Answer:
Yes
Step-by-step explanation:
These are the rules regarding polynomial graphs and their real zeros (solutions)
- If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.
- If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. That means it will have 2, 4 repeated solutions
- If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity i.e. multiplicity of 3, 5 etc
- The sum of the multiplicities cannot be greater than the degree of the polynomial
In the above graph we see that the graph touches the x-axis at x = 3. This means (x - 3)ᵃ is a factor of the polynomial where p is an even number (2, 4, 6 etc)
That means x = 3 is a repeated real solution. It appears that the multiplicity is 2 so one factor of the polynomial is (x - 3)²
More inf o(ignore if you want)
The other real roots are x = -1 and x = 5 because the graph crosses the x axis at these two values of x. So the multiplicity of both factors ( x + 1) and (x - 5) is odd. Since at the crossover point, the graph is almost linear, we can safely assume that the multiplicity of the graph is odd at these points and equal to 1
Therefore the polynomial can be factored as
(x + 1)(x - 5)(x - 3)²
The degree of the polynomial is the sum of the multiplicities and is equal to 1 + 1 + 2 = 4 so it is a quartic polynomial of the form ax⁴ + bx³ + cx² + dx + e