This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). The Zero Product Property is super useful: it tells us that if a product equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 (giving x = 3) or x + 5 = 0 (giving x = -5)—those are the zeros! Let's factor x² + 2x - 8: we need two numbers that multiply to -8 and add to 2. Those numbers are 4 and -2 because 4·-2 = -8 and 4 + -2 = 2. So the factored form is (x + 4)(x - 2). Now we use the Zero Product Property: set each factor to zero and solve. x + 4 = 0 gives x = -4, and x - 2 = 0 gives x = 2. Those are the zeros! Choice A is correct because it properly identifies the zeros as x = -4 and x = 2 using the Zero Product Property. Nicely done! Choice B gets the factors right but makes a mistake with the Zero Product Property: from (x + 4) = 0, we get x = -4, not x = 4. When you solve x + 4 = 0, you subtract 4 from both sides. The signs flip! Quick pattern to remember: in the factored form (x - p)(x - q), the zeros are x = p and x = q. Notice how the signs flip! If you have (x - 3), the zero is x = 3 (positive), and if you have (x + 5), the zero is x = -5 (negative). The sign in the factor is opposite the sign of the zero.

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). To factor a quadratic like $x^2 + bx + c$, we need to find two numbers that multiply to give $c$ (the constant) and add to give $b$ (the coefficient of x), then write the factored form as $(x + \text{[first number]})(x + \text{[second number]})$. Let's factor $P(x) = -x^2 + 10x - 16$: First, factor out $-1$ to get $-(x^2 - 10x + 16)$. Now we need two numbers that multiply to $16$ and add to $-10$. Those numbers are $-2$ and $-8$ because $(-2) \cdot(-8) = 16$ and $(-2) + (-8) = -10$. So the factored form is $-(x - 2)(x - 8)$. Now we use the Zero Product Property: set each factor to zero and solve. $x - 2 = 0$ gives $x = 2$, and $x - 8 = 0$ gives $x = 8$. Those are the zeros! Choice A is correct because it properly identifies the zeros as $x = 2$ and $x = 8$ using the Zero Product Property. Nicely done! Choice B has a sign error in the factoring—a really common slip! When the constant is positive and the middle term is negative, we need two negative numbers that multiply to positive. It's easy to get signs mixed up, so always check by FOILing your factors back! Quick pattern to remember: in the factored form $(x - p)(x - q)$, the zeros are $x = p$ and $x = q$. Notice how the signs flip! If you have $(x - 3)$, the zero is $x = 3$ (positive), and if you have $(x + 5)$, the zero is $x = -5$ (negative). The sign in the factor is opposite the sign of the zero.

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). The Zero Product Property is super useful: it tells us that if a product equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 (giving x = 3) or x + 5 = 0 (giving x = -5)—those are the zeros! This is a difference of squares, which has a special pattern! For difference of squares: x² - 9 = x² - 3² = (x + 3)(x - 3), giving zeros at x = -3 and x = 3. Recognizing these patterns saves time! Choice B is correct because it properly factors the quadratic as (x + 3)(x - 3) and identifies the zeros as x = -3 and x = 3 using the Zero Product Property. Nicely done! Choice A only lists one zero when this quadratic actually has two! From (x + 3)(x - 3) = 0, we get both x = -3 AND x = 3. Don't forget to set each factor equal to zero! Watch out for perfect patterns: x² - 9 is a difference of squares = (x + 3)(x - 3) with zeros at ±3, and x² + 6x + 9 is a perfect square = (x + 3)² with one repeated zero at -3. Recognizing these patterns will speed you up!

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² - 5x + 6 factors to (x - 2)(x - 3), which immediately tells us the zeros are x = 2 and x = 3 because those values make each factor (and therefore the whole expression) equal to zero. Let's factor x² + x - 12: we need two numbers that multiply to -12 and add to 1. Those numbers are 4 and -3 because 4·-3 = -12 and 4 + -3 = 1. So the factored form is (x + 4)(x - 3). Now we use the Zero Product Property: set each factor to zero and solve. x + 4 = 0 gives x = -4, and x - 3 = 0 gives x = 3. Those are the zeros! Choice B is correct because it properly identifies the zeros as x = 3 and x = -4 using the Zero Product Property. Nicely done! Choice A has a sign error in the factoring—a really common slip! When the constant is negative and the middle term is positive, we need one positive and one negative number, with the larger one positive. It's easy to get signs mixed up, so always check by FOILing your factors back! To check if x = 3 is really a zero, substitute it into the original: (3)² + (3) - 12 = 9 + 3 - 12 = 0. Yes! Since we get 0, this confirms x = 3 is indeed a zero. This is a great way to catch factoring mistakes!

