Algebra 3 Quiz: Solving Polynomial Equations
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Solving Polynomial EquationsQuestion 1 of 12

A rectangular box has a volume of 84 cubic inches. Its height is 2 inches less than its width ww, and its length is 3 inches more than its width ww.

After writing the volume equation and simplifying, a student gets w3+w2−6w−84=0w^3 + w^2 - 6w - 84 = 0. Between which two consecutive integers is the positive solution ww?

3 and 4
4 and 5
5 and 6
6 and 7
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Algebra 3 Quiz

Algebra 3 Quiz: Solving Polynomial Equations

Practice Solving Polynomial Equations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Polynomial Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A rectangular box has a volume of 84 cubic inches. Its height is 2 inches less than its width ww, and its length is 3 inches more than its width ww.

After writing the volume equation and simplifying, a student gets w3+w2−6w−84=0w^3 + w^2 - 6w - 84 = 0. Between which two consecutive integers is the positive solution ww?

  1. 3 and 4
  2. 4 and 5 (correct answer)
  3. 5 and 6
  4. 6 and 7
Explanation: When you see a volume equation, start by writing the product of the actual dimensions. Here the width is ww, the height is w−2w-2, and the length is w+3w+3, so the volume equation is w(w−2)(w+3)=84.w(w-2)(w+3)=84. Expanding gives w3+w2−6w−84=0w^3+w^2-6w-84=0, which matches the student's equation. To locate the positive solution, test consecutive integer boundaries. At w=4w=4, the volume is 4(2)(7)=564(2)(7)=56, still below 84. At w=5w=5, the volume is 5(3)(8)=1205(3)(8)=120, already above 84. Since the volume grows as ww increases for positive ww, the crossing must happen between 4 and 5. The interval "3 and 4" cannot be correct because even at w=4w=4 the volume is too small. "5 and 6" and "6 and 7" are both past the crossing: at w=5w=5 the volume already exceeds 84, and it continues increasing, so the root cannot lie there. A common trap is comparing 84 to w3w^3 alone, since 843≈4.4\sqrt[3]{84}\approx 4.4; but the adjusted dimensions shift the true root. On exam day, when asked for an interval containing a polynomial root, evaluate the equation at the integer endpoints and look for a sign change. That will quickly point you to the correct consecutive integers.

Question 2

Let p(x)=x3(x−2)2(x+1)p(x)=x^3(x-2)^2(x+1). At which xx-values does the graph of y=p(x)y=p(x) cross the xx-axis, rather than only touch it?

  1. −1-1 and 00 (correct answer)
  2. −1-1 and 22
  3. 00 and 22
  4. −1-1, 00, and 22
Explanation: When you see a factored polynomial and a question about the graph crossing or touching the xx-axis, the key is the multiplicity of each factor. A root with odd multiplicity makes the graph cross the axis; a root with even multiplicity makes it only touch or bounce off the axis. Here, the factors are already written as x3(x−2)2(x+1)x^3(x-2)^2(x+1). The root x=0x=0 comes from x3x^3, so its multiplicity is 33, which is odd — the graph crosses at 00. The root x=−1x=-1 comes from (x+1)(x+1), so its multiplicity is 11, also odd — the graph crosses at −1-1. The root x=2x=2 comes from (x−2)2(x-2)^2, so its multiplicity is 22, even — the graph only touches at 22. Therefore the correct set of crossing points is −1-1 and 00. Now look at the wrong answers. The choice "−1-1 and 22" incorrectly treats 22 as a crossing point and leaves out 00. The choice "00 and 22" also includes 22 while missing −1-1. The choice "−1-1, 00, and 22" assumes every root crosses, ignoring the even multiplicity at 22. Each wrong answer is a version of forgetting to check the exponent on the factor. Your takeaway: whenever a polynomial is factored, don't just list the zeros — check each exponent. Odd exponent means "through" the axis; even exponent means "bounce" off it. This distinction is a favorite on algebra exams.

Question 3

How many distinct rational solutions does 2x3+3x2−8x+3=02x^3 + 3x^2 - 8x + 3 = 0 have?

