Core practice ← Back to lesson

Vieta's Formulas: 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 Quadratic specifications

Write the quadratic polynomial with leading coefficient $4$ whose two roots, counted with multiplicity, have sum $3$ and product $-2$ . Give expanded form.
1. Hint 1
  The sum and product determine the two remaining coefficients.
2. Hint 2
  Expand $4(x-r_1)(x-r_2)$ before inserting the given values.
Answer

$4x^2-12x-8$ .

Full solution

For two roots the expansion is
$\begin{aligned} p(x) &= 4x^2-4(r_1+r_2)x \\ &\quad +4r_1r_2 \end{aligned}$

The stated values give
$p(x)=4x^2-12x-8$
Its root sum is $-(-12)/4=3$ and product is $-8/4=-2$ , checking both conditions.

Answer

$4x^2-12x-8$ .

Key idea

The quadratic sum and product rules come from matching the expansion of two linear factors.
Problem 1: Mark complete
Problem 2 A fifth-degree coefficient

The five complex roots of $2x^5+3x^4-4x^3+6x^2-x+7$ are listed with multiplicity. Find the sum of all products of three different entries of this list.
1. Hint 1
  A product of three root entries belongs to the coefficient three powers below the leading term.
2. Hint 2
  Account for both the leading coefficient and the odd number of minus signs.
Answer

$-3$ .

Full solution

The requested quantity is $e_3$ .
It is attached to the $x^2$ coefficient, since $5-3=2$ .
The formula gives
$e_3=-\frac{6}{2}$ $e_3=-3$

The sign is negative because taking a root from three linear factors contributes three minus signs.

Answer

$-3$ .

Key idea

The coefficient position and the number of chosen roots determine both the symmetric sum and its sign.
Problem 2: Mark complete
Problem 3 A root of zero, expanded

A polynomial has leading coefficient $3$ and complete root list $0,2,-5$ . Write the polynomial in expanded standard form $ax^3+bx^2+cx+d$ .
1. Hint 1
  Write the factored form first: the leading coefficient times one linear factor $(x-r)$ per root, including the factor for the root $0$ .
2. Hint 2
  Expand the two factors that do not involve the root $0$ first, then multiply the result by the remaining linear factor.
Answer

$3x^3+9x^2-30x$ (that is, $a=3$ , $b=9$ , $c=-30$ , $d=0$ ).

Full solution

The factored form uses the leading coefficient and one linear factor per root, and the factors for the roots $0$ and $-5$ simplify to $x$ and $x+5$ :
$p(x)=3x(x-2)(x+5)$

Expanding the two factors that do not involve the root $0$ first, $(x-2)(x+5)=x^2+3x-10$ , and multiplying by the remaining factor $x$ gives
$x(x^2+3x-10)=x^3+3x^2-10x$
Multiplying by the leading coefficient gives
$p(x)=3x^3+9x^2-30x$

As a check, the symmetric sums from the root list agree with the expanded coefficients: $e_1=0+2+(-5)=-3$ matches $-b/a=-9/3$ , $e_2=(0)(2)+(0)(-5)+(2)(-5)=-10$ matches $c/a=-30/3$ , and $e_3=(0)(2)(-5)=0$ matches $-d/a=-0/3$ .

Answer

$3x^3+9x^2-30x$ (that is, $a=3$ , $b=9$ , $c=-30$ , $d=0$ ).

Key idea

A root of $0$ still contributes its own linear factor $x$ , and shows up as the polynomial's missing constant term.
Problem 3: Mark complete
Problem 4 Repeated imaginary roots

A polynomial has leading coefficient $2$ and complete root list $i,i,-i,-i$ . Find $e_1,e_2,e_3,e_4$ and the expanded polynomial.
1. Hint 1
  Repeated entries are separate choices in each symmetric sum.
2. Hint 2
  Pair $i$ with $-i$ to expand the polynomial, then compare all four coefficients.
Answer

$e_1=0$ , $e_2=2$ , $e_3=0$ , $e_4=1$ ; $p(x)=2x^4+4x^2+2$ .

Full solution

Two matched conjugate pairs give
$p(x)=2(x^2+1)^2$ $p(x)=2x^4+4x^2+2$

Reading all coefficients, including the missing ones, gives $e_1=0$ , $e_2=4/2=2$ , $e_3=0$ , and $e_4=2/2=1$ .
Directly, the six pair products comprise $i^2=-1$ , $(-i)^2=-1$ , and four copies of $i(-i)=1$ , whose sum is $2$ .
That checks the repeated-entry count.

Answer

$e_1=0$ , $e_2=2$ , $e_3=0$ , $e_4=1$ ; $p(x)=2x^4+4x^2+2$ .

Key idea

Repeated and nonreal roots both remain full entries when coefficients are matched.
Problem 4: Mark complete
Problem 5 Shifted root values

Let $r_1,r_2,r_3$ be all roots of $3x^3+6x^2-3x+8$ , listed with multiplicity. Without finding the roots, find $(r_1+1)^2+(r_2+1)^2+(r_3+1)^2$ .
1. Hint 1
  Expand the squares to express the target using a root sum and a sum of squares.
2. Hint 2
  The sum of squares is $e_1^2-2e_2$ .
Answer

$5$ .

Full solution

The coefficients give $e_1=-6/3=-2$ and
$e_2=-3/3=-1.$
If $S$ is the sum of the root squares,
$S=e_1^2-2e_2$ $S=6$

Expanding each shifted square makes the requested total $S+2e_1+3$ , so
$S+2e_1+3=5$
This expression treats every root identically, so the labeling of roots does not matter.

Answer

$5$ .

