Factor Quadratics to Find Zeros - Algebra 2

The expression $2x^2 - 8x - 42$ can be factored to reveal its zeros. Which of the following correctly identifies all zeros of the function $f(x) = 2x^2 - 8x - 42$ ?
Answer: $x = -3$ and $x = 7$ First factor out 2: 2x² - 8x - 42 = 2(x² - 4x - 21). Then factor the quadratic: x² - 4x - 21 = (x + 3)(x - 7). So f(x) = 2(x + 3)(x - 7), giving zeros at x = -3 and x = 7. Choice B would result from sign errors in factoring. Choice C represents the zeros of x² - 8x - 84. Choice D also comes from sign confusion.
A quadratic function $y = ax^2 + bx + c$ has zeros at $x = m$ and $x = n$ , where $m < n$ . If the factored form is $y = 2(x + 3)(x - 5)$ , which statement about the zeros and their relationship is correct?
Answer: The zeros are $x = -3$ and $x = 5$ , and they are symmetric about $x = 1$ When you encounter a quadratic function in factored form like $y = 2(x + 3)(x - 5)$ , finding the zeros is about determining where the function equals zero. This happens when either factor equals zero, since any number times zero equals zero. To find the zeros, set each factor equal to zero: $x + 3 = 0$ gives $x = -3$ , and $x - 5 = 0$ gives $x = 5$ . The coefficient 2 doesn't affect the zeros' locations. Now let's check their symmetry: the axis of symmetry for any parabola lies exactly halfway between its zeros. The midpoint between $x = -3$ and $x = 5$ is $\frac{-3 + 5}{2} = 1$ , so the zeros are symmetric about $x = 1$ . This confirms answer choice C. Looking at the wrong answers: Choice A correctly identifies the zeros as $x = -3$ and $x = 5$ , but their product is $(-3)(5) = -15$ , not $-15$ as stated—wait, that's actually correct, but the symmetry information is missing. Choice B incorrectly finds the zeros as $x = 3$ and $x = -5$ by forgetting that $(x + 3) = 0$ means $x = -3$ , not $x = 3$ . Choice D makes the same sign error and incorrectly calculates the product as positive 15. Remember this key strategy: when finding zeros from factored form, set each factor equal to zero and solve. The zeros are always symmetric about the axis of symmetry, which you can find by averaging the two zero values.
A quadratic function $f(x) = x^2 - 6x + k$ has zeros at $x = 2$ and $x = 4$ . What is the factored form of this expression?
Answer: $(x - 2)(x - 4)$ Since the zeros are at x = 2 and x = 4, the factored form is (x - 2)(x - 4). We can verify: (x - 2)(x - 4) = x² - 6x + 8, and indeed k = 8. Choice B gives zeros at x = -2 and x = -4. Choice C gives zeros at x = 2 and x = -4. Choice D gives zeros at x = -2 and x = 4.
A quadratic expression in the form $x^2 + bx + c$ has been factored as $(x + 4)(x - 7)$ . What is the sum of the coefficients $b + c$ ?
Answer: $b + c = -31$ When you encounter a factored quadratic that needs to be expanded back to standard form, you're working with the reverse of the factoring process. The key is systematically applying the distributive property (FOIL) to find the original coefficients. To expand $(x + 4)(x - 7)$ , multiply each term in the first binomial by each term in the second: - First terms: $x \cdot x = x^2$ - Outer terms: $x \cdot (-7) = -7x$ - Inner terms: $4 \cdot x = 4x$ - Last terms: $4 \cdot (-7) = -28$ Combining these gives: $x^2 - 7x + 4x - 28 = x^2 - 3x - 28$ Therefore, $b = -3$ and $c = -28$ , so $b + c = -3 + (-28) = -31$ . Looking at the wrong answers: Choice A gives $b + c = 31$ , which incorrectly makes both coefficients positive. This suggests adding the absolute values rather than considering the actual signs. Choice B yields $b + c = -25$ , likely from incorrectly calculating $c = -21$ instead of $c = -28$ . Choice C gives $b + c = 25$ , which combines the sign error from A with a calculation mistake. The most reliable strategy here is to always expand the factored form completely before identifying coefficients. Don't try to shortcut by guessing the relationship between the factors and coefficients—work through the multiplication step-by-step to avoid sign errors and computational mistakes.
Use structure to factor $q(x)=3x^2-12$ and then find the zeros (the x-values where the graph crosses the x-axis).
