What is Polynomials & factoring on the Algebra 2 Regents?

Polynomials & factoring is a Polynomials topic on the Algebra 2 Regents. RegentsLab has 41 practice questions on it, each with a full worked solution.

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Polynomials & factoring: Algebra 2 Regents Practice

Practice Polynomials & factoring for the Algebra 2 Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 1 · 2 pts

Factor completely: $x^3 + 2x^2 - 9x - 18$.

(A) $(x+2)(x^2-9)$
(B) $(x+2)(x-3)(x+3)$ ✓
(C) $(x-2)(x-3)(x+3)$
(D) $(x+2)(x-3)^2$

Worked solution

$x^3 + 2x^2 - 9x - 18 = (x^3 + 2x^2) + (-9x - 18)$ Group the four terms into two pairs.

$= x^2(x+2) - 9(x+2)$ Factor the GCF out of each pair.

$= (x+2)(x^2 - 9)$ Factor out the common binomial (x+2).

$= (x+2)(x-3)(x+3)$ x^2 - 9 is a difference of squares.

Answer: $(x+2)(x-3)(x+3)$

Factoring by grouping exposes a common binomial, and any resulting difference of squares must be factored further to be complete.

Question 2 · 2 pts

Factor completely: $x^3 + 8$.

(A) $(x+2)^3$
(B) $(x+2)(x^2-2x+4)$ ✓
(C) $(x+2)(x^2+2x+4)$
(D) $(x+2)(x^2-4x+4)$

Worked solution

$x^3 + 8 = x^3 + 2^3$ Recognize a sum of two cubes with a = x, b = 2.

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ Apply the sum-of-cubes formula.

$= (x+2)(x^2 - 2x + 4)$ The quadratic factor does not factor further over the reals.

Answer: $(x+2)(x^2-2x+4)$

A sum of cubes factors as $(a+b)(a^2-ab+b^2)$, where the trinomial's middle term is $-ab$, not $-2ab$.

Question 12 · 2 pts

What is the remainder when $P(x) = x^3 - 4x^2 + 2x + 5$ is divided by $(x - 3)$?

(A) $-64$
(B) $2$ ✓
(C) $0$
(D) $-3$

Worked solution

$\text{Remainder Theorem: remainder} = P(3)$

$P(3) = 3^3 - 4(3)^2 + 2(3) + 5$

$= 27 - 36 + 6 + 5$

$= 2$

Answer: $2$

The Remainder Theorem says the remainder when $P(x)$ is divided by $(x - a)$ equals $P(a)$, so you just evaluate the polynomial at $a = 3$.

Question 16 · 2 pts

Describe the end behavior of $f(x) = -2x^3 + 5x^2 - x + 4$.

(A) As $x \to -\infty,\ f(x) \to +\infty$; as $x \to +\infty,\ f(x) \to -\infty$ ✓
(B) As $x \to -\infty,\ f(x) \to +\infty$; as $x \to +\infty,\ f(x) \to +\infty$
(C) As $x \to -\infty,\ f(x) \to -\infty$; as $x \to +\infty,\ f(x) \to +\infty$
(D) As $x \to -\infty,\ f(x) \to -\infty$; as $x \to +\infty,\ f(x) \to -\infty$

Worked solution

$\text{Leading term: } -2x^3$ Degree 3 (odd), leading coefficient -2 < 0.

$\text{Odd degree} \Rightarrow \text{the two ends go in opposite directions}$

$\text{Negative leading coefficient} \Rightarrow \text{left end up, right end down}$

$x \to -\infty,\ f(x) \to +\infty;\quad x \to +\infty,\ f(x) \to -\infty$

Answer: As $x \to -\infty,\ f(x) \to +\infty$ and as $x \to +\infty,\ f(x) \to -\infty$

End behavior is set by the leading term: an odd degree makes the two ends point opposite ways, and a negative leading coefficient flips the standard rising-to-the-right pattern.

Question 2 · 2 pts

If $x-2$ is a factor of $p(x)=x^3-5x^2+kx+6$, what is the value of $k$?

