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The Fundamental Theorem of Algebra: 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 Reading roots from a factorization

The polynomial $p(x)=4(x+3)(x-2)^2(x+1)^3$ is already a complete factorization over the complex numbers. State every root of $p$ and its multiplicity, then give the number of distinct roots and the number of roots counted with multiplicity.
1. Hint 1
  Each linear factor names one root, and a repeated factor raises that root's multiplicity.
2. Hint 2
  Add the multiplicities together and compare the total to the degree.
Answer

$-3$ : multiplicity $1$ ; $2$ : multiplicity $2$ ; $-1$ : multiplicity $3$ ; $3$ distinct roots; $6$ roots counted with multiplicity.

Full solution

Each linear factor contributes one root, repeated as many times as its exponent shows.
$p(x)=4(x+3)(x-2)^2(x+1)^3$
So $-3$ has multiplicity $1$ , $2$ has multiplicity $2$ , and $-1$ has multiplicity $3$ .

Those are three different numbers, so $p$ has $3$ distinct roots.

Counted with multiplicity, the roots number $1+2+3=6$ , which matches the degree of $p$ .

Answer

$-3$ : multiplicity $1$ ; $2$ : multiplicity $2$ ; $-1$ : multiplicity $3$ ; $3$ distinct roots; $6$ roots counted with multiplicity.

Key idea

A complete factorization reads off every root and its multiplicity directly, and the multiplicities always sum to the degree.
Problem 1: Mark complete
Problem 2 A division record

Dividing $p(x)$ by $x+4$ gives quotient $x^2+5x+4$ and remainder $0$ . Determine the multiplicity of $-4$ in $p$ .
1. Hint 1
  A zero remainder starts the count but does not finish it.
2. Hint 2
  Divide the quotient by the same linear expression, then test what remains.
Answer

$-4$ has multiplicity $2$ .

Full solution

The first division gives one copy of $x+4$ .
Dividing its quotient gives
$x^2+5x+4=(x+4)(x+1)$

A further division of $x+1$ by $x+4$ has remainder $-3$ .
Thus
$p(x)=(x+4)^2(x+1)$
The remaining factor is nonzero at $-4$ , so the multiplicity is exactly $2$ .

Answer

$-4$ has multiplicity $2$ .

Key idea

The first nonzero remainder ends the count of repeated linear factors.
Problem 2: Mark complete
Problem 3 A quadratic with a complex coefficient

Write $3x^2+9ix-6$ as a constant times linear factors over the complex numbers.
1. Hint 1
  Separate the nonzero leading constant first.
2. Hint 2
  Find two imaginary numbers whose sum is $3i$ and whose product is $-2$ .
Answer

$3(x+i)(x+2i)$ .

Full solution

Pull out $3$ .
The product of $i$ and $2i$ is $-2$ , so
$x^2+3ix-2=(x+i)(x+2i)$

Therefore the requested factorization is $3(x+i)(x+2i)$ .
Expanding gives middle term $9ix$ and constant term $6i^2=-6$ , as required.

Answer

$3(x+i)(x+2i)$ .

Key idea

Complex coefficients still allow a complete factorization into linear factors.
Problem 3: Mark complete
Problem 4 Zeros at two inputs

For $p(x)=x^5+2x^4+x^3$ , establish the multiplicities of $0$ and $-1$ by repeated division, then give the distinct-root count and the count with multiplicity.
1. Hint 1
  Start with the root whose factor can be removed without synthetic division.
2. Hint 2
  After removing the powers of $x$ , divide the remaining quadratic by $x+1$ until a nonzero remainder appears.
Answer

$0$ : multiplicity $3$ ; $-1$ : multiplicity $2$ ; $2$ distinct roots and $5$ with multiplicity.

Full solution

Dividing by $x$ successively gives $x^4+2x^3+x^2$ , then $x^3+2x^2+x$ , then $x^2+2x+1$ .
Each division has remainder $0$ .
The last quotient has value $1$ at $0$ , so the multiplicity of $0$ is exactly $3$ .

The remaining quadratic divides by $x+1$ to give $x+1$ , and again to give $1$ .
A further division has nonzero remainder $1$ .
Thus
$p(x)=x^3(x+1)^2$

At $-1$ , the other factor is $(-1)^3=-1$ , which is nonzero.
The root multiplicities are therefore $3$ and $2$ .
Their sum $5$ matches the degree, and their two distinct values exhaust the root list.