Answer: $q(x)=3(x-2)(x+2)$ ; zeros: $x=2,-2$ ; x-intercepts: $(2,0),(-2,0)$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For q(x) = 3x² - 12, factor out 3: 3(x² - 4) = 3(x - 2)(x + 2); zeros x = 2 and x = -2, x-intercepts (2, 0) and (-2, 0)—common factor first! Choice A correctly factors and identifies the zeros. Choice D mistakes it for a perfect square, but x² - 4 isn't squared—check inside after factoring out the constant. Always factor out GCF first, then difference of squares; symmetric zeros are a clue—you're mastering this!
A profit function is $P(x)=x^2-8x+15$ , where zeros represent break-even points. Factor $P(x)$ and find the break-even x-values.
Answer: $P(x)=(x-3)(x-5)$ ; break-even x-values: $x=3,5$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations; here, zeros are break-even points. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For P(x) = x² - 8x + 15, find numbers multiplying to 15 adding to -8: -3 and -5, so (x - 3)(x - 5); break-even at x = 3 and x = 5—positive zeros make sense for profits! Choice A correctly factors and identifies the break-even points. Choice B flips signs, giving positive factors that expand to x² + 8x + 15—signs are key, opposite for negative sum! Apply this to real-world models: factors reveal intercepts directly; practice sign relationships to avoid common errors—you're building strong skills!
A quadratic expression $ax^2 + bx + c$ has zeros at $x = -\frac{2}{3}$ and $x = \frac{1}{4}$ . If $a = 12$ , what is the value of $c$ ?
Answer: $c = -2$ With zeros at x = -2/3 and x = 1/4, the factored form is a(x + 2/3)(x - 1/4). With a = 12: 12(x + 2/3)(x - 1/4) = 12(x² + 2x/3 - x/4 - 1/6) = 12(x² + 5x/12 - 1/6) = 12x² + 5x - 2. Therefore c = -2. Choice B has the wrong sign. Choice C would result from calculation errors with the fractions. Choice D combines both sign and calculation errors.
To sketch the parabola $y=x^2-25$ , first factor to reveal the zeros. What are the zeros and corresponding x-intercepts?
Answer: Zeros: $x=5,-5$ ; x-intercepts: $(5,0),(-5,0)$ This question tests your ability to recognize and factor a difference of squares, then use the factored form to find zeros and x-intercepts for sketching the parabola. Factoring a quadratic like x² - c (difference of squares) into (x - √c)(x + √c) reveals the zeros immediately via the zero product property. Setting each factor to zero gives the roots, which are symmetric around zero for this form. These zeros are the x-intercepts, key points for graphing the parabola that opens upward from the origin. Great job spotting this special case—it simplifies everything! To factor x² - 25: recognize it as (x)² - (5)² = (x - 5)(x + 5). Set to zero: x - 5 = 0 so x = 5, or x + 5 = 0 so x = -5. Zeros are x = 5 and x = -5, with x-intercepts (5, 0) and (-5, 0). This symmetry makes sense since the vertex is at x=0. Choice B correctly identifies the zeros and intercepts from the proper factoring. Choice A mistakenly uses ±25, perhaps confusing with x² - 625 = (x - 25)(x + 25), but always check by expanding back: (x - 5)(x + 5) = x² - 25, perfect match! For difference of squares a² - b² = (a - b)(a + b), zeros at x = b and x = -b. Example: x² - 16 = (x - 4)(x + 4), zeros ±4. If it's x² + c, no real factors, but here it's minus. Remember to factor out common factors first if present, like 4x² - 9 = (2x - 3)(2x + 3). Keep practicing these patterns—they speed up solving!
Show that $x=2$ and $x=-3$ are zeros of $f(x)=x^2+x-6$ by factoring $f(x)$ . Which factored form correctly reveals the zeros?
Answer: $f(x)=(x-2)(x+3)$ This question verifies given zeros by factoring, confirming they satisfy f(x)=0—a great check for understanding roots. Factoring should include the given roots as (x - 2)(x - (-3)) = (x-2)(x+3). This matches the quadratic. Zeros are where factors are zero. Nice verification practice! For x² + x - 6: numbers multiply -6, add 1: 3 and -2. (x + 3)(x - 2). Plug in x=2: (2)+3=5, (2)-2=0, product 0. x=-3: -3+3=0, -3-2=-5, product 0. Yes! Choice B correctly shows the factored form revealing the zeros. Choice A swaps signs: (x+2)(x-3)=x² -x -6, wrong middle term; test by expanding! To write from zeros r,s: f(x)=(x-r)(x-s). Example: zeros 1,4: (x-1)(x-4)=x²-5x+4. Verify by plugging back. You're mastering this!