(A) $-3$
(B) $11$
(C) $-11$
(D) $3$ ✓

Worked solution

$p(2)=0$ By the Factor Theorem, $x-2$ is a factor exactly when $p(2)=0$.

$2^3 - 5(2)^2 + 2k + 6 = 0$

$8 - 20 + 2k + 6 = 0$

$2k - 6 = 0$

$k = 3$

Answer: $3$

The Factor Theorem says $x-a$ is a factor of $p(x)$ if and only if $p(a)=0$. Substituting $x=2$ and solving gives $k=3$. Substituting $x=-2$ by mistake leads to $k=-11$.

Question 13 · 2 pts

Which binomial is a factor of $x^3 + 2x^2 - 9x - 18$?

(A) $x + 6$
(B) $x - 2$
(C) $x + 2$ ✓
(D) $x - 9$

Worked solution

$x^3 + 2x^2 - 9x - 18 = x^2(x+2) - 9(x+2)$ Factor by grouping

$= (x+2)(x^2 - 9)$

$= (x+2)(x-3)(x+3)$ Difference of squares

$\text{Of the choices, only } x+2 \text{ appears as a factor.}$

Answer: $x + 2$

Grouping the first two terms and the last two terms produces a common binomial factor of $x+2$, so the polynomial factors as $(x+2)(x-3)(x+3)$. You can confirm with the factor theorem: substituting $x=-2$ gives $-8+8+18-18=0$, so $x+2$ is a factor.

Question 26 · 4 pts · constructed response

Factor the expression $x^3 - 4x^2 - 9x + 36$ completely.

Model answer: $(x-4)(x-3)(x+3)$

Worked solution

$x^3 - 4x^2 - 9x + 36 = x^2(x-4) - 9(x-4)$ Factor by grouping: group the first two terms and the last two terms.

$= (x-4)(x^2 - 9)$ Factor out the common binomial $(x-4)$.

$= (x-4)(x-3)(x+3)$ Factor the difference of two perfect squares.

Answer: $(x-4)(x-3)(x+3)$

Grouping works because the first pair of terms shares the factor $x^2$ and the second pair shares $-9$, leaving the common binomial $x-4$. The remaining factor $x^2-9$ is a difference of squares, so the complete factorization is $(x-4)(x-3)(x+3)$.

Question 34 · 4 pts · constructed response

Given the function $p(x) = x^3 - 5x^2 + 2x + 8$, use the remainder theorem to show that $(x - 4)$ is a factor of $p(x)$. Then write $p(x)$ as a product of three linear factors.

Model answer: $p(4) = 0$, so $(x-4)$ is a factor by the remainder theorem, and $p(x) = (x-4)(x-2)(x+1)$.

Worked solution

$p(4) = 4^3 - 5(4)^2 + 2(4) + 8 = 64 - 80 + 8 + 8 = 0$ Since the remainder when dividing by (x - 4) equals p(4), and p(4) = 0, (x - 4) is a factor.

$p(x) \div (x - 4) = x^2 - x - 2$ Divide using long division or synthetic division with 4: coefficients 1, -5, 2, 8 give 1, -1, -2, remainder 0.

$x^2 - x - 2 = (x - 2)(x + 1)$ Factor the quadratic quotient.

$p(x) = (x - 4)(x - 2)(x + 1)$

Answer: $p(4) = 0$, so $(x-4)$ is a factor by the remainder theorem, and $p(x) = (x-4)(x-2)(x+1)$.

The remainder theorem says the remainder when a polynomial is divided by $(x - a)$ equals $p(a)$. Because $p(4) = 0$, the remainder is zero, so $(x-4)$ divides $p(x)$ evenly. Dividing gives the quadratic $x^2 - x - 2$, which factors into $(x-2)(x+1)$, producing the complete linear factorization.

Question 1 · 2 pts

Which binomial is a factor of $p(x) = x^3 + 2x^2 - 5x - 6$?

(A) $x + 1$ ✓
(B) $x - 3$
(C) $x + 2$
(D) $x - 1$

Worked solution

$p(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6$ By the Factor Theorem, $x + 1$ is a factor if $p(-1) = 0$.