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). To factor a quadratic like x² + bx + c, we need to find two numbers that multiply to give c (the constant) and add to give b (the coefficient of x), then write the factored form as (x + [first number])(x + [second number]). For 2x² + 7x + 3, we can try grouping or trial: it factors to (2x + 1)(x + 3), since 2x·x = 2x², 2x·3 + 1·x = 7x, and 1·3 = 3. Now we use the Zero Product Property: set each factor to zero and solve. 2x + 1 = 0 gives x = -1/2, and x + 3 = 0 gives x = -3. Those are the zeros! Choice A is correct because it properly factors the quadratic as (2x + 1)(x + 3) and identifies the zeros as x = -3 and x = -1/2 using the Zero Product Property. Nicely done! Choice B gets the factors right but makes a mistake with the Zero Product Property: from (x + 3) = 0, we get x = -3, not x = 3. When you solve x + 3 = 0, you subtract 3 from both sides. The signs flip! Quick pattern to remember: in the factored form (x - p)(x - q), the zeros are x = p and x = q. Notice how the signs flip! If you have (x - 3), the zero is x = 3 (positive), and if you have (x + 5), the zero is x = -5 (negative). The sign in the factor is opposite the sign of the zero.

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² - 5x + 6 factors to (x - 2)(x - 3), which immediately tells us the zeros are x = 2 and x = 3 because those values make each factor (and therefore the whole expression) equal to zero. Let's factor x² - 6x + 5: we need two numbers that multiply to 5 and add to -6. Those numbers are -1 and -5 because -1·-5 = 5 and -1 + -5 = -6. So the factored form is (x - 1)(x - 5). Now we use the Zero Product Property: set each factor to zero and solve. x - 1 = 0 gives x = 1, and x - 5 = 0 gives x = 5. Those are the zeros! Choice A is correct because it properly identifies the zeros as x = 1 and x = 5 using the Zero Product Property. Nicely done! Choice B has a sign error in the factoring—a really common slip! When the constant is positive and the middle term is negative, we need both numbers negative. It's easy to get signs mixed up, so always check by FOILing your factors back! The Zero Product Property is your friend: whenever you have something like (x - 3)(x + 5) = 0, you know that one of those factors must be zero, so x - 3 = 0 OR x + 5 = 0. Solve each little equation separately, and boom—you've got your zeros! They're x = 3 and x = -5.

This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² + 6x + 9 factors to (x + 3)², which immediately tells us the zero is x = -3 (repeated) because that value makes the factor (and therefore the whole expression) equal to zero. This is a perfect square trinomial, which has a special pattern! For perfect square: x² + 6x + 9 = (x + 3)², giving one zero (repeated) at x = -3. Recognizing these patterns saves time! To check if x = -3 is really a zero, substitute it into the original: (-3)² + 6(-3) + 9 = 9 - 18 + 9 = 0. Yes! Since we get 0, this confirms x = -3 is indeed a zero. This is a great way to catch factoring mistakes! Choice A is correct because it properly identifies the zero as x = -3. Nicely done! Choice C only lists one zero when this quadratic actually has one (repeated), but it incorrectly suggests two distinct zeros! From (x + 3)² = 0, we get x = -3 (twice). Don't forget it's repeated! Watch out for perfect patterns: x² - 9 is a difference of squares = (x + 3)(x - 3) with zeros at ±3, and x² + 6x + 9 is a perfect square = (x + 3)² with one repeated zero at -3. Recognizing these patterns will speed you up! To connect factoring to graphing: once you know the zeros, you know exactly where the parabola crosses the x-axis! If the zero is x = -3 (repeated), the graph touches the x-axis at (-3, 0) and bounces back, like a U-shape just grazing the axis.

Expand $$(x + 3)(x - 5)$$: $(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$. Comparing with $$f(x) = x^2 + bx + c$$, we have $$b = -2$$ and $$c = -15$$. Therefore $$b - c = (-2) - (-15) = -2 + 15 = 13$$. Choice A results from calculating $$c - b$$ instead of $$b - c$$. Choice B comes from sign errors in the expansion. Choice D results from incorrectly computing $$b - c$$ as $$-2 - 15 = -17$$ and then making a sign error.

Factor $$f(x) = x^2 - 8x + 15$$. We need two numbers that multiply to 15 and add to 8. These are 3 and 5. So $$f(x) = (x - 3)(x - 5)$$, giving zeros $$x = 3$$ and $$x = 5$$. Factor $$g(x) = x^2 - 8x + 16$$. We need two numbers that multiply to 16 and add to 8. These are 4 and 4. So $$g(x) = (x - 4)(x - 4) = (x - 4)^2$$, giving a double zero at $$x = 4$$. The zeros of $$f(x)$$ are 3 and 5. The zero of $$g(x)$$ is 4. These functions share no common zeros. Choice B suggests they share one zero, which would occur if one of 3, 5 equaled 4. Choice C suggests they share two zeros, which is impossible since $$g(x)$$ has only one distinct zero. Choice D is impossible since there are only 3 distinct zeros total between both functions.

When a quadratic expression is multiplied by a constant, the zeros remain the same. This is because if $$f(x) = 0$$ when $$x = a$$, then $$2f(x) = 2 \cdot 0 = 0$$ when $$x = a$$. The original expression $$x^2 + px + q$$ has zeros at $$x = -3$$ and $$x = 2$$, so the new expression $$2x^2 + 2px + 2q = 2(x^2 + px + q)$$ also has zeros at $$x = -3$$ and $$x = 2$$. Choice A results from incorrectly doubling the zeros. Choice C comes from incorrectly halving the zeros. Choice D results from completely misunderstanding the relationship between scaling and zeros.