PSAT Math Quiz: Polynomial Equations

What this quiz covers

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

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.

Question 1

1. As $x\to\infty$, $q(x)\to\infty$
2. As $x\to-\infty$, $q(x)\to\infty$
3. As $x\to\infty$, $q(x)\to-\infty$ (correct answer)
4. Both ends go up

Explanation: To determine end behavior of $q(x)=-2(x-3)(x+1)^2$, we need the degree and leading coefficient. Expanding the highest degree terms: $-2 cdot x cdot x^2 = -2x^3$, so the degree is 3 and the leading coefficient is $-2$ (negative). For odd-degree polynomials with negative leading coefficients: as $x o infty$, $q(x) o -infty$, and as $x o -infty$, $q(x) o infty$. The key is recognizing that the negative coefficient flips the typical odd-degree behavior.

Question 2

1. $\{1,2,3\}$ (correct answer)
2. $\{-1,-2,-3\}$
3. $\{1,2,-3\}$
4. $\{0,2,3\}$

Explanation: The question asks for the set listing all zeros of f(x) = x³ - 6x² + 11x - 6, which are the x-intercepts. Using the rational root theorem, possible roots are ±1, 2, 3, 6; testing x=1 yields 1 - 6 + 11 - 6 = 0. Synthetic division divides f(x) by (x-1), resulting in x² - 5x + 6, which factors to (x-2)(x-3). Thus, the zeros are 1, 2, 3, and the factored form (x-1)(x-2)(x-3) shows the roots clearly. This form connects the algebraic structure to the graph's x-intercepts. Common errors include mistakes in synthetic division or overlooking a root during testing.

Question 3

1. $(x-3)(x+2)$
2. $(x+3)(x-2)$ (correct answer)
3. $(x+3)(x+2)$
4. $(x-3)(x-2)$

Explanation: If a polynomial has x-intercepts at $x=-3$ and $x=2$, and crosses at both points, then $(x-(-3))=(x+3)$ and $(x-2)$ are factors. The complete factorization must include $(x+3)(x-2)$ as a divisor. Be careful with signs: an x-intercept at $x=-3$ means the factor is $(x+3)$, not $(x-3)$. When given intercepts, always convert carefully to factor form.

Question 4

1. $0$
2. $1$
3. $2$ (correct answer)
4. $3$

Explanation: The question requires counting the real zeros of a polynomial based on its graph, which crosses the x-axis at x = -1 and x = 3 with no other intercepts visible. Each crossing indicates a real zero (at least multiplicity 1, odd), so there are two distinct real zeros, assuming no hidden multiplicities or touches elsewhere. This connects graph intercepts to the polynomial's real roots, where crossings confirm real solutions to p(x) = 0. Errors might involve assuming extra zeros off the graph or counting multiplicities without evidence, leading to higher counts like 3. If the graph showed touches, those would still count as real but even multiplicity. In graph-based questions, tally distinct x-intercepts as the number of real zeros, verifying if any indicate higher multiplicity through bounce or flatness.

Question 5

1. $(x+1)(x-2)(x-3)$
2. $(x-1)(x-2)(x-3)$ (correct answer)
3. $(x-1)(x^2-5x+6)$
4. $(x-1)(x-2)(x+3)$

Explanation: We're given that $g(x)=x^3-6x^2+11x-6$ has zeros at $x=1,2,3$, and we need to write it in factored form. If a polynomial has zeros at these values, then $(x-1)$, $(x-2)$, and $(x-3)$ are factors. Therefore, $g(x)=(x-1)(x-2)(x-3)$. We can verify by expanding: $(x-1)(x-2)=x^2-3x+2$, then $(x^2-3x+2)(x-3)=x^3-3x^2-3x^2+9x+2x-6=x^3-6x^2+11x-6$. A common sign error is writing $(x+1)$ instead of $(x-1)$ for a zero at $x=1$. Remember: if $a$ is a zero, then $(x-a)$ is the corresponding factor.

