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

Explanation: End behavior that goes down on the left and up on the right tells you two things at once: the degree must be odd, so the two ends head in opposite directions, and the leading coefficient must be positive, so the right end rises. The leading term 2x^5 has both features, which is why it fits. The term -2x^5 has odd degree, so its ends do go opposite ways, but the negative coefficient flips them, giving up on the left and down on the right, the reverse of what is described. The term 2x^4 has even degree, so both ends rise together, and the term -2x^4 has even degree with a negative coefficient, so both ends fall together. Neither even-degree option can ever produce opposite behavior at the two ends.

Question 2

1. $x^3-2x^2-13x+12$
2. $x^3+2x^2-13x-12$
3. $x^3+2x^2-11x-12$
4. $x^3-2x^2-11x+12$ (correct answer)

Explanation: We need to expand $g(x)=(x-1)(x+3)(x-4)$ into standard form. First, multiply any two factors: $(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$. Now multiply this result by the third factor: $(x^2 + 2x - 3)(x-4) = x^3 - 4x^2 + 2x^2 - 8x - 3x + 12$. Combining like terms: $x^3 + (-4+2)x^2 + (-8-3)x + 12 = x^3 - 2x^2 - 11x + 12$. The most common error is sign mistakes when distributing negative terms. When multiplying three linear factors, work systematically by first multiplying two factors, then multiplying that result by the third.

Question 3

1. $-\sqrt{97}$
2. $-7$ (correct answer)
3. $7$
4. $\sqrt{97}$

Explanation: To find the solutions of a quadratic or polynomial, begin by setting the equation equal to zero. Therefore $x^3+7x^2=-12x$ becomes: $x^3+7x^2 +12x=0$ From here, you can factor out an $x$: $x(x^2+7x +12)=0$ That means that one solution is $x=0$. You can then factor the quadratic in parentheses to: $x^2 + 7x + 12 = (x+6)(x+1)$ So we now have: $x(x+6)(x+1)=0$ Remember in quadratic problems to always set each term to zero to find your solutions, so the solution set here is: $x = 0, -6, -1$ Therefore the sum of solutions is -7.

Question 4

1. $k = 2$
2. $k = 3$
3. $k = 4$ (correct answer)
4. $k = 6$

Explanation: While this problem is certainly doable with pure algebra, many complicated polynomial questions like this one are much more quickly solved by picking numbers. There are lots of x terms in this problem, but the problem specifically states that the equation must hold true for ALL values of x so you can turn a lot of complicated algebra into some rather quick arithmetic by picking an easy value for x. If you pick an easy number to work with like $x=1$ you will have very small numbers to calculate. Doing so will leave you with: $(k + 2)(3 -1) = 12 + 2 - 2$ $2(k+2) = 12$ $k+2 = 6$ $k = 4$ Pro tip: when a question says that an equation holds true for all values of a variable, you can then choose a simple number to stand in for that variable and turn several steps of algebra into a few steps of basic arithmetic.

Question 5

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

Explanation: We need to identify which factored form matches a polynomial that crosses at $x=-2$ and touches (bounces) at $x=3$. When a polynomial crosses the x-axis at a zero, that factor has odd multiplicity (usually 1); when it touches and bounces, the factor has even multiplicity (usually 2). Since it crosses at $x=-2$, we need factor $(x+2)$ with odd power; since it touches at $x=3$, we need factor $(x-3)$ with even power. With positive leading coefficient, the polynomial is $(x+2)(x-3)^2$. The key insight is that multiplicity determines crossing versus touching behavior. Remember: odd multiplicity means the graph crosses through; even multiplicity means it touches and turns around.

Question 6

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

Explanation: We need a cubic polynomial with x-intercepts at $x=-4$, $x=1$, and $x=2$ with leading coefficient 2. X-intercepts correspond to zeros, so the polynomial has factors $(x-(-4))=(x+4)$, $(x-1)$, and $(x-2)$. With leading coefficient 2, the polynomial is $2(x+4)(x-1)(x-2)$. A common error is writing $(x+4)$ as $(x-4)$ when the zero is at $x=-4$, or forgetting to include the leading coefficient. Remember: a zero at $x=a$ gives factor $(x-a)$, and the leading coefficient multiplies the entire factored form.

