Question

what could be the equation of p?
choose 1 answer:
a ( p(x) = (x + 3)^2(x + 1)^2(2x - 1) )
b ( p(x) = (x + 3)^2(x + 1)(2x - 1) )
c ( p(x) = (x + 3)(x + 1)^2(2x - 1) )
d ( p(x) = (x + 3)(x + 1)(2x - 1)^2 )

Explanation:

Step1: Analyze the roots and multiplicities

Assume the graph of \( p(x) \) has a root at \( x = - 3 \) with even multiplicity (since the graph touches the x - axis and turns around, indicating a double root), a root at \( x=-1\) with even multiplicity (same reason as \( x = - 3 \)) and a root at \( x=\frac{1}{2}\) (from \( 2x - 1=0\)) with odd multiplicity (the graph crosses the x - axis here).

Step2: Match with the options

• For option A: \( p(x)=(x + 3)^{2}(x + 1)^{2}(2x-1) \). The root \( x=-3 \) has multiplicity 2 (even), \( x = - 1 \) has multiplicity 2 (even) and \( x=\frac{1}{2}\) has multiplicity 1 (odd), which matches the expected behavior of the graph (touching at \( x=-3\) and \( x = - 1\), crossing at \( x=\frac{1}{2}\)).
• For option B: \( p(x)=(x + 3)^{2}(x + 1)(2x-1) \). The root \( x=-1\) has multiplicity 1 (odd), which would mean the graph crosses the x - axis at \( x=-1\), but we expect it to touch and turn around (even multiplicity), so this is incorrect.
• For option C: \( p(x)=(x + 3)(x + 1)^{2}(2x-1) \). The root \( x=-3\) has multiplicity 1 (odd), which would mean the graph crosses the x - axis at \( x=-3\), but we expect it to touch and turn around (even multiplicity), so this is incorrect.
• For option D: \( p(x)=(x + 3)(x + 1)(2x-1)^{2} \). The root \( x=\frac{1}{2}\) has multiplicity 2 (even), which would mean the graph touches the x - axis at \( x=\frac{1}{2}\), but we expect it to cross the x - axis at \( x=\frac{1}{2}\) (odd multiplicity), and the roots \( x=-3\) and \( x=-1\) have multiplicity 1 (odd), which would mean the graph crosses the x - axis at \( x=-3\) and \( x=-1\) instead of touching, so this is incorrect.

Answer:

A. \( p(x)=(x + 3)^{2}(x + 1)^{2}(2x-1) \)