Question 6

1. $(x+2)(x-1)$
2. $(x+2)(x-1)^2$ (correct answer)
3. $(x+2)^2(x-1)$
4. $-(x+2)(x-1)^2$

Explanation: The question asks for the least-degree polynomial with positive leading coefficient that crosses the x-axis at x=-2 and touches it at x=1. Crossing indicates odd multiplicity (at least 1) at x=-2, while touching indicates even multiplicity (at least 2) at x=1. For least degree, use multiplicity 1 at x=-2 and 2 at x=1, giving (x+2)(x-1)² with degree 3 and positive x³ leading term. This form links multiplicities to graph behavior: odd for crossing, even for touching. A common error is swapping the multiplicities, as in choice C. A test-taking strategy is to confirm end behavior from the leading term and match described root behaviors.

Question 7

1. $2x^3+5x^2-22x+15$ (correct answer)
2. $2x^3+8x^2-10x-3$
3. $2x^3+5x^2-23x+15$
4. $2x^3-5x^2-22x+15$

Explanation: The question asks for the expanded standard form of the product (2x - 3)(x² + 4x - 5) after multiplying and combining like terms. To solve, distribute each term in the first binomial to the trinomial: multiply 2x by x² to get 2x³, by 4x to get 8x², and by -5 to get -10x; then multiply -3 by x² to get -3x², by 4x to get -12x, and by -5 to get 15. Combine like terms: 2x³ + (8x² - 3x²) + (-10x - 12x) + 15 simplifies to 2x³ + 5x² - 22x + 15. A common error is mishandling the signs during distribution, such as forgetting the negative from -3, which could lead to incorrect coefficients like in choices B or C. Another mistake might be subtracting instead of adding coefficients, as seen in choice D's negative x² term. When expanding polynomials, always double-check the distribution of negative signs to ensure accurate combination of like terms.

Question 8

1. $(x+1)(x-2)(x-4)$
2. $-(x+1)(x-2)(x-4)$ (correct answer)
3. $-(x-1)(x-2)(x-4)$
4. $(x+1)(x-2)^2(x-4)$

Explanation: The question asks for a possible factored form of a polynomial that rises to the left, falls to the right, and crosses the x-axis at x=-1, 2, 4. This end behavior indicates an odd degree with negative leading coefficient. With three distinct crossings, the least-degree form is degree 3 with multiplicity 1 at each root: -(x+1)(x-2)(x-4), yielding a -x³ leading term matching the behavior. The factored form connects the roots and leading sign to the graph's intercepts and ends. Common errors include choosing a positive leading coefficient or incorrect roots, like x=1 in choice C. A test-taking strategy is to determine the leading coefficient's sign from end behavior and ensure factors align with given roots.

Question 9

1. $\{-1,0,4\}$
2. $\{-1,-1,0,4\}$ (correct answer)
3. $\{-4,0,1,1\}$
4. $\{-1,0,4,4\}$

Explanation: The question seeks the complete set of zeros for f(x) = x(x - 4)(x + 1)², including multiplicities, as repeated factors indicate multiple instances of the same zero. The zeros are found from each factor: x = 0 from the first (multiplicity 1), x = 4 from the second (multiplicity 1), and x = -1 from the third (multiplicity 2, since squared). Thus, the full list is {-1, -1, 0, 4}, reflecting how multiplicity affects the polynomial's behavior, like touching the x-axis at repeated zeros. Errors often occur by omitting multiplicities, such as listing only distinct zeros like in choice A, or miscounting the repetition as in choices C or D. Another mistake is inverting signs, leading to wrong zeros like -4 or 1. In multiple-choice, verify by plugging zeros back into the factored form to confirm they satisfy f(x) = 0 with the given multiplicities.

Question 10

1. $x^2+x-4$
2. $(x-4)(x+5)$ (correct answer)
3. $(x-4)(x+1)$
4. $(x+3)(x+2)$

Explanation: We need to factor the expression $(x-4)(x+3)+2(x-4)$. Notice that $(x-4)$ appears in both terms, making it a common factor. Factoring out $(x-4)$ gives us $(x-4)[(x+3)+2] = (x-4)(x+5)$. A common error is to distribute first instead of recognizing the common factor, which makes the problem unnecessarily complex. When you see repeated factors, always look to factor first before expanding.