A parabola has x-intercepts at $x = 1 - 2\sqrt{3}$ and $x = 1 + 2\sqrt{3}$ . Which factored form represents a quadratic expression with these zeros?
Answer: $(x - 1 + 2\sqrt{3})(x - 1 - 2\sqrt{3})$ When you're given the x-intercepts of a parabola, you need to work backwards to find the factored form. The key insight is understanding the relationship between zeros and factors: if $x = a$ is a zero, then $(x - a)$ is a factor. Given the x-intercepts $x = 1 - 2\sqrt{3}$ and $x = 1 + 2\sqrt{3}$ , the corresponding factors must be $(x - (1 - 2\sqrt{3}))$ and $(x - (1 + 2\sqrt{3}))$ . Simplifying these expressions: the first factor becomes $x - 1 + 2\sqrt{3}$ , and the second becomes $x - 1 - 2\sqrt{3}$ . This matches choice C exactly. Let's examine why the other options fail. Choice A, $(x - 1)^2 - 12$ , isn't even in factored form—it's a difference of squares that would expand differently. Choice B has the wrong signs: $(x + 1 - 2\sqrt{3})(x + 1 + 2\sqrt{3})$ would give zeros at $x = -1 + 2\sqrt{3}$ and $x = -1 - 2\sqrt{3}$ , not the given intercepts. Choice D, $(x + 1)^2 + 12$ , represents a sum of positive terms, which can never equal zero, meaning this quadratic has no real zeros at all. You can verify choice C by setting each factor to zero: $x - 1 + 2\sqrt{3} = 0$ gives $x = 1 - 2\sqrt{3}$ , and $x - 1 - 2\sqrt{3} = 0$ gives $x = 1 + 2\sqrt{3}$ —exactly our given intercepts. Study tip: Always remember that if $x = a$ is a zero, then $(x - a)$ is the corresponding factor. Pay careful attention to signs when converting between zeros and factors.
Factor $s(x)=4x^2-9$ to reveal its zeros. Then identify the x-intercepts of the graph of $y=s(x)$ .
Answer: $s(x)=(2x-3)(2x+3)$ ; zeros: $x=\tfrac{3}{2},-\tfrac{3}{2}$ ; x-intercepts: $(\tfrac{3}{2},0),(-\tfrac{3}{2},0)$ This question tests factoring difference of squares with coefficient—reveals zeros quickly. s(x) = (2x - 3)(2x + 3), zeros 3/2, -3/2. Intercepts: (3/2, 0), (-3/2, 0). Recognize 4x² - 9 = (2x)² - 3². Great pattern spotting! Choice B correct. Choice C thinks perfect square, but no middle term. Verify expansion! Strategy: a²x² - b² = (ax - b)(ax + b). Example: 9x² - 16 = (3x - 4)(3x + 4), zeros ±4/3.
Factor the quadratic function $f(x)=x^2+7x+12$ and use the factored form to find the zeros (and thus the x-intercepts) of the graph.
Answer: $f(x)=(x+3)(x+4)$ ; zeros: $x=-3,-4$ ; x-intercepts: $(-3,0),(-4,0)$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. To factor x² + 7x + 12 and find zeros: look for two numbers that multiply to 12 and add to 7 (3 and 4), so (x + 3)(x + 4); set to zero for x = -3 or x = -4, with x-intercepts (-3, 0) and (-4, 0)—factored form makes this straightforward! Choice A correctly factors the quadratic and identifies both zeros by setting each factor equal to zero and solving. Choice B has a sign error in the factors: using negative numbers for positive coefficients leads to incorrect factors like (x - 3)(x - 4), which expands to x² - 7x + 12, flipping the middle sign—always check by expanding! Factoring trinomial x² + bx + c: find two numbers that multiply to c and add to b, then form (x + first)(x + second); for zeros, remember the sign flips, as in (x + 3) = 0 gives x = -3—practice this to master it quickly!
To sketch $f(x)=x^2+2x-8$ , first factor it to find the zeros. Which choice gives the correct factored form and x-intercepts?
Answer: $f(x)=(x+4)(x-2)$ ; zeros: $x=-4,2$ ; x-intercepts: $(-4,0),(2,0)$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For f(x) = x² + 2x - 8, numbers multiply to -8, add to 2: 4 and -2, so (x + 4)(x - 2); zeros x = -4 and x = 2, x-intercepts (-4, 0) and (2, 0)—one positive, one negative! Choice A correctly factors and finds the zeros. Choice B swaps signs, expanding to x² - 2x - 8—close but middle term wrong; always FOIL to check! For mixed signs, choose pairs with opposite signs; this helps sketch parabolas quickly—keep going, you're doing great!
