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The Factor Theorem: 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 A missing coefficient

For which value of $k$ is $x+2$ a factor of $P(x)=3x^3+kx^2-7x-10$ ?
1. Hint 1
  A factor vanishes at a particular input.
2. Hint 2
  Set $P(-2)=0$ and solve the resulting linear equation for $k$ .
Answer

$k=5$ .

Full solution

The factor $x+2$ vanishes at $-2$ .
Substituting $x=-2$ into each term gives $3(-2)^3=-24$ , $k(-2)^2=4k$ , and $-7(-2)=14$ .
$\begin{aligned}P(-2)&=-24+4k+14-10\\&=4k-20\end{aligned}$
Setting $4k-20=0$ gives $k=5$ .

Answer

$k=5$ .

Key idea

When the unknown sits inside the polynomial itself, the factor condition becomes a linear equation to solve for it.
Problem 1: Mark complete
Problem 2 A candidate expression

Determine whether $3x-2$ is a factor of $P(x)=9x^2-4$ . Justify the decision with one evaluation.
1. Hint 1
  Test the input that makes the candidate expression zero.
2. Hint 2
  The relevant input is $\frac23$ , not $2$ or $3$ .
Answer

Yes; $P(\frac23)=0$ .

Full solution

The expression $3x-2$ vanishes at $x=\frac23$ .
$P\left(\frac23\right)=9\left(\frac49\right)-4$
This equals zero, so the Factor Theorem confirms that $3x-2$ is a factor.
Indeed $(3x-2)(3x+2)$ expands to the given polynomial.

Answer

Yes; $P(\frac23)=0$ .

Key idea

A nonmonic linear factor is tested at its own zero.
Problem 2: Mark complete
Problem 3 A recorded identity

Suppose $P(8)=-11$ , and dividing $P(x)$ by $x-8$ gives quotient $Q(x)=2x^2-5x+1$ . What remainder does this division leave?
1. Hint 1
  The remainder of dividing by $x-c$ is always a single number, and the master identity names exactly which one.
2. Hint 2
  The formula for $Q(x)$ does not change the value you already know for $P(8)$ .
Answer

Remainder $-11$ .

Full solution

The master identity for division by $x-8$ is
$P(x)=(x-8)\,Q(x)+P(8)$
so whatever $Q(x)$ turns out to be, the remainder is always the single number $P(8)$ .

Since $P(8)=-11$ , the remainder is $-11$ , and substituting the given quotient gives
$P(x)=(x-8)\left(2x^2-5x+1\right)-11$
The quotient's own coefficients play no role in fixing the remainder; only the value $P(8)$ does.

Answer

Remainder $-11$ .

Key idea

The master identity writes a linear-division remainder as an evaluation.
Problem 3: Mark complete
Problem 4 A supplied pair

The polynomial $P(x)=x^4-2x^3-2x^2+8$ has $2$ and $-2$ as candidate roots. Determine which candidate is a root, then use it to find every complex root of $P$ .
1. Hint 1
  Test the candidates, and work inside the quotient after a successful test.
2. Hint 2
  After removing a root, test that same root again before finishing the quadratic.
Answer

Only $2$ is a root among the candidates. All roots: $2$ twice, $-1+i$ and $-1-i$ .

Full solution

Evaluation gives $P(2)=0$ and $P(-2)=32$ .
Dividing by $x-2$ produces $x^3-2x-4$ , which also vanishes at $2$ .
A second division by $x-2$ gives $x^2+2x+2$ .

The remaining quadratic has roots $-1\pm i$ , since
$x^2+2x+2=(x+1)^2+1.$
Thus
$P(x)=(x-2)^2(x^2+2x+2)$
Expanding restores $x^4-2x^3-2x^2+8$ , checking the full root list.

Answer

Only $2$ is a root among the candidates. All roots: $2$ twice, $-1+i$ and $-1-i$ .

Key idea

Each successful root test reduces the remaining factorization to a smaller polynomial.
Problem 4: Mark complete
Problem 5 A real-coefficient construction

Find the monic polynomial of least degree with real coefficients whose roots include $-3$ , $1$ , and $2+3i$ . Give it in standard form.
1. Hint 1
  Real coefficients require the conjugate of the nonreal root.
2. Hint 2
  Multiply the conjugate pair of factors first, then attach the two real-root factors.
Answer

$x^4-2x^3+2x^2+38x-39$ .

Full solution

The root $2-3i$ is also required, so the least degree is $4$ .
The pair gives $(x-2)^2+9=x^2-4x+13$ , and multiplying $(x+3)(x-1)$ gives $x^2+2x-3$ .

Multiplying $x^2+2x-3$ by $x^2-4x+13$ and collecting like powers gives
$P(x)=x^4-2x^3+2x^2+38x-39$
The three factors together supply all four required roots: $-3$ , $1$ , and the conjugate pair $2\pm3i$ .

Answer

$x^4-2x^3+2x^2+38x-39$ .