  1. 1
  2. 2
  3. 3 (correct answer)
  4. 4
Explanation: Whenever you see a cubic and the question asks for rational solutions, your first move should be the rational root theorem. For 2x3+3x2−8x+3=02x^3+3x^2-8x+3=0, the possible rational roots are ±1,±3,±12,±32\pm1,\pm3,\pm\frac12,\pm\frac32. Testing them gives: x=1:2+3−8+3=0x=1:\quad 2+3-8+3=0 and x=−3:−54+27+24+3=0.x=-3:\quad -54+27+24+3=0. Now divide or factor there: 2x3+3x2−8x+3=(x−1)(x+3)(2x−1).2x^3+3x^2-8x+3=(x-1)(x+3)(2x-1). So the roots are 11, −3-3, and 12\frac12. That is three distinct rational solutions, which matches the choice labeled 3. The choice 1 is a trap if you test only x=1x=1 and stop too early. The choice 2 appears if you find 11 and −3-3but miss the fractional root 12\frac12; remember to keep factoring all the way down. The choice 4 is impossible: a cubic can have at most three real roots total, so you can never have four distinct solutions. . For your study habit: after finding one rational root, factor it out immediately. Then you are left with a quadratic, which is easy to solve — and that prevents you from losing the remaining roots. Also, always check fractional candidates from the rational root theorem, especially when the leading coefficient isn't 11.

Question 4

Which of the following is NOT a possible rational root of 6x4−5x3+2x−7=06x^4 - 5x^3 + 2x - 7 = 0?

  1. 72\frac{7}{2}
  2. 13\frac{1}{3}
  3. 56\frac{5}{6} (correct answer)
  4. 77
Explanation: When you see a polynomial like 6x4−5x3+2x−7=06x^4 - 5x^3 + 2x - 7 = 0 and are asked about possible rational roots, the Rational Root Theorem is your tool: any rational root must be of the form pq\frac{p}{q}, where pp is a factor of the constant term and qq is a factor of the leading coefficient. Here, the constant term is −7-7, so pp can only be ±1\pm 1 or ±7\pm 7. The leading coefficient is 66, so qq can be ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6. That means possible roots include 77 (since 7=717 = \frac{7}{1}), 72\frac{7}{2} (since 77 divides 77 and 22 divides 66), and 13\frac{1}{3} (since 11 divides 77 and 33 divides 66). But 56\frac{5}{6} cannot be a rational root because its numerator, 55, is not a factor of 77. Even though the denominator 66 is a factor of the leading coefficient, the numerator must divide the constant term — that's the trap this question sets. Remember to check both parts: the numerator must divide the constant, and the denominator must divide the leading coefficient. If either fails, that number is not a possible rational root. So when you see fractions, quickly test numerator against the constant term before doing any polynomial arithmetic.

Question 5

Which set contains all real solutions of 4x4−17x2+4=04x^4 - 17x^2 + 4 = 0?

  1. {−2,2}\{-2, 2\}
  2. {−12,12}\{-\frac{1}{2}, \frac{1}{2}\}
  3. {−2,−14,14,2}\{-2, -\frac{1}{4}, \frac{1}{4}, 2\}
  4. {−2,−12,12,2}\{-2, -\frac{1}{2}, \frac{1}{2}, 2\} (correct answer)
Explanation: Whenever you see an equation with only even powers of xx—like x4x^4 and x2x^2—think "quadratic in disguise." Let u=x2u = x^2. Then the equation becomes 4u2−17u+4=0.4u^2 - 17u + 4 = 0. Factor: (4u−1)(u−4)=0.(4u - 1)(u - 4) = 0. So u=14u = \frac14 or u=4u = 4. Since u=x2u = x^2, solve each piece: x2=14  ⟹  x=±12,x^2 = \frac14 \implies x = \pm \frac12, x2=4  ⟹  x=±2.x^2 = 4 \implies x = \pm 2. Thus the full real solution set is {−2,−12,12,2}.\left\{-2, -\frac12, \frac12, 2\right\}. The choice containing exactly these four values is the correct one. Now look at the wrong choices. The set {−2,2}\{-2, 2\} misses the two solutions −12-\frac12 and 12\frac12; this happens if you only solve x2=4x^2 = 4 and forget that x2=14x^2 = \frac14 also gives real roots. The set {−12,12}\{-\frac12, \frac12\} makes the opposite mistake: it includes only the smaller roots and misses ±2\pm 2. The set {−2,−14,14,2}\{-2, -\frac14, \frac14, 2\} includes ±2\pm 2 but has ±14\pm \frac14 instead of ±12\pm \frac12; this is a square-root error—if x2=14x^2 = \frac14, then x=±12x = \pm \frac12, not ±14\pm \frac14. Your study tip: for even-powered equations, substitute u=x2u = x^2, solve for uu, then take both positive and negative square roots. Always ask yourself, "Did I include both signs for every positive uu?" That single habit prevents the most common trap on this type of question.