To sketch $f(x)=x^2+2x-8$ , first factor to find the zeros. Which option gives the correct factored form and zeros (x-intercepts)?
Answer: $f(x)=(x+4)(x-2)$ ; zeros: $x=-4,2$ ; x-intercepts: $(-4,0),(2,0)$ This question tests factoring to sketch graphs by finding zeros (x-intercepts)—fundamental for visualization. Factored form shows zeros directly via zero product property. X-intercepts help plot the parabola. Keep practicing! To factor $x^2 + 2x - 8$ : numbers multiply to -8, add to 2: 4 and -2. $(x + 4)(x - 2)$ . Zeros: $-4, 2$ . X-intercepts: $(-4, 0), (2, 0)$ . Excellent! Choice A correctly factors and lists zeros. Choice B swaps signs; $(x - 4)(x + 2) = x^2 - 2x - 8$ , wrong middle term. Check addition! Strategy: list factor pairs of c, find pair adding to b considering signs. Example: $x^2 - x - 12$ , pairs of -12 adding to -1: 3 and -4. $(x + 3)(x - 4)$ . Zeros: $-3, 4$ .
Use factoring to determine where the function equals zero: $g(x)=x^2-25$ . What are the zeros, and what x-intercepts does this give for the graph?
Answer: $g(x)=(x-5)(x+5)$ ; zeros: $x=5,-5$ ; x-intercepts: $(5,0),(-5,0)$ This question tests your ability to factor quadratic expressions, specifically difference of squares, and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form g(x) = (x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s. These are the x-intercepts of the parabola—points (r, 0) and (s, 0) where the graph crosses the x-axis. Factoring transforms the quadratic from a form where zeros are hidden (standard form) to a form where they're obvious (factored form)! To factor x² - 25 and find zeros: recognize it as a difference of squares, a² - b² = (a - b)(a + b), where a = x, b = 5. So (x - 5)(x + 5). Find zeros: x - 5 = 0 gives x = 5, x + 5 = 0 gives x = -5. X-intercepts: (5, 0), (-5, 0). Great job spotting the pattern! Choice B correctly factors using difference of squares and identifies the zeros accurately. Choice C mistakes it for a perfect square trinomial, but x² - 25 isn't (x - 5)² = x² - 10x + 25; there's no middle term, so it's difference of squares, not a repeated factor. Always check if it's a² - b²! Factoring difference of squares is a quick shortcut: x² - c = (x - √c)(x + √c) if c > 0. Example: x² - 16 = (x - 4)(x + 4), zeros ±4. For sums like x² + 25, it doesn't factor over reals. Remember the pattern to save time!
Factor $r(x)=4x^2-9$ to reveal its zeros. Then identify the x-intercepts of the graph.
Answer: $r(x)=(2x-3)(2x+3)$ ; zeros: $x=\tfrac{3}{2},-\tfrac{3}{2}$ ; x-intercepts: $(\tfrac{3}{2},0),(-\tfrac{3}{2},0)$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For r(x) = 4x² - 9, difference of squares: (2x - 3)(2x + 3); zeros x = 3/2 and x = -3/2, x-intercepts (3/2, 0) and (-3/2, 0)—fractional but symmetric! Choice A correctly factors and finds the zeros. Choice D ignores the 4, treating as x² - 9, zeros too large—scale factors affect roots; always include coefficients. Difference of squares with coefficients: (√a x - √c)(√a x + √c); practice yields quick results—you can do this!
Solve the equation by factoring and using the zero product property: $3x^2-12=0$ . Then list the zeros of the related function $y=3x^2-12$ and the x-intercepts.
Answer: Zeros: $x=\pm 2$ ; x-intercepts: $(-2,0),(2,0)$ This question tests solving equations by factoring and relating to function zeros (x-intercepts)—builds equation-solving skills. Factor 3x² - 12 = 0 → 3(x² - 4) = 0 → 3(x - 2)(x + 2) = 0. Zeros: ±2 (3≠0). X-intercepts: (-2, 0), (2, 0). You're doing great! Choice A correctly identifies zeros. Choice B mistakes the constant; -12/3 = -4, but √4=2, not 4. Factor completely! Difference of squares after factoring out common factor. Example: 2x² - 8 = 0 → 2(x² - 4)=0 → 2(x-2)(x+2)=0, zeros ±2.
Factor the quadratic function $f(x)=x^2+7x+12$ . Then use the zero product property to find the zeros and identify the x-intercepts of the graph.