Answer

$0$ : multiplicity $3$ ; $-1$ : multiplicity $2$ ; $2$ distinct roots and $5$ with multiplicity.

Key idea

An exhausted factorization checks both multiplicities and completeness of the root list.
Problem 4: Mark complete
Problem 5 A pair of unknown counts

A degree $7$ polynomial with complex coefficients has exactly two distinct complex roots, $a$ and $b$ . The multiplicity of $a$ is one greater than the multiplicity of $b$ . Find both multiplicities and write every possible complete factorization.
1. Hint 1
  Every factor must belong to one of the two stated roots.
2. Hint 2
  Let the smaller multiplicity be $m$ and make the multiplicities add to the degree.
Answer

$a$ : multiplicity $4$ ; $b$ : multiplicity $3$ ; $c(x-a)^4(x-b)^3$ , where $c$ may be any nonzero complex number.

Full solution

Let $m$ be the multiplicity of $b$ .
The full root count gives
$m+(m+1)=7$ $2m=6$ $m=3$

So $a$ has multiplicity $4$ .
With nonzero leading coefficient $c$ , the factorization is
$p(x)=c(x-a)^4(x-b)^3$
The distinctness of $a$ and $b$ ensures that no extra copy is hidden in the other factor.

Answer

$a$ : multiplicity $4$ ; $b$ : multiplicity $3$ ; $c(x-a)^4(x-b)^3$ , where $c$ may be any nonzero complex number.

Key idea

The degree is the sum of the multiplicities of all distinct complex roots.
Problem 5: Mark complete
Problem 6 Changing a constant

Let $p(x)=x^4+8x^3+26x^2+40x+25$ . Compare the distinct complex roots of $p(x)$ with those of $p(x)-25$ , giving every root and its multiplicity.
1. Hint 1
  Write $p$ as the square of a quadratic first.
2. Hint 2
  Once $p(x)=u(x)^2$ for a quadratic $u$ , subtracting $25$ is subtracting $5^2$ , so it factors as a difference of squares in $u$ .
Answer

$p$ : $-2+i,-2-i$ , each multiplicity $2$ ; $p-25$ : $0,-4,-2+i\sqrt6,-2-i\sqrt6$ , all simple.

Full solution

Complete the square inside: $x^2+4x+5=(x+2)^2+1$ , so
$p(x)=\left((x+2)^2+1\right)^2$
Setting $(x+2)^2+1=0$ gives $(x+2)^2=-1$ , so $x+2=\pm i$ and $x=-2\pm i$ .
Its two distinct roots are $-2+i$ and $-2-i$ , both repeated twice.

Let $u=(x+2)^2+1$ , so $p(x)=u^2$ .
Subtracting $25=5^2$ factors as a difference of squares:
$p(x)-25=(u-5)(u+5)$

Since $u-5=(x+2)^2-4$ is a difference of squares,
$u-5=x(x+4)$
And $u+5=(x+2)^2+6$ , which is the quadratic $x^2+4x+10$ .

Solve $x^2+4x+10=0$ by completing the square: $(x+2)^2+6=0$ , so $(x+2)^2=-6$ and
$x=-2\pm i\sqrt6.$
So $p(x)-25$ has four linear factors, whose roots are $0$ , $-4$ , $-2+i\sqrt6$ , and $-2-i\sqrt6$ , all four simple.
Both lists have four entries when repetition is included, matching the unchanged degree.

Answer

$p$ : $-2+i,-2-i$ , each multiplicity $2$ ; $p-25$ : $0,-4,-2+i\sqrt6,-2-i\sqrt6$ , all simple.

Key idea

A coefficient change can change the number of distinct roots while leaving the degree unchanged.
Problem 6: Mark complete
Problem 7 The last quotient

Two exact divisions of a polynomial give $p(x)=(x-3)q(x)$ and $q(x)=(x+2)(ix+2)$ . Write $p$ as a constant times monic linear factors, and explain why these give its entire root list.
1. Hint 1
  The last linear factor must be normalized without changing its root.
2. Hint 2
  Solve $ix+2=0$ and keep its leading coefficient outside the product.
Answer

$p(x)=i(x-3)(x+2)(x-2i)$ ; roots $3,-2,2i$ , all simple; no other complex number satisfies $p$ .

Full solution

Since $i(x-2i)=ix+2$ , substituting the second division record into the first gives
$p(x)=i(x-3)(x+2)(x-2i)$

The roots $3,-2,2i$ are distinct.
At any other complex number all three factors are nonzero, so their product is nonzero.
There are three linear factors and degree $3$ , which also confirms that the list is complete.