Question 6

A cubic polynomial has zeros −2-2, 11, and 33. If f(0)=12f(0)=12, what is f(2)f(2)?

  1. −8-8 (correct answer)
  2. −4-4
  3. 44
  4. 88
Explanation: When a cubic polynomial has known zeros, the most reliable move is to write it in factored form with an unknown leading coefficient:
f(x)=a(x+2)(x−1)(x−3).f(x)=a(x+2)(x-1)(x-3).
The zeros tell you the factors; the extra point determines aa. Plug in x=0x=0:
12=a(2)(−1)(−3)=6a,12=a(2)(-1)(-3)=6a,
so a=2a=2. Now evaluate at x=2x=2:
f(2)=2(4)(1)(−1)=−8.f(2)=2(4)(1)(-1)=-8.
Thus the correct value is −8-8. The distractor −4-4 is what you get if you evaluate (x+2)(x−1)(x−3)(x+2)(x-1)(x-3) at x=2x=2 but forget to multiply by the leading coefficient a=2a=2. The choice 44 comes from the same kind of slip plus losing the negative sign: (4)(1)(1)=4(4)(1)(1)=4. The choice 88 arises when you remember to multiply by a=2a=2 but accidentally treat 2−32-3 as 11 instead of −1-1, giving 2(4)(1)(1)=82(4)(1)(1)=8. Always start with factored form when zeros are given, then use the extra point to find the leading coefficient. Finally, evaluate each factor carefully at the target input — especially watch the sign of any negative factor like (x−3)(x-3).

Question 7

One solution of 2x3−4x2−10x+12=02x^3 - 4x^2 - 10x + 12 = 0 is x=1x=1. What is the product of the other two solutions?

  1. −6-6 (correct answer)
  2. −2-2
  3. 33
  4. 66
Explanation: Whenever a cubic equation gives you one root, you can use Vieta's formulas to avoid factoring everything by hand. For a cubic ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0, the product of all three roots is −da-\frac{d}{a}. Here that product is −122=−6-\frac{12}{2}=-6. Since one root is x=1x=1, the product of the other two roots must be −6÷1=−6-6 \div 1 = -6. You can confirm by synthetic division: dividing 2x3−4x2−10x+122x^3-4x^2-10x+12 by x−1x-1 gives 2x2−2x−122x^2-2x-12, which factors as 2(x−3)(x+2)2(x-3)(x+2). So the other roots are 33 and −2-2, and their product is −6-6. The choice −2-2 is just one of the other roots, not the product. The choice 33 is the other root alone. The choice 66 is the common sign mistake: it ignores the negative sign in Vieta's product formula or forgets that the leading coefficient affects the product. So when one root is given, remember: total product of all roots divided by the known root gives you the product of the remaining roots. That turns this into a one-step calculation.

Question 8

For what value of kk is x−3x-3 a factor of x3−5x2+kx−12=0x^3 - 5x^2 + kx - 12 = 0?

  1. −10-10
  2. 22
  3. 66
  4. 1010 (correct answer)
Explanation: Whenever you are asked whether a linear expression like x−3x-3 is a factor, the Factor Theorem is your key: x−cx-c is a factor exactly when plugging x=cx=c into the polynomial gives 00. Here c=3c=3, so substitute 33 into x3−5x2+kx−12x^3 - 5x^2 + kx - 12: 27−45+3k−12=−30+3k27 - 45 + 3k - 12 = -30 + 3k. For x−3x-3 to be a factor, this remainder must be 00, so −30+3k=0-30 + 3k = 0, giving k=10k=10. Thus 1010 is the value that makes the polynomial equal 00 at x=3x=3. The choice −10-10 is a classic sign trap: it comes from solving −30−3k=0-30 - 3k = 0 instead of −30+3k=0-30 + 3k = 0, or from thinking the root is x=−3x=-3. Choices 22 and 66 fail the test: with k=2k=2, the polynomial evaluates to −24-24, and with k=6k=6, it evaluates to −12-12; neither is zero, so neither produces a factor. A quick way to avoid these errors is to always write the remainder expression before solving, then check by evaluating at x=3x=3. Remember: for x−3x-3, you plug in positive 33, and set the whole result to zero.