$p(-1) = -1 + 2 + 5 - 6 = 0$ Since the remainder is 0, $x + 1$ is a factor.

$p(1) = 1 + 2 - 5 - 6 = -8 \neq 0, \quad p(-2) = -8 + 8 + 10 - 6 = 4 \neq 0, \quad p(3) = 27 + 18 - 15 - 6 = 24 \neq 0$ The other choices leave nonzero remainders.

Answer: $x + 1$

By the Factor Theorem, a binomial $x - a$ is a factor of a polynomial exactly when substituting $a$ gives 0. Testing $x = -1$ gives $p(-1) = 0$, so $x + 1$ is a factor, while the other candidates all produce nonzero remainders.

Question 13 · 2 pts

Which binomial is a factor of $x^3 - 3x^2 - 4x + 12$?

(A) $x + 2$ ✓
(B) $x + 1$
(C) $x - 4$
(D) $x + 3$

Worked solution

$x^3 - 3x^2 - 4x + 12 = x^2(x - 3) - 4(x - 3)$ Factor by grouping.

$= (x - 3)(x^2 - 4)$

$= (x - 3)(x - 2)(x + 2)$ Difference of squares.

$\text{The factors are } x - 3,\ x - 2,\ x + 2$

Answer: $x + 2$

Grouping the first two and last two terms gives a common factor of $(x-3)$, and $x^2-4$ factors as a difference of squares into $(x-2)(x+2)$. Only $x+2$ appears among the choices; you can confirm with the factor theorem since substituting $x=-2$ gives $-8-12+8+12=0$.

Question 20 · 2 pts

When $p(x) = x^3 - 2x^2 + 5x - 6$ is divided by $x - 2$, the remainder is

(A) $-32$
(B) $4$ ✓
(C) $-4$
(D) $0$

Worked solution

$\text{Remainder Theorem: the remainder is } p(2)$

$p(2) = 2^3 - 2(2)^2 + 5(2) - 6$

$= 8 - 8 + 10 - 6 = 4$

Answer: $4$

By the Remainder Theorem, dividing by $x - 2$ leaves remainder $p(2)$. Evaluating gives $8 - 8 + 10 - 6 = 4$. Substituting $x = -2$ instead produces $-32$, and assuming $x - 2$ is a factor gives the distractor 0.

Question 3 · 2 pts

If $x-2$ is a factor of $p(x)=x^3-3x^2+kx-2$, then the value of $k$ is

(A) $-3$
(B) $6$
(C) $3$ ✓
(D) $1$

Worked solution

$x-2 \text{ is a factor} \Rightarrow p(2)=0$ Factor Theorem.

$p(2)=2^3-3(2)^2+2k-2=8-12+2k-2$

$2k-6=0$

$k=3$

Answer: $3$

By the Factor Theorem, $x-2$ is a factor exactly when $p(2)=0$. Substituting $x=2$ gives $8-12+2k-2=2k-6$, and setting this equal to zero yields $k=3$.

Question 3 · 2 pts

Which binomial is a factor of $x^3+2x^2-5x-6$?

(A) $x+1$ ✓
(B) $x-1$
(C) $x-3$
(D) $x+2$

Worked solution

$p(-1)=(-1)^3+2(-1)^2-5(-1)-6$ By the factor theorem, $(x+1)$ is a factor if $p(-1)=0$.

$p(-1)=-1+2+5-6=0$

$x^3+2x^2-5x-6=(x+1)(x+3)(x-2)$ Full factorization confirms the result.

Answer: $x+1$

By the factor theorem, $(x-a)$ is a factor exactly when $p(a)=0$. Testing $x=-1$ gives $-1+2+5-6=0$, so $(x+1)$ is a factor. The complete factorization is $(x+1)(x+3)(x-2)$, so none of the other listed binomials work.

Question 16 · 2 pts

Which binomial is a factor of $p(x)=x^3-4x^2+x+6$?

