Answer

$q(-i)=3-i$ ; $\overline{q(i)}=1-i$ ; they are unequal.

Key idea

Conjugating an input alone does not preserve the output rule when coefficients are nonreal.
Problem 2: Mark complete
Problem 3 A rational quadratic

A monic quadratic with rational coefficients has a root $-3+\sqrt2$ . Find its constant coefficient.
1. Hint 1
  Rational coefficients determine the other root.
2. Hint 2
  The constant coefficient of a monic quadratic is the product of its roots.
Answer

$7$ .

Full solution

The surd conjugate $-3-\sqrt2$ must be the other root.
Their product is
$(-3+\sqrt2)(-3-\sqrt2)=7$

The middle terms cancel and the constant is $9-2=7$ .
The corresponding quadratic is $x^2+6x+7$ , which checks the required rational coefficients.

Answer

$7$ .

Key idea

A surd conjugate pair has a rational product.
Problem 3: Mark complete
Problem 4 A quartic value

A monic polynomial $p$ of degree $4$ has real coefficients, and $2+4i$ is a root of multiplicity $2$ . Find $p(1)$ and give $p$ as a product of real factors.
1. Hint 1
  The conjugate root has the same multiplicity.
2. Hint 2
  The two copies of each conjugate use every degree; pair their linear factors before substituting.
Answer

$p(x)=(x^2-4x+20)^2$ ; $p(1)=289$ .

Full solution

The other root is $2-4i$ , also twice.
Together they use the four roots counted with multiplicity.
Let $Q$ be the product of one pair of linear factors.
The difference of squares gives
$Q=(x-2)^2-(4i)^2$ $Q=(x-2)^2+16$

So
$p(x)=(x^2-4x+20)^2.$
At $x=1$ , the quadratic factor has value
$Q(1)=17$
giving
$p(1)=17^2$ $p(1)=289$
The monic factorization has degree $4$ and exactly the required multiplicities.

Answer

$p(x)=(x^2-4x+20)^2$ ; $p(1)=289$ .

Key idea

A repeated nonreal root of a real polynomial forces equally many copies of its conjugate.
Problem 4: Mark complete
Problem 5 A cubic with a recorded value

A cubic $p$ has rational coefficients, leading coefficient $3$ , root $2+\sqrt{19}$ , and $p(0)=45$ . Find its remaining roots and give a factorization over the rationals.
1. Hint 1
  First account for the root forced by rational coefficients.
2. Hint 2
  After pairing those roots, use the constant value to identify the remaining linear factor.
Answer

Remaining roots: $2-\sqrt{19}$ and $1$ ; $p(x)=3(x^2-4x-15)(x-1)$ .

Full solution

Rational coefficients force $2-\sqrt{19}$ .
Their monic quadratic is
$(x-2)^2-19=x^2-4x-15$

The remaining factor is linear.
Write
$p(x)=3(x^2-4x-15)(x-r).$
Substitution at $0$ gives
$45r=45$ $r=1$

The factorization has the correct leading coefficient, and $3(-15)(-1)=45$ checks the recorded value.
Its remaining roots are $2-\sqrt{19}$ and $1$ .

Answer

Remaining roots: $2-\sqrt{19}$ and $1$ ; $p(x)=3(x^2-4x-15)(x-1)$ .

Key idea

A forced root pair can leave one linear factor that a polynomial value determines.
Problem 5: Mark complete
Problem 6 Odd degree forces a real root

A real polynomial has odd degree $11$ and splits over the reals into linear factors and quadratic factors with negative discriminant, counting repeated factors separately, with two more linear factors than quadratic factors. First explain in general why any real polynomial of odd degree must have at least one real root. Then apply that reasoning here: how many factors of each kind are present, and how many real roots are there with multiplicity?
1. Hint 1
  Every real polynomial splits into linear factors and quadratic factors with negative discriminant, and a quadratic factor always contributes an even amount to the degree.
2. Hint 2
  Write both the degree equation and the given difference between factor counts to find the specific factor counts.
Answer

Odd degree forces an odd, hence nonzero, count of linear factors, so a real root always exists; here, $5$ linear factors, $3$ quadratic factors, and $5$ real roots with multiplicity.

Full solution

Every real polynomial factors into real linear factors and real quadratic factors with negative discriminant.
If $k$ counts the linear factors and $m$ the quadratics, the degree is
$n=k+2m$
Since $2m$ is always even, $k$ has the same parity as $n$ .
An odd degree $n$ forces $k$ to be odd, and an odd number is never $0$ , so at least one linear factor exists, and it supplies a real root.

Apply this with $n=11$ .
Let $k$ count linear factors and $m$ count quadratics.
Then
$k+2m=11$ $k=m+2$
Substitution gives $3m+2=11$ , so $m=3$ and $k=5$ .

Each linear factor supplies one real root, so this polynomial has $5$ real roots counted with multiplicity, and $5$ is indeed odd, matching the general argument.
The count $5+2(3)=11$ also checks the total degree.

Answer

Odd degree forces an odd, hence nonzero, count of linear factors, so a real root always exists; here, $5$ linear factors, $3$ quadratic factors, and $5$ real roots with multiplicity.