Question 9

A monic cubic polynomial p(x)=x3+ax2+bx+cp(x)=x^3+ax^2+bx+c satisfies p(1)=0p(1)=0, p(2)=0p(2)=0, and p(3)=2p(3)=2. What is the value of aa?

  1. −7-7
  2. −6-6
  3. −4-4
  4. −5-5 (correct answer)
Explanation: When you see a monic cubic like x3+ax2+bx+cx^3+ax^2+bx+c and are told values of the polynomial, your first move should be to use the known roots. Since p(1)=0p(1)=0 and p(2)=0p(2)=0, both 11 and 22 are roots, so the polynomial factors as p(x)=(x−1)(x−2)(x−r)p(x)=(x-1)(x-2)(x-r) where rr is the third root. Expanding gives p(x)=x3−(r+3)x2+(3r+2)x−2r,p(x)=x^3-(r+3)x^2+(3r+2)x-2r, so a=−(r+3)a=-(r+3). Now use p(3)=2p(3)=2: (3−1)(3−2)(3−r)=2  ⟹  2(1)(3−r)=2  ⟹  3−r=1  ⟹  r=2.(3-1)(3-2)(3-r)=2 \implies 2(1)(3-r)=2 \implies 3-r=1 \implies r=2. Thus the third root is also 22, so p(x)=(x−1)(x−2)2=x3−5x2+8x−4,p(x)=(x-1)(x-2)^2=x^3-5x^2+8x-4, and a=−5a=-5. Why the others fail: choosing −7-7 would mean r=4r=4, but then p(3)=2(1)(−1)=−2p(3)=2(1)(-1)=-2, not 22. Choosing −6-6 would mean r=3r=3, giving p(3)=0p(3)=0. Choosing −4-4 would mean r=1r=1, giving p(3)=2(2)(1)=4p(3)=2(2)(1)=4, not 22. Each of these mistakes comes from misidentifying the third root or plugging in incorrectly. Study tip: whenever a polynomial is monic and has given roots, write it in factored form first. Then use the remaining condition to solve for the unknown root before expanding. This avoids guessing among coefficients.

Question 10

Which statement correctly describes the real solutions of (x−1)2(x+2)3(x2+1)=0(x-1)^2(x+2)^3(x^2+1)=0?

  1. There are 2 distinct real solutions and 5 real solutions counting multiplicity. (correct answer)
  2. There are 5 distinct real solutions and 5 real solutions counting multiplicity.
  3. There are 2 distinct real solutions and 2 real solutions counting multiplicity.
  4. There are 7 distinct real solutions and 7 real solutions counting multiplicity.
Explanation: When you see a factored polynomial equation like this, your first move is to identify the roots from each factor and then check multiplicities. Also remember that factors like x2+1x^2+1 have no real roots, since x2=−1x^2=-1 has no real solution. Here, the factor (x−1)2(x-1)^2 gives the real root x=1x=1, and the exponent 22 means it contributes 22 to the real-solution count with multiplicity. The factor (x+2)3(x+2)^3 gives the real root x=−2x=-2, contributing 33 with multiplicity. So there are 22 distinct real solutions: x=1x=1 and x=−2x=-2. Counting multiplicity, the total is 2+3=52+3=5 real solutions. The statement claiming "5 distinct real solutions and 5 real solutions counting multiplicity" is wrong because there are only two distinct real roots; the 55 is the multiplicity count, not the number of distinct roots. The statement claiming "2 distinct real solutions and 2 real solutions counting multiplicity" misses the multiplicities entirely—it counts each distinct root only once. The statement claiming "7 distinct real solutions and 7 real solutions counting multiplicity" confuses the total degree of the polynomial with real solutions; the x2+1x^2+1 factor contributes nonreal solutions ±i\pm i, which do not count as real solutions. Strategy: when counting real solutions, first list the real roots from each factor, then add their multiplicities. Ignore nonreal factors like x2+1x^2+1 for real-solution counts.

