To determine the possible multiplicity of the factor \((x+1)\) in the polynomial \(m(x)\), we can analyze the behavior of the graph at the point \(x = -1\).
Here are the key points to consider regarding the multiplicity of a factor:
- If \((x + 1)\) has an odd multiplicity, the graph will cross the x-axis at \(x = -1\).
- If \((x + 1)\) has an even multiplicity, the graph will touch the x-axis at \(x = -1\) but will not cross it.
From your description of the graph:
- The graph starts in quadrant 2, which means it is above the x-axis initially.
- It then descends to the x-axis, indicating it crosses the x-axis (x = -1) at this point.
- Given that the graph ascends after reaching the x-axis at some point, it suggests that this location is indeed a crossing point rather than just touching.
Since the factor \((x + 1)\) results in crossing the x-axis at \(x = -1\), this corresponds to an odd multiplicity. Possible odd multiplicities include 1, 3, 5, etc.
Among the options given:
- Option #1: \(-1\) (not valid)
- Option #2: \(0\) (not valid)
- Option #3: \(1\) (valid; odd multiplicity)
- Option #4: \(2\) (not valid; even multiplicity)
Thus, the only option that gives a possible multiplicity of \((x + 1)\) is:
Option #3: 1.