Answer

$p(x)=i(x-3)(x+2)(x-2i)$ ; roots $3,-2,2i$ , all simple; no other complex number satisfies $p$ .

Key idea

A nonzero product of linear factors has no roots beyond the zeros of those factors.
Problem 7: Mark complete
Problem 8 A polynomial built from another

A polynomial $q$ has degree $3$ . How many roots does $q(x)^2-q(x)$ have over the complex numbers, counted with multiplicity?
1. Hint 1
  Find the degree before counting roots.
2. Hint 2
  The highest power in the square has greater degree than any term being subtracted.
Answer

$6$ roots, counted with multiplicity.

Full solution

If the leading term of $q$ is $ax^3$ with $a\ne0$ , the leading term of its square is $a^2x^6$ .
Subtracting a degree $3$ polynomial does not remove it.
$\deg(q^2-q)=6$

The root count over the complex numbers, with multiplicity included, equals that degree.

Answer

$6$ roots, counted with multiplicity.

Key idea

A lower-degree subtraction does not remove a polynomial product's leading term.
Problem 8: Mark complete
Problem 9 Finding the non-algebraic step

A student sketches a proof of the Fundamental Theorem of Algebra in three steps. Step 1: on a circle of very large radius, the polynomial's leading term dominates its other terms in size, so the image of that circle loops around the origin. Step 2: as the radius shrinks to $0$ , the image shrinks to the single point $p(0)$ , which sits away from the origin whenever $p(0)\ne0$ . Step 3: a loop cannot stop enclosing a point without sweeping across it, so some circle in between has an image passing through the origin, which is exactly where a root of $p$ sits. One of these three steps depends on a property beyond ordinary algebra: that a continuously changing quantity cannot skip past a value without passing through it. Which step is that? Then state the Fundamental Theorem of Algebra.
1. Hint 1
  Two of the three steps are settled by comparing sizes of numbers or by a direct substitution; the remaining step describes something moving continuously and concludes it must pass through a particular value.
2. Hint 2
  A statement built from inequalities between absolute values can be checked by arithmetic directly; a statement that a path cannot skip over a point needs the idea that change happens continuously.
Answer

Step 3 depends on continuity: a continuously changing quantity cannot skip past a value without passing through it. Every nonconstant polynomial with complex coefficients has at least one complex root.

Full solution

Steps 1 and 2 rest on comparing the sizes of specific terms and on evaluating $p$ directly at a particular radius; each conclusion follows from an ordinary absolute-value inequality or a substitution.

Step 3 is different.
Saying a loop "cannot stop enclosing a point without sweeping across it" is a claim about continuous motion: a quantity that changes without a jump cannot skip past a value without passing through it exactly.

That continuity property holds for the real and complex numbers but fails for the rational numbers, where a continuously moving quantity can slip past a target value through a gap with no rational number sitting in it.
Arithmetic rules alone, without continuity, cannot supply step 3.

The correct statement is the Fundamental Theorem of Algebra: every polynomial of degree at least $1$ with complex coefficients has at least one complex root.

Answer

Step 3 depends on continuity: a continuously changing quantity cannot skip past a value without passing through it. Every nonconstant polynomial with complex coefficients has at least one complex root.

Key idea

The one step a proof of this theorem cannot get from algebra alone is the continuity step, which is exactly why no purely algebraic proof exists.
Problem 9: Mark complete
Problem 10 Two polynomial stages

Let $p$ and $q$ be nonconstant polynomials with complex coefficients. Is it guaranteed that the equation $q(p(z))=0$ has a complex solution, without enlarging the number system again? Explain.
1. Hint 1
  Apply the root-existence guarantee one stage at a time.
2. Hint 2
  After finding a root $w$ of $q$ , consider the polynomial $p(z)-w$ .
Answer

Yes, at least one complex solution is guaranteed.

Full solution

The Fundamental Theorem of Algebra gives a complex number $w$ for which
$q(w)=0$

The polynomial $p(z)-w$ is still nonconstant and has complex coefficients.
The theorem therefore gives a complex $z$ with
$p(z)=w$
Substituting into $q$ gives $q(p(z))=0$ .
Both stages remain within the complex numbers.

Answer

Yes, at least one complex solution is guaranteed.

Key idea

A polynomial equation with complex coefficients does not require roots outside the complex numbers.
Problem 10: Mark complete

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