Key idea

Odd degree forces an odd, hence nonzero, count of linear factors, so a real root always exists.
Problem 6: Mark complete
Problem 7 Completing the root picture

The figure shows two nonreal roots of a real monic quartic. All four roots are distinct. Give the coordinates of the other two roots in the complex plane and write the polynomial as a product of real monic quadratics.

Two nonreal roots of a real monic quartic.

Text description of this figure

A grid with the Real axis from -4 to 3 and the Imaginary axis from -4 to 4, gridlines and number labels at every integer. A solid point labeled A is plotted two units left of the origin and three units above it. An open point labeled B, drawn with a dashed outline, is plotted three units left of the origin and three units below it. Neither point's conjugate is shown.
1. Hint 1
  Reflect each shown root across the real axis.
2. Hint 2
  Convert each reflected pair into a real quadratic, then multiply the quadratics.
Answer

Other roots: $(-2,-3)$ and $(-3,3)$ ; $p(x)=(x^2+4x+13)(x^2+6x+18)$ .

Full solution

The plotted inputs are $-2+3i$ and $-3-3i$ .
Their conjugates are $-2-3i$ and $-3+3i$ , so the missing coordinates are $(-2,-3)$ and $(-3,3)$ .

The first pair gives $(x+2)^2+9$ , and the second gives $(x+3)^2+9$ .
Thus
$\begin{aligned} p(x) &= (x^2+4x+13) \\ &\quad \times (x^2+6x+18) \end{aligned}$
This product is monic of degree $4$ , with four distinct roots and real coefficients.

Answer

Other roots: $(-2,-3)$ and $(-3,3)$ ; $p(x)=(x^2+4x+13)(x^2+6x+18)$ .

Key idea

Reflection in the real axis represents the root pairing forced by real coefficients.
Problem 7: Mark complete
Problem 8 Equal outputs

A polynomial $p$ has real coefficients. A student claims: "If $p(4+3i)=p(4-3i)$ , their common value must be real." Is the claim true? Explain.
1. Hint 1
  The outputs at conjugate inputs are themselves conjugates.
2. Hint 2
  Ask which complex numbers equal their own conjugates.
Answer

True; the common value is real.

Full solution

Let $w=p(4+3i)$ .
Since the coefficients are real,
$p(4-3i)=\overline w$

The claimed equality therefore says
$w=\overline w.$
If $w=a+bi$ with real $a,b$ , matching imaginary parts gives $b=-b$ , so $b=0$ .
Thus the common output is real.

Answer

True; the common value is real.

Key idea

Equal conjugate outputs must have zero imaginary part.
Problem 8: Mark complete
Problem 9 A real coefficient claim

Let $p(x)=(x-\sqrt3)(x^2+2)$ . A student says that its real coefficients and root $\sqrt3$ force $-\sqrt3$ to be a root. Decide whether this conclusion is valid and verify your decision by evaluating $p(-\sqrt3)$ .
1. Hint 1
  The two root-pairing theorems have different coefficient hypotheses.
2. Hint 2
  The square of either proposed input is $3$ .
Answer

Invalid; $p(-\sqrt3)=-10\sqrt3\ne0$ .

Full solution

The product has real coefficients, but its expanded coefficients include irrational values.
A surd partner is guaranteed by rational coefficients, a stronger condition.
Direct substitution gives
$p(-\sqrt3)=(-2\sqrt3)(3+2)$ $p(-\sqrt3)=-10\sqrt3$

The value is nonzero.
The complex conjugate theorem is untroubled: $\sqrt3$ is real and is its own complex conjugate.

Answer

Invalid; $p(-\sqrt3)=-10\sqrt3\ne0$ .

Key idea

Real coefficients force complex conjugates, while a surd sign change requires rational coefficients.
Problem 9: Mark complete
Problem 10 Scaling a root pair

Let $c$ be a nonzero complex number and $p(x)=c(x^2+6x+10)$ . A student says its paired nonreal roots force $c$ to be real. Is that true? Determine exactly which choices of $c$ make every coefficient real.
1. Hint 1
  Multiplication by a nonzero constant changes coefficients but not roots.
2. Hint 2
  The leading coefficient is already one of the coefficients that must be real.
Answer

False; the roots are $-3+i,-3-i$ for every $c\ne0$ . All coefficients are real exactly when $c$ is real.

Full solution

The quadratic is $(x+3)^2+1$ , so its roots are $-3+i$ and $-3-i$ .
Multiplying by $c\ne0$ leaves those roots unchanged.

Expanding the coefficients gives
$p(x)=cx^2+6cx+10c$
If all coefficients are real, the leading one, $c$ , is real.
Conversely, real $c$ makes $c,6c,10c$ all real.
Thus paired roots alone do not establish the hypothesis.

Answer

False; the roots are $-3+i,-3-i$ for every $c\ne0$ . All coefficients are real exactly when $c$ is real.

Key idea

Conjugate root pairing by itself does not determine whether a polynomial's leading constant is real.
Problem 10: Mark complete

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