This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = (x + 4)(x + 1)(x - $2)^2$, zeros are x = -4 (multiplicity 1, crosses), x = -1 (multiplicity 1, crosses), x = 2 (multiplicity 2, touches), degree 4 even positive, both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B misidentifies zeros like x = 4 instead of x = -4—always solve factors with attention to signs! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = (x - $1)^4$(x + 3), zeros are x = 1 (multiplicity 4, touches), x = -3 (multiplicity 1, crosses), degree 5 odd positive, left down right up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice D gets end behavior wrong by assuming even degree, but total degree is 5 (odd)—add all exponents to confirm! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = 3(x + $2)^2$(x - 1)(x - 4), zeros are x = -2 (multiplicity 2, touches), x = 1 (multiplicity 1, crosses), x = 4 (multiplicity 1, crosses), degree 4 even positive (leading 3 > 0), both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice C mistakes the touching zero, but multiplicity 2 at x = -2 means touch, not at x = 1 which is odd—check exponents for each factor! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For q(x) = -2 (x - $2)^2$ (x + 4), the zeros are x = 2 (multiplicity 2, from (x - $2)^2$ = 0) and x = -4 (multiplicity 1, from x + 4 = 0); the graph touches at x = 2 due to even multiplicity and crosses at x = -4 due to odd multiplicity, with degree 3 (odd) and negative leading coefficient (-2) indicating left up and right down. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing, and has correct end behavior. Choice D fails because it incorrectly states both ends up, which would be for a positive leading coefficient with even degree, but here it's odd degree and negative. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = -x(x-3)²(x+2), we identify zeros at x = 0 (multiplicity 1, from -x), x = 3 (multiplicity 2, from (x-3)²), and x = -2 (multiplicity 1, from (x+2)). The graph crosses at x = 0 and x = -2 (odd multiplicities) and touches at x = 3 (even multiplicity). The polynomial has degree 1+2+1=4 (even) with negative leading coefficient (the -1 from -x), so both ends go down: as x→-∞, p(x)→-∞ and as x→∞, p(x)→-∞. Choice A correctly identifies all zeros with their multiplicities, states that the graph crosses at 0 and -2 while touching at 3, and has the correct end behavior for an even-degree polynomial with negative leading coefficient. Choice B has the same zero information but incorrectly shows both ends going up, which would be true for a positive leading coefficient. Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero.

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For $$p(x) = 3(x-2)(x+1)(x-4)^2$$, the degree is $1+1+2=4$ (even). When expanded, the leading term will be $3x^4$, which has positive coefficient. For even degree with positive leading coefficient, both ends go up: as $x \to -\infty$, $p(x) \to \infty$ and as $x \to \infty$, $p(x) \to \infty$. Choice C correctly states this end behavior. Choice A shows the end behavior for odd degree (opposite ends), Choice B shows odd degree with negative leading coefficient, and Choice D shows even degree with negative leading coefficient. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor ($x - r$) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = -3x⁵(x-2)², the leading term comes from -3x⁵ · x² = -3x⁷ when fully expanded. The polynomial has degree 5+2=7 (odd) with negative leading coefficient (-3). For odd degree with negative leading coefficient, as x→-∞, f(x)→+∞ (left up) and as x→+∞, f(x)→-∞ (right down). Choice B correctly shows as x→-∞, f(x)→+∞ and as x→+∞, f(x)→-∞, which matches odd-degree polynomial with negative leading coefficient. Choice A incorrectly shows left down, right up, which would be for positive leading coefficient with odd degree. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For f(x) = (x-3)(x+1)(x-2)², we find zeros: x-3=0 gives x=3 (multiplicity 1), x+1=0 gives x=-1 (multiplicity 1), and (x-2)²=0 gives x=2 (multiplicity 2). The graph crosses at x=3 and x=-1 (odd multiplicities) and touches at x=2 (even multiplicity). The polynomial has degree 1+1+2=4 (even) with positive leading coefficient (1), so both ends go up: as x→±∞, f(x)→+∞. Choice A correctly identifies zeros x=3 (mult. 1), x=-1 (mult. 1), x=2 (mult. 2), shows crossing at x=3 and x=-1, touching at x=2, and has both ends going to +∞. Choice B incorrectly claims the graph crosses at all three zeros, missing that x=2 with even multiplicity should touch, not cross. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For p(x) = -(x+4)²(x-1)³, we find zeros: (x+4)²=0 gives x=-4 (multiplicity 2), and (x-1)³=0 gives x=1 (multiplicity 3). At x=-4, multiplicity 2 is even, so the graph touches and bounces. At x=1, multiplicity 3 is odd, so the graph crosses with flattening. The polynomial has degree 2+3=5 (odd) with negative leading coefficient (-1), so as x→-∞, p(x)→+∞ (left up) and as x→+∞, p(x)→-∞ (right down). Choice A correctly identifies zeros x=-4 (mult. 2), x=1 (mult. 3), shows touching at x=-4 and crossing (flattened) at x=1, with end behavior as x→-∞, p(x)→+∞ and as x→+∞, p(x)→-∞. Choice D incorrectly identifies x=4 instead of x=-4 as a zero, missing the sign in the factor (x+4). The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!