Question 11

1. $2x^3+x^2-23x+20$ (correct answer)
2. $2x^3+x^2-22x+20$
3. $2x^3+x^2-23x-20$
4. $2x^3-x^2-23x+20$

Explanation: To find the product $(2x-5)(x^2+3x-4)$, we distribute each term in the first polynomial to every term in the second. This gives us $2x(x^2+3x-4) - 5(x^2+3x-4) = 2x^3+6x^2-8x-5x^2-15x+20$. Combining like terms: $2x^3+(6x^2-5x^2)+(-8x-15x)+20 = 2x^3+x^2-23x+20$. The most common error is sign mistakes when distributing the $-5$, forgetting that $-5 imes -4 = +20$. Always double-check signs when distributing negative terms.

Question 12

1. $(x-9)(x+2)$ (correct answer)
2. $(x-6)(x-3)$
3. $(x+9)(x-2)$
4. $(x+6)(x-3)$

Explanation: To factor $x^2-7x-18$, we need two numbers that multiply to $-18$ and add to $-7$. Testing factor pairs of $-18$: $(-9)(2) = -18$ and $-9 + 2 = -7$, which works! So $x^2-7x-18 = (x-9)(x+2)$. A common error is finding factors that multiply correctly but forgetting to check if they add to the middle coefficient. When the constant term is negative, one factor must be positive and one negative.

Question 13

1. $(x+2)(x-3)$
2. $(x+2)^2(x-3)$
3. $(x+2)(x-3)^2$ (correct answer)
4. $(x+2)^2(x-3)^2$

Explanation: When a polynomial crosses the x-axis at a point, that zero has odd multiplicity; when it touches and bounces, the zero has even multiplicity. Since $g(x)$ crosses at $x=-2$ (multiplicity 1) and bounces at $x=3$ (multiplicity 2), the factored form is $(x+2)(x-3)^2$. This gives the least possible degree of 3 with a positive leading coefficient. Students often confuse crossing with bouncing behavior or forget that bouncing requires even multiplicity.

Question 14

1. 1; $x=3$
2. 2; $x=-4,3$ (correct answer)
3. 3; $x=-4,3,0$
4. 2; $x=-4,6$

Explanation: The question asks for the number of distinct x-intercepts of k(x) = (x-3)²(x+4) and their x-values. The factored form shows zeros at x=3 with multiplicity 2 and x=-4 with multiplicity 1. Despite the even multiplicity at x=3 (touching), it is still an x-intercept, resulting in two distinct intercepts at x=-4 and x=3. This algebraic form highlights distinct roots regardless of multiplicity for intercept counting. A key error is assuming even multiplicity does not count as a distinct intercept. A test-taking strategy is to list unique root values from the factors, focusing on distinct x-values for intercepts.

Question 15

1. 1
2. 2 (correct answer)
3. 3
4. 4

Explanation: The polynomial $k(x)=(x-1)^2(x+4)$ has zeros where each factor equals zero. Setting $(x-1)^2=0$ gives $x=1$ (with multiplicity 2), and $(x+4)=0$ gives $x=-4$. Though $x=1$ appears with multiplicity 2, it's still just one distinct zero. Therefore, $k(x)$ has exactly 2 distinct real zeros: $x=1$ and $x=-4$. Don't confuse multiplicity with the number of distinct zeros—a repeated root counts as one distinct zero.

Question 16

1. $-(t-7)(t+1)$ (correct answer)
2. $-(t-1)(t-7)$
3. $(t-7)(t+1)$
4. $-(t+7)(t-1)$

Explanation: The question seeks the factored form of h(t) = -t² + 6t + 7 that makes the zeros easiest to identify. Factor out the negative: h(t) = -(t² - 6t - 7). The quadratic t² - 6t - 7 factors to (t-7)(t+1), as the factors of -7 summing to -6 are 1 and -7. Thus, h(t) = -(t-7)(t+1), revealing zeros at t=7 and t=-1 directly from the factors. This factored form emphasizes how roots correspond to when each linear factor is zero. A key error is selecting incorrect factor pairs or mishandling the negative sign. A test-taking strategy is to expand the choices back to standard form to verify the match.