(A) $x-1$
(B) $x+2$
(C) $x+3$
(D) $x-2$ ✓

Worked solution

$p(2)=2^3-4(2)^2+2+6=8-16+2+6=0$ By the factor theorem, $x-a$ is a factor when $p(a)=0$.

$p(2)=0 \;\Rightarrow\; (x-2) \text{ is a factor}$

$p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$ Full factorization confirms the result.

Answer: $x-2$

By the factor theorem, $x-2$ is a factor of $p(x)$ exactly when $p(2)=0$. Substituting gives $8-16+2+6=0$, so $x-2$ is a factor. The other choices give nonzero remainders when tested.

Question 9 · 2 pts

If $P(x) = x^3 - 2x^2 + x - 5$, what is $P(2)$ ?

(A) $-3$ ✓
(B) $-7$
(C) $13$
(D) $3$

Worked solution

$P(2) = 8 - 2(4) + 2 - 5$ Substitute x = 2.

$= 8 - 8 + 2 - 5 = -3$

Answer: -3

Evaluate by substitution; the remainder theorem says P(2) is also the remainder when dividing by (x − 2).

Question 3 · 2 pts

Factor completely: $6x^2 + 7x - 20$.

(A) $(2x+5)(3x-4)$ ✓
(B) $(3x+5)(2x-4)$
(C) $(6x-5)(x+4)$
(D) $(2x-5)(3x+4)$

Worked solution

$ac = (6)(-20) = -120$ Use the ac-method on the trinomial.

$15 \cdot (-8) = -120, \quad 15 + (-8) = 7$ Find two numbers with product -120 and sum 7.

$6x^2 + 15x - 8x - 20$ Split the middle term using those numbers.

$3x(2x+5) - 4(2x+5) = (2x+5)(3x-4)$ Factor by grouping.

Answer: $(2x+5)(3x-4)$

The ac-method rewrites the middle term as a sum so the trinomial can be factored by grouping; check by expanding.

Question 4 · 2 pts

When the polynomial $P(x)=x^3-4x^2+2x+7$ is divided by $x-3$, what is the remainder?

(A) $4$ ✓
(B) $-3$
(C) $0$
(D) $76$

Worked solution

$\text{Remainder Theorem: remainder} = P(3)$ Divisor x - 3 means evaluate at x = 3.

$P(3) = (3)^3 - 4(3)^2 + 2(3) + 7$ Substitute x = 3.

$= 27 - 36 + 6 + 7 = 4$

Answer: $4$

By the Remainder Theorem, dividing $P(x)$ by $x-c$ leaves a remainder of $P(c)$; here $c=3$.

Question 5 · 2 pts

Find all real solutions of $x^3 - 5x^2 - 6x = 0$.

(A) $x = -1,\ 0,\ 6$ ✓
(B) $x = -6,\ 0,\ 1$
(C) $x = -1,\ 6$
(D) $x = 0,\ 2,\ 3$

Worked solution

$x^3 - 5x^2 - 6x = 0$

$x(x^2 - 5x - 6) = 0$ Factor out the GCF x; do not divide it away.

$x(x-6)(x+1) = 0$ Factor the quadratic (product -6, sum -5).

$x = 0, \quad x = 6, \quad x = -1$ Zero-product property.

Answer: $x = -1,\ 0,\ 6$

Always factor out a common variable rather than dividing by it, so the root $x=0$ is not lost.

Question 9 · 2 pts

Factor completely: $x^3 + 3x^2 - 4x - 12$.

(A) $(x + 3)(x^2 - 4)$
(B) $(x + 3)(x^2 + 4)$
(C) $(x - 3)(x - 2)(x + 2)$
(D) $(x + 3)(x - 2)(x + 2)$ ✓

Worked solution

$x^3 + 3x^2 - 4x - 12 = (x^3 + 3x^2) + (-4x - 12)$ Group into pairs.

$= x^2(x + 3) - 4(x + 3)$ Factor the GCF from each pair.

$= (x + 3)(x^2 - 4)$ Factor out the common (x + 3).

$= (x + 3)(x - 2)(x + 2)$ Factor the difference of squares.