Answer: $f(x)=(x+3)(x+4)$ ; zeros: $x=-3,-4$ ; x-intercepts: $(-3,0),(-4,0)$ This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s. These are the x-intercepts of the parabola—points (r, 0) and (s, 0) where the graph crosses the x-axis. Factoring transforms the quadratic from a form where zeros are hidden (standard form) to a form where they're obvious (factored form)! To factor x² + 7x + 12 and find zeros: (1) Look for two numbers that multiply to 12 (constant term) and add to 7 (middle coefficient): factors of 12 are 1 and 12 (sum 13, no), 2 and 6 (sum 8, no), 3 and 4 (sum 7, yes!). (2) Write factored form: (x + 3)(x + 4). (3) Find zeros using zero product property: set (x + 3)(x + 4) = 0, so x + 3 = 0 giving x = -3, or x + 4 = 0 giving x = -4. The zeros are x = -3 and x = -4. (4) X-intercepts are (-3, 0) and (-4, 0). The factored form makes zeros immediate—no quadratic formula needed! Choice A correctly factors the quadratic and identifies both zeros by setting each factor equal to zero and solving. Choice B has a sign error in the zeros: from factor (x + 3), the zero is x = -3 (not x = 3). The sign flips! From (x - r), zero is x = r, so the sign in the factor is opposite to the zero. Think: what value makes (x + 3) equal zero? x = -3. This sign relationship trips everyone up initially—practice makes it automatic! Factoring trinomial x² + bx + c: find two numbers that (1) multiply to c (constant term), (2) add to b (middle coefficient). Those numbers go in factors: (x + first number)(x + second number). Example: x² - 5x + 6, find factors of 6 that add to -5: that's -2 and -3 (multiply to 6, add to -5). Factored: (x - 2)(x - 3). Zeros: x = 2, 3. For leading coefficient not 1 (like 2x² + 7x + 3), use grouping or trial combinations. From factored form to zeros: the sign trick is crucial. (x - r) gives zero x = r (opposite sign), (x + s) = (x - (-s)) gives zero x = -s (opposite sign). From (x - 5), zero is x = 5. From (x + 2), zero is x = -2. Set each factor equal to zero: (x - 5) = 0 means x = 5, (x + 2) = 0 means x = -2. The zeros always have opposite sign from what appears in the factored form (unless the factor is written differently). Master this sign relationship!
The quadratic function $g(x) = x^2 + px + q$ can be written as $g(x) = (x - r)(x - s)$ where $r$ and $s$ are the zeros. If $r + s = 8$ and $rs = 15$ , what are the zeros of $g(x)$ ?
Answer: $x = 3$ and $x = 5$ From Vieta's formulas, if the zeros are r and s, then r + s = -p and rs = q. We need r + s = 8 and rs = 15. This gives us the system where r and s satisfy t² - 8t + 15 = 0, which factors as (t - 3)(t - 5) = 0. So r = 3 and s = 5. Choice B would be correct if r + s = -8. Choice C gives rs = 15 but r + s = 16. Choice D gives rs = 15 but r + s = -16.
The quadratic $4x^2 - 4x - 15$ factors into the form $(2x + a)(2x + b)$ where $a$ and $b$ are integers. What are the zeros revealed by this factorization?
Answer: $x = -\frac{3}{2}$ and $x = \frac{5}{2}$ When you encounter a quadratic that needs to be factored into a specific form, you're looking at a strategic factoring problem that connects factorization to finding zeros. Let's factor $4x^2 - 4x - 15$ into the form $(2x + a)(2x + b)$ . When we expand this form, we get $4x^2 + 2bx + 2ax + ab = 4x^2 + 2(a+b)x + ab$ . Comparing this to our original quadratic, we need $2(a+b) = -4$ and $ab = -15$ . This gives us $a+b = -2$ and $ab = -15$ . We need two integers that multiply to $-15$ and add to $-2$ . Testing factor pairs of $-15$ : $3$ and $-5$ work because $3 \times (-5) = -15$ and $3 + (-5) = -2$ . So our factorization is $(2x + 3)(2x - 5)$ . To find the zeros, set each factor equal to zero: $2x + 3 = 0$ gives $x = -\frac{3}{2}$ , and $2x - 5 = 0$ gives $x = \frac{5}{2}$ . This matches answer D. Choice A gives the values of $a$ and $b$ themselves, not the zeros. Choice B represents a sign error—these would be zeros if the factorization were $(2x - 3)(2x + 5)$ . Choice C has both the wrong signs and wrong reasoning, possibly confusing the relationship between factors and zeros. Remember: when finding zeros from factored form, set each factor equal to zero and solve for $x$ . Don't confuse the constants in the factors with the actual zeros.

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