Question 17

1. $(x-4)(x+4)$
2. $(x^2-4)(x^2+4)$
3. $(x-2)(x+2)(x^2+4)$ (correct answer)
4. $(x-2)^2(x+2)^2$

Explanation: The question seeks the complete integer factorization of g(x) = x⁴ - 16, noting that (x² - 4)(x² + 4) is partial and difference of squares applies further. Apply it to x⁴ - 16 = (x²)² - 4² = (x² - 4)(x² + 4), then to x² - 4 = (x - 2)(x + 2), yielding (x - 2)(x + 2)(x² + 4). This shows how repeated difference of squares breaks down higher powers, with x² + 4 irreducible over integers. Errors include stopping at the partial factorization (B) or over-factoring into non-integers. Choice D squares factors incorrectly, while A is quadratic only. In tests, repeatedly apply patterns like difference of squares until no further integer factors remain to ensure completeness.

Question 18

1. $(x-9)(x+2)$ (correct answer)
2. $(x-6)(x+3)$
3. $(x+9)(x-2)$
4. $(x-3)(x+6)$

Explanation: The question requires factoring x² - 7x - 18 to find its zeros, ensuring the factors' constants multiply to -18 and their cross terms add to -7x. Test pairs: -9 and 2 work since -9 * 2 = -18 and -9x + 2x = -7x, giving (x - 9)(x + 2). This quadratic form reveals zeros at x = 9 and x = -2, linking factorization to root-finding. Common errors include pairs like -6 and 3 (adding to -3x, as in B) or sign swaps like 9 and -2 (adding to 7x, as in C). Choice D's -3 and 6 add to 3x, missing the negative. When factoring quadratics, list factor pairs of the constant and check the middle term sum to avoid mismatches.

Question 19

1. $(x+2)(x-3)^2$ (correct answer)
2. $(x+2)^2(x-3)$
3. $-(x+2)(x-3)^2$
4. $(x+2)(x-3)$

Explanation: The question involves identifying a cubic polynomial with x-intercepts at x = -2 (crossing) and x = 3 (touching, or bouncing), assuming a positive leading coefficient. For a touch at x = 3, the factor (x - 3) must have even multiplicity (typically 2 for cubics), indicating a repeated root, while x = -2 has multiplicity 1 for crossing. Thus, the form is (x + 2)(x - 3)², where the quadratic factor at 3 causes the bounce, and the overall positive leading coefficient matches the upward end behavior for odd degree. Common errors include swapping multiplicities, like in choice B, which would bounce at -2 instead, or adding a negative sign as in choice C, flipping the leading coefficient. Omitting multiplicity entirely, as in D, would make both cross. When analyzing graphs, link intercept behavior to factor multiplicities: even for touching, odd for crossing, to build the factored form efficiently.

Question 20

1. $(x+3)(x-1)(x-2)^2$ (correct answer)
2. $(x+3)^2(x-1)(x-2)$
3. $(x-3)(x+1)(x-2)^2$
4. $-(x+3)(x-1)(x-2)^2$

Explanation: The question asks for a possible factored form of a polynomial whose graph crosses the x-axis at x = -3 and x = 1, touches at x = 2, with a positive leading coefficient. Crossing indicates odd multiplicity (usually 1), while touching suggests even multiplicity (like 2), so factors are (x + 3)(x - 1)(x - 2)², resulting in degree 4 with positive leading term. This form connects algebraic structure to graph behavior, where even multiplicity at 2 causes the bounce, and odd at others cause crossings. Errors might involve misplaced multiplicities, like B putting even at -3, or sign errors in factors as in C. Choice D's negative sign would invert end behavior, unsuitable for positive leading. For such problems, sketch a quick graph from choices to match described intercepts and behaviors.