Key idea

Prescribed nonreal roots may force additional factors when coefficients must be real.
Problem 5: Mark complete
Problem 6 A coefficient constraint

A polynomial of degree $3$ has zeros $0$ , $-2$ , and $3$ , and its coefficient of $x^2$ is $4$ . Find the polynomial in standard form and its value at $x=1$ .
1. Hint 1
  The three distinct roots give the polynomial up to a nonzero multiplier.
2. Hint 2
  Compare the $x^2$ coefficient to determine that multiplier.
Answer

$P(x)=-4x^3+4x^2+24x$ ; $P(1)=24$ .

Full solution

Write $P(x)=a x(x+2)(x-3)$ with $a\ne0$ .
The monic product is
$x(x+2)(x-3)=x^3-x^2-6x$
Its $x^2$ coefficient after scaling is $-a=4$ , so $a=-4$ .
Therefore $P=-4x^3+4x^2+24x$ , and
$P(1)=-4+4+24=24.$
Its factored form checks all three zeros.

Answer

$P(x)=-4x^3+4x^2+24x$ ; $P(1)=24$ .

Key idea

A specified nonleading coefficient can fix the scale of a polynomial built from its roots.
Problem 6: Mark complete
Problem 7 A reduced equation

The polynomial $P(x)=5x^3+7x^2-58x-24$ is divisible by $5x+2$ . Find all its roots and give its factorization over the real numbers.
1. Hint 1
  The known divisor removes one linear factor.
2. Hint 2
  Divide to get a quadratic and then solve that smaller equation.
Answer

Roots $-\frac25,3,-4$ ; $P(x)=(5x+2)(x-3)(x+4)$ .

Full solution

Division by $5x+2$ gives quotient $x^2+x-12$ and remainder zero.
The quotient factors as
$(x-3)(x+4)=x^2+x-12$
Hence $P=(5x+2)(x-3)(x+4)$ , with roots $-\frac25,3,-4$ .
Multiplying the quadratic by $5x+2$ restores $5x^3+7x^2-58x-24$ .

Answer

Roots $-\frac25,3,-4$ ; $P(x)=(5x+2)(x-3)(x+4)$ .

Key idea

A known nonmonic linear factor can reduce a cubic equation to a quadratic.
Problem 7: Mark complete
Problem 8 Agreement at three inputs

Polynomials $A$ and $B$ each have degree at most $2$ , and they agree at the three distinct inputs $-1$ , $2$ , and $4$ . Must they agree for every input? Explain using a root bound.
1. Hint 1
  Look at the polynomial formed by subtracting the two rules.
2. Hint 2
  Its degree is at most two unless it is the zero polynomial.
Answer

Yes; $A(x)=B(x)$ for every input.

Full solution

Define $H=A-B$ .
The three agreements mean
$H(-1)=0$
together with $H(2)=0$ and $H(4)=0$ .
A nonzero polynomial of degree at most $2$ has at most two distinct roots.
Hence $H$ is zero, so $A=B$ everywhere.

Answer

Yes; $A(x)=B(x)$ for every input.

Key idea

Enough matching values force two bounded-degree polynomials to be identical.
Problem 8: Mark complete
Problem 9 Two roots and a value

A student says the roots $-1$ and $2$ together with the condition $P(0)=-2$ determine exactly one polynomial. Is the statement correct when the degree is not specified? Give two different polynomials, each having exactly the real roots $-1$ and $2$ , if the statement is false.
1. Hint 1
  More than one polynomial can share the same roots and the same value at zero.
2. Hint 2
  A repeated factor raises the degree without introducing a new root.
Answer

False; $(x+1)(x-2)$ and $(x+1)^2(x-2)$ both meet the conditions.

Full solution

The first polynomial $A(x)=(x+1)(x-2)$ has both required roots and
$A(0)=(1)(-2)=-2.$
Let
$B(x)=(x+1)^2(x-2)$
It has the same real roots, $-1$ (now repeated) and $2$ , and also satisfies $B(0)=(1)^2(-2)=-2$ , yet has degree $3$ rather than $2$ .
Both share exactly the same real-root list even though they are different polynomials.

Answer

False; $(x+1)(x-2)$ and $(x+1)^2(x-2)$ both meet the conditions.

Key idea

Root and value data may fail to determine a unique polynomial when its degree is unrestricted.
Problem 9: Mark complete
Problem 10 A repeated test

If $x-c$ is a factor of $P(x)$ , a student claims it must also be a factor of $P(x)+(x-c)^2$ . Is the claim true? Explain without assuming any particular degree of $P$ .
1. Hint 1
  A common factor can be pulled out of a sum.
2. Hint 2
  Write the given factorization as $P(x)=(x-c)Q(x)$ .
Answer

True.

Full solution

The hypothesis gives $P=(x-c)Q$ for a polynomial $Q$ .
Set $T(x)=Q(x)+x-c$ , which is also a polynomial.
Pulling out the common factor gives
$P(x)+(x-c)^2=(x-c)T(x)$
Thus $x-c$ divides the whole sum.
Evaluating at $c$ also gives zero.

Answer

True.

Key idea

Adding another multiple of a linear factor preserves divisibility by that factor.
Problem 10: Mark complete

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