Answer: $(x + 3)(x - 2)(x + 2)$

Factoring by grouping pairs terms that share a common binomial factor; always check whether a remaining factor such as a difference of squares can be factored further for a complete factorization.

Question 13 · 2 pts

For what value of $k$ is $(x - 2)$ a factor of $P(x) = x^3 + kx^2 - 5x - 2$?

(A) $k = 0$
(B) $k = 4$
(C) $k = -4$
(D) $k = 1$ ✓

Worked solution

$\text{Factor Theorem: } (x - 2) \text{ a factor} \iff P(2) = 0$

$P(2) = 2^3 + k(2)^2 - 5(2) - 2 = 0$

$8 + 4k - 10 - 2 = 0$

$4k - 4 = 0 \;\Rightarrow\; k = 1$

Answer: $k = 1$

By the Factor Theorem, $(x - a)$ is a factor of $P(x)$ exactly when $P(a) = 0$; set $P(2) = 0$ and solve the resulting equation for $k$.

Question 15 · 2 pts

Solve for all real values of $x$: $x^3 - 5x^2 - 24x = 0$.

(A) $x = 8,\ -3$
(B) $x = 0,\ -8,\ -3$
(C) $x = 0,\ -8,\ 3$
(D) $x = 0,\ 8,\ -3$ ✓

Worked solution

$x^3 - 5x^2 - 24x = 0$

$x(x^2 - 5x - 24) = 0$ Factor out the GCF x.

$x(x - 8)(x + 3) = 0$ -8 and +3 multiply to -24, add to -5.

$x = 0,\quad x = 8,\quad x = -3$ Zero Product Property.

Answer: $x = 0,\ x = 8,\ x = -3$

Always factor out the GCF first; dividing by the variable would discard a valid root, so factoring out $x$ keeps the solution $x = 0$.

Question 10 · 2 pts

What is the remainder when $x^3 + 3x^2 - 2x + 7$ is divided by $x+2$?

(A) $15$ ✓
(B) $7$
(C) $3$
(D) $23$

Worked solution

$\text{Remainder Theorem: remainder} = p(-2)$ Dividing by $x+2$ means evaluating at $x=-2$.

$p(-2) = (-2)^3 + 3(-2)^2 - 2(-2) + 7$

$= -8 + 12 + 4 + 7$

$= 15$

Answer: $15$

By the Remainder Theorem, the remainder on division by $x+2$ equals $p(-2)$. Careful sign work gives $-8+12+4+7=15$. Evaluating at $x=2$ instead gives the distractor 23, and sign slips on the middle terms give 7 or 3.

Question 19 · 2 pts

When $p(x) = x^3 - 4x^2 + x - 6$ is divided by $x - 2$, the remainder is

(A) $12$
(B) $0$
(C) $-12$ ✓
(D) $-32$

Worked solution

$\text{Remainder Theorem: the remainder is } p(2)$

$p(2) = 2^3 - 4(2)^2 + 2 - 6$

$= 8 - 16 + 2 - 6 = -12$

Answer: $-12$

By the Remainder Theorem, dividing a polynomial by $x - 2$ leaves a remainder equal to $p(2)$. Evaluating gives $8 - 16 + 2 - 6 = -12$. The distractor $-32$ comes from evaluating $p(-2)$ instead, a common sign error with the theorem.

Question 2 · 2 pts

What is the remainder when $x^3 - 4x^2 + 2x - 5$ is divided by $x - 3$?

(A) $8$
(B) $-5$
(C) $-8$ ✓
(D) $-74$

Worked solution

$\text{Remainder Theorem: the remainder when } p(x) \text{ is divided by } x - 3 \text{ is } p(3)$

$p(3) = 3^3 - 4(3)^2 + 2(3) - 5$

$p(3) = 27 - 36 + 6 - 5 = -8$

Answer: $-8$

The Remainder Theorem says the remainder on dividing $p(x)$ by $x - 3$ equals $p(3)$. Evaluating gives $27 - 36 + 6 - 5 = -8$. Evaluating at $-3$ instead of $3$, or dropping a sign, produces the distractors.

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