Question

what could be the equation of p? choose 1 answer: a $p(x) = x^2(2x - 5)(x + 2)$ b $p(x) = x(2x - 5)(x + 2)^2$ c $p(x) = x^2(2x - 5)^2(x + 2)$ d $p(x) = x(2x - 5)^2(x + 2)$

Explanation:

Step1: Analyze the graph's behavior at roots

The graph has a "bounce" at \( x = -2 \) (even multiplicity) and a "cross" at \( x = 0 \) and \( x=\frac{5}{2} \) (odd multiplicity). Also, the root \( x = 0 \) has multiplicity 2 (since it's a double root, maybe from \( x^2 \))? Wait, no—wait, the key is: when a root has even multiplicity, the graph touches the axis (bounce), odd multiplicity crosses. From the options, let's check multiplicities:

• For root \( x = -2 \): in option B, \( (x + 2)^2 \) (multiplicity 2, even) → bounce.
• For root \( x = 0 \): in option B, \( x \) (multiplicity 1, odd) → cross.
• For root \( 2x - 5 = 0 \) → \( x=\frac{5}{2} \): in option B, \( (2x - 5) \) (multiplicity 1, odd) → cross.

Wait, but the original graph (from the arrow and the "V" shape? Wait, maybe the graph has a double root at \( x = -2 \) (bounce) and single roots at \( x = 0 \) and \( x=\frac{5}{2} \) (cross). Let's check each option:

Option A: \( p(x)=x^2(2x - 5)(x + 2) \). Multiplicities: \( x=0 \) (2, even → bounce), \( x=\frac{5}{2} \) (1, odd → cross), \( x=-2 \) (1, odd → cross). But we need a bounce at \( x=-2 \), so A is out.

Option B: \( p(x)=x(2x - 5)(x + 2)^2 \). Multiplicities: \( x=0 \) (1, odd → cross), \( x=\frac{5}{2} \) (1, odd → cross), \( x=-2 \) (2, even → bounce). This matches the bounce at \( x=-2 \) and cross at 0 and \( \frac{5}{2} \).

Option C: \( p(x)=x^2(2x - 5)^2(x + 2) \). Multiplicities: \( x=0 \) (2, even → bounce), \( x=\frac{5}{2} \) (2, even → bounce), \( x=-2 \) (1, odd → cross). Doesn't match.

Option D: \( p(x)=x(2x - 5)^2(x + 2) \). Multiplicities: \( x=0 \) (1, odd → cross), \( x=\frac{5}{2} \) (2, even → bounce), \( x=-2 \) (1, odd → cross). Doesn't match.

So the correct option is B.

Answer:

B. \( p(x) = x(2x - 5)(x + 2)^2 \)