Question 7

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

Explanation: The graph has x-intercepts at $x=-4$, $x=0$, and $x=2$, with a negative leading coefficient. The factors corresponding to these zeros are $(x-(-4)) = (x+4)$, $(x-0) = x$, and $(x-2)$. With a negative leading coefficient, the polynomial is $f(x) = -(x+4)(x)(x-2)$. We can verify the sign by checking end behavior: as $x o infty$, this cubic with negative leading coefficient goes to $-infty$, which matches what we expect. The factored form can also be written as $-(x+4)x(x-2)$. When determining polynomial equations from graphs, use the x-intercepts to identify factors and the end behavior to determine the sign of the leading coefficient.

Question 8

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

Explanation: We need to expand $f(x)=(x^2-1)(x^2+3x)$ to standard form. Using the distributive property: $(x^2-1)(x^2+3x)=x^2(x^2+3x)-1(x^2+3x)=x^4+3x^3-x^2-3x$. Breaking it down: $x^2·x^2=x^4$, $x^2·3x=3x^3$, $(-1)·x^2=-x^2$, $(-1)·3x=-3x$. The expanded form is $x^4+3x^3-x^2-3x$. A common error is to expand only one binomial product or to forget the middle terms. When multiplying polynomials, ensure every term in the first polynomial multiplies every term in the second.

Question 9

1. $-2x^2-6x+16$ (correct answer)
2. $2x^2-6x+16$
3. $-2x^2+2x-2$
4. $2x^2+2x-2$

Explanation: We need to simplify $(3x^2-2x+7)-(5x^2+4x-9)$ by subtracting the second polynomial from the first. When subtracting, distribute the negative sign to every term in the second polynomial: $3x^2-2x+7-5x^2-4x+9$. Now combine like terms: $(3-5)x^2 + (-2-4)x + (7+9) = -2x^2 - 6x + 16$. The most common error is forgetting to change the sign of every term in the subtracted polynomial, especially the constant term which becomes +9 not -9. When subtracting polynomials, rewrite the subtraction as addition of the opposite before combining like terms.

Question 10

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

Explanation: The question asks for the product of $(x + 2)$ and $(x - 3)$. Using the FOIL method: First terms give $x \cdot x = x^2$, Outer terms give $x \cdot (-3) = -3x$, Inner terms give $2 \cdot x = 2x$, and Last terms give $2 \cdot (-3) = -6$. Combining these: $x^2 - 3x + 2x - 6 = x^2 - x - 6$. A common error is incorrectly combining the middle terms or making sign errors during multiplication. Always double-check your arithmetic when expanding polynomial products, especially with negative coefficients.

Question 11

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

Explanation: The question asks for $f(2)$ when $f(x) = x^2 - 4x + 4$. Substitute $x = 2$: $f(2) = (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0$. Notice that this polynomial can be factored as $(x-2)^2$, which confirms that $x = 2$ is a zero (making $f(2) = 0$). A common error is making arithmetic mistakes during substitution or incorrectly handling negative signs. When evaluating polynomial functions, substitute carefully and double-check your arithmetic, especially with multiple operations.

Question 12

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

Explanation: The question asks for the number of distinct real zeros of $r(x) = (x^2 - 9)(x^2 + 4x + 4)$. Factor: $x^2 - 9 = (x - 3)(x + 3)$, $x^2 + 4x + 4 = (x + 2)^2$. Zeros are $x = 3$, $-3$, and $-2$ (repeated). Distinct real zeros: 3. This factorization reveals roots and multiplicities, linking algebraic form to unique solutions despite repetition. A common error counts the repeated zero as two distinct or includes complex ones. Some miscount by not fully factoring. Strategically, solve each factor separately and use a set to count uniques.

Question 13

1. Both ends rise to $\infty$
2. Left falls, right rises
3. Left rises, right falls
4. Both ends fall to $-\infty$ (correct answer)

Explanation: The question asks for the end behavior of h(x) = −3x⁴ + 2x² − 7 as x approaches ±∞. The leading term −3x⁴ dominates, with even degree 4 and negative coefficient, so both ends fall to −∞. Lower terms like 2x² become negligible for large |x|. This illustrates how degree and leading sign dictate long-term trends, connecting polynomial form to asymptotic behavior. A common mistake is treating the degree as odd or ignoring the negative sign, predicting rising ends. Some focus on positive terms only. In tests, isolate the leading term to quickly determine end behavior without full graphing.

Question 14

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

Explanation: This question asks us to identify the degree of a polynomial, which is the highest power of the variable. In the polynomial 4x³ - 2x² + 7x - 5, we examine each term's exponent: the x³ term has degree 3, the x² term has degree 2, the x term has degree 1, and the constant term has degree 0. The degree of the entire polynomial is the highest of these individual degrees, which is 3. A common mistake is confusing the degree with the number of terms or the leading coefficient. To find polynomial degree, always look for the term with the largest exponent on the variable.