Key idea

Expanding a symmetric expression can reduce it to coefficient information.
Problem 5: Mark complete
Problem 6 Differences between roots

The roots of $x^3-x^2-2x+5$ are $r,s,t$ , including any repeats. Find $(r-s)^2+(r-t)^2+(s-t)^2$ without finding the roots.
1. Hint 1
  Expand the three squares and collect the same kinds of terms.
2. Hint 2
  Each root square appears twice and each pair product appears with coefficient $-2$ .
Answer

$14$ .

Full solution

The root sum is $e_1=1$ and pair sum is $e_2=-2$ .
The sum of squares is
$r^2+s^2+t^2=1^2-2(-2)$ $r^2+s^2+t^2=5$

If $D$ is the requested total, expansion gives
$D=2(r^2+s^2+t^2)-2e_2$ $D=2(5)-2(-2)$ $D=14$
The calculation includes the full complex root list and does not require the roots themselves.

Answer

$14$ .

Key idea

A symmetric combination of root differences can be computed by expanding it into symmetric sums.
Problem 6: Mark complete
Problem 7 A coefficient condition

For $p(x)=2x^4-3x^3+5x^2+cx+6$ , the sum of the reciprocals of all four roots equals twice the sum of the roots. Find $c$ .
1. Hint 1
  The nonzero constant term makes each reciprocal meaningful.
2. Hint 2
  For four roots, their reciprocal sum is $e_3/e_4$ .
Answer

$c=-18$ .

Full solution

No root is zero because $p(0)=6$ .
The root sum is $e_1=3/2$ , the triple-product sum is $e_3=-c/2$ , and the product is $e_4=3$ .
Hence the reciprocal sum is $-c/6$ .

The required equality is
$-\frac{c}{6}=2\left(\frac32\right)$ $c=-18$
Substitution gives reciprocal sum $3$ and twice the root sum $3$ , checking the condition.

Answer

$c=-18$ .

Key idea

A nonzero constant term permits reciprocal-root conditions to be read through symmetric sums.
Problem 7: Mark complete
Problem 8 Two equal entries

A cubic has complete root list $u,u,v$ . Find $e_2$ by listing every pair of entries in the list.
1. Hint 1
  Symmetric sums choose list positions, not distinct numerical values.
2. Hint 2
  There are three ways to choose two positions from a three-entry list.
Answer

$e_2=u^2+2uv$ .

Full solution

The first and second entries produce $u^2$ .
The first and third produce $uv$ , and the second and third produce another $uv$ .
Therefore
$e_2=u^2+2uv$

If $v=u$ , those three pair products all equal $u^2$ , giving $3u^2$ .
Equal values do not merge their positions in the root list.

Answer

$e_2=u^2+2uv$ .

Key idea

A repeated value still occupies each of its positions in a symmetric root sum.
Problem 8: Mark complete
Problem 9 Interpreting a negative total

A student reads the roots $r,s$ of $x^2+2x+5$ and says: "Their squares add to a negative number, so the calculation must be wrong." Decide whether the objection is valid without solving for the roots, and state $r^2+s^2$ .
1. Hint 1
  Squares of real numbers are nonnegative, but the full root list need not be real.
2. Hint 2
  Compute the sum of squares from the sum and product of the roots.
Answer

Invalid; $r^2+s^2=-6$ .

Full solution

The coefficients give $r+s=-2$ and $rs=5$ .
Therefore
$r^2+s^2=(r+s)^2-2rs$ $r^2+s^2=-6$

The negative total is not a contradiction.
The discriminant is $2^2-4(1)(5)=-16$ , so the roots are nonreal, and their squares are not constrained to be nonnegative real numbers.

Answer

Invalid; $r^2+s^2=-6$ .

Key idea

A symmetric sum of squares over complex roots need not be nonnegative.
Problem 9: Mark complete
Problem 10 A formula for all degrees

Let $p(x)=a_nx^n+\cdots+a_0$ with $a_n\ne0$ have complete root list $r_1,\ldots,r_n$ , and let $e_k$ be the sum of products of $k$ different entries of that list. For which $k$ with $1\le k\le n$ is $e_k=-a_{n-k}/a_n$ , and for which is it $+a_{n-k}/a_n$ instead? Explain how the expansion of $(x-r_1)\cdots(x-r_n)$ decides every case.
1. Hint 1
  Count the minus signs selected from the linear factors.
2. Hint 2
  Compare an odd number of selected root terms with an even number.
Answer

For $1\le k\le n$ : $e_k=-a_{n-k}/a_n$ when $k$ is odd, and $e_k=+a_{n-k}/a_n$ when $k$ is even.

Full solution

Taking $k$ roots from the factors $x-r_j$ and the $x$ from the rest contributes $(-1)^k$ to that term.
Dividing $p(x)$ by $a_n$ makes both sides monic, so matching the coefficient of $x^{n-k}$ gives, for every $k$ with $1\le k\le n$ ,
$e_k=(-1)^k\frac{a_{n-k}}{a_n}$

For odd $k$ , $(-1)^k=-1$ , so
$e_k=-a_{n-k}/a_n.$
For even $k$ , $(-1)^k=1$ , so $e_k=+a_{n-k}/a_n$ instead.
The sign depends only on the parity of the number of chosen factors, not on whether the expression is described as a sum or a product.

Answer

For $1\le k\le n$ : $e_k=-a_{n-k}/a_n$ when $k$ is odd, and $e_k=+a_{n-k}/a_n$ when $k$ is even.

Key idea

The number of selected root terms, rather than a memorized minus sign, controls every Vieta formula for $1\le k\le n$ .
Problem 10: Mark complete

← Back to lesson