Question 11

What is the sum of the squares of all distinct real solutions of x3−3x2−4x+12=0x^3 - 3x^2 - 4x + 12 = 0?

  1. 33
  2. 99
  3. 1717 (correct answer)
  4. 2525
Explanation: Whenever you see a cubic like this, try factoring by grouping before doing anything else. Group the first two terms and the last two terms: x3−3x2−4x+12=x2(x−3)−4(x−3)=(x2−4)(x−3)=(x−2)(x+2)(x−3).x^3-3x^2-4x+12=x^2(x-3)-4(x-3)=(x^2-4)(x-3)=(x-2)(x+2)(x-3). So the distinct real solutions are x=−2x=-2, x=2x=2, and x=3x=3. The sum of their squares is: (−2)2+22+32=4+4+9=17.(-2)^2+2^2+3^2=4+4+9=17. Thus 1717 is the correct choice. The other choices are traps from rushing. 33 is simply the sum of the roots, using Vieta's formula from the coefficient of x2x^2, but the question asks for the sum of their squares, not the sum itself. 99 is the square of that root-sum; it answers (∑roots)2\left(\sum roots\right)^2, not ∑roots2\sum roots^2. 2525 often comes from squaring coefficients such as (−3)2+(−4)2=25(-3)^2+(-4)^2=25 and treating coefficients as the solutions—but coefficients are not roots. A safe strategy: factor completely, list the actual distinct solutions, then square each one. If you prefer Vieta, remember: r12+r22+r32=(r1+r2+r3)2−2(r1r2+r1r3+r2r3).r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_1r_3+r_2r_3). But here, factoring by grouping is the fastest and clearest path to the answer.

Question 12

Suppose 22 and −1-1 are two of the solutions of x3+ax2−3x+b=0x^3+ax^2-3x+b=0. What is the remaining solution, counting multiplicity?

  1. −2-2
  2. −1-1 (correct answer)
  3. 11
  4. 22
Explanation: Whenever you see a cubic equation with given roots, think factor theorem: each solution rr corresponds to a factor (x−r)(x-r). Since 22 and −1-1 are solutions, the polynomial must be divisible by (x−2)(x+1)=x2−x−2.(x-2)(x+1)=x^2-x-2. Let the remaining solution be rr. Then x3+ax2−3x+b=(x2−x−2)(x−r).x^3+ax^2-3x+b=(x^2-x-2)(x-r). Expanding the right side: x3+(−r−1)x2+(r−2)x+2r.x^3+(-r-1)x^2+(r-2)x+2r. Now match the coefficient of xx: the original coefficient is −3-3,so r−2=−3  ⟹  r=−1.r-2=-3 \implies r=-1. Thus the remaining solution is −1-1. Indeed, x3−3x−2=(x−2)(x+1)2,x^3-3x-2=(x-2)(x+1)^2, so −1-1 is a double root, counted twice. Why are the others wrong? If the remaining solution were −2-2, then factoring would give (x−2)(x+1)(x+2)=x3+x2−4x−4(x-2)(x+1)(x+2)=x^3+x^2-4x-4, whose xx-coefficient is −4-4, not −3-3. If it were 11, the product would be (x−2)(x+1)(x−1)=x3−2x2−x+2(x-2)(x+1)(x-1)=x^3-2x^2-x+2, with xx-coefficient −1-1. If it were 22, you would have (x−2)2(x+1)=x3−3x2+0x+4(x-2)^2(x+1)=x^3-3x^2+0x+4, with xx-coefficient 00. None match. Study tip: Use Vieta's formulas: for x3+ax2+px+qx^3+ax^2+px+q, the sum of pairwise products of roots equals pp. Here p=−3p=-3,so with roots 2,−1,r2,-1,r, 2(−1)+2r+(−1)r=−2+r=−3  ⟹  r=−1.2(-1)+2r+(-1)r=-2+r=-3 \implies r=-1. And always remember: "counting multiplicity" means repeated roots count as separate solutions.