Question 15

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

Explanation: The question asks which expression is a factor of $x^3 - 4x^2 + 4x$. First, factor out the common factor of $x$: $x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x-2)^2$. The complete factorization shows that the factors are $x$, $(x-2)$, and $(x-2)$ again. Among the choices, only $(x)$ appears as a factor. A common error is not recognizing the need to factor out the greatest common factor first. When looking for factors, always start by factoring out any common terms before attempting other factoring methods.

Question 16

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

Explanation: We need to find an integer zero of $t(x)=x^3-6x^2+11x-6$. By the Rational Root Theorem, possible integer zeros are factors of the constant term (-6): ±1, ±2, ±3, ±6. Testing $x=1$: $t(1)=1^3-6(1)^2+11(1)-6=1-6+11-6=0$. Since $t(1)=0$, $x=1$ is a zero. We can verify by synthetic division or by checking that $(x-1)$ is a factor. Common errors include arithmetic mistakes when evaluating or testing the wrong values first. When finding integer zeros, test small values like ±1, ±2 first as they're most common.

Question 17

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

Explanation: The question asks for the remaining zero of $p(x) = x^3 + a x^2 + b x + 6$ with zeros at $-2$ and $1$, and $p(0) = 6$. Since it's cubic and monic, factor as $(x + 2)(x - 1)(x - k)$. The constant term is $( - (-2) ) ( -1 ) ( -k ) = - (2) (-1) (-k)$ wait, better: constant = $-$(product of roots)$$ = 6, so product = $-6$; $(-2)(1)k = -2k = -6$, k=3. Alternatively, expand $(x + 2)(x - 1)(x - 3) = x^3 - 2x^2 - 5x + 6$, but wait, constant 6 yes. This uses the factor theorem and root product, connecting coefficients to zeros. A common error assumes the third zero is 0 due to the constant. Some misapply Vieta's formulas with signs. Strategically, use the product of roots for quick constant-based solving.

Question 18

1. 4
2. 5 (correct answer)
3. 6
4. 7

Explanation: The question seeks the degree of the polynomial p(x) = (x³ − 4x)(2x² + 7). To determine the degree, identify the highest power terms from each factor: x³ from the first and 2x² from the second, multiplying to give 2x⁵ as the leading term. The other terms produce lower degrees, such as x³ · 7 = 7x³, −4x · 2x² = −8x³, and −4x · 7 = −28x, confirming the highest degree is 5. This process highlights how the degree of a product is the sum of the degrees of the factors, linking algebraic structure to the polynomial's overall order. A common mistake is adding the visible exponents instead of considering the product of leading terms. Some might overlook that constants don't affect the degree. For test-taking, remember that degree questions often focus on leading terms, so prioritize those in expansions.

Question 19

1. $(x+2)(x-10)$
2. $(x-2)(x-10)$
3. $(x-4)(x-5)$ (correct answer)
4. $(x+4)(x+5)$

Explanation: We need to factor $x^2-9x+20$ completely over the integers. For a quadratic $x^2+bx+c$, we look for two numbers that multiply to $c=20$ and add to $b=-9$. The factor pairs of 20 are: (1,20), (2,10), (4,5), and their negatives. We need the pair that adds to -9, which is (-4,-5) since $(-4)+(-5)=-9$ and $(-4)×(-5)=20$. Therefore, $x^2-9x+20=(x-4)(x-5)$. Common errors include using the wrong signs or finding pairs that multiply to 20 but don't add to -9. Always verify by expanding your factored form back to the original.

Question 20

1. $2x(x^3-4x^2+3x)$
2. $2x^2(x^2-8x+6)$
3. $2x^2(x^2-4x+3)$ (correct answer)
4. $x^2(2x^2-8x+6)$

Explanation: To factor $m(x)=2x^4-8x^3+6x^2$, we first identify the greatest common factor (GCF) of all terms. Each term has a factor of $x^2$: $2x^4 = 2x^2 \cdot x^2$, $-8x^3 = -8x^2 \cdot x$, $6x^2 = 6x^2 \cdot 1$. The coefficients 2, -8, and 6 have GCF of 2. Therefore, the overall GCF is $2x^2$. Factoring this out: $m(x) = 2x^2(x^2 - 4x + 3)$. We can verify by expanding: $2x^2(x^2 - 4x + 3) = 2x^4 - 8x^3 + 6x^2$ ✓. Always factor out the GCF first when working with polynomials, as it simplifies further factoring steps.