Graphs of Polynomials | Multiplicities

Question 1: Graph of a Zero with Even Multiplicity (from file image_d10f7c.png)

The Question: If $p(x)$ be a polynomial function and 1 is a zeros of the given polynomial with even multiplicity then which of the following graph is the best representation of the graph of $p(x)$.

Core Concept: As defined above, a zero with even multiplicity means the graph must touch the x-axis at that point and “bounce” off, without crossing to the other side.

Detailed Solution:

1. We are looking for a graph that has a zero at $x=1$.
2. The multiplicity at this zero is even. This means the graph must be tangent to the x-axis at $x=1$.
3. Let’s examine the options:
   • The first two graphs both show a parabola whose vertex (turning point) is exactly at $(1, 0)$. This “touch and turn” behavior is the classic sign of a zero with even multiplicity.
   • The third graph shows a line crossing the axis, which would be a zero of odd multiplicity (specifically, 1).
   • The fourth graph shows a parabola that does not have a zero at $x=1$.

Final Answer: The first (and second) option correctly represents a polynomial with a zero at $x=1$ having even multiplicity.

Question 2: Zeros and Intercepts from a Graph (from file image_d10ee1.png)

The Question: Choose the set of correct options from the graph of a polynomial $p(x)$.

• Zeros of the polynomial are -3, -2, 1
• Zeros of the polynomial are -3, -1, 0
• y-intercept of the polynomial is 1
• y-intercept of the polynomial is 0

Core Concept:

• Zeros (x-intercepts): The points where the graph crosses or touches the horizontal x-axis.
• y-intercept: The point where the graph crosses the vertical y-axis (where $x=0$).

Detailed Solution:

1. Find the Zeros: Look at the provided graph. The curve intersects the x-axis at three points: $x = -3$, $x = -1$, and $x = 0$.
2. Find the y-intercept: Look at where the curve intersects the y-axis. It passes through the origin, $(0, 0)$. Therefore, the y-intercept is 0.

Final Answer: The correct options are:

• Zeros of the polynomial are -3, -1, 0
• y-intercept of the polynomial is 0

Question 3: Multiplicities from a Graph (from file image_d10bf1.png)

The Question: Choose the correct options regarding the multiplicities of the zeros (from the graph in Question 2).

Core Concept: We apply the graphical rules for multiplicity to the zeros we identified in the previous question.

Detailed Solution:

1. At the zero $x = -3$: The graph crosses the x-axis. This indicates an odd multiplicity.
2. At the zero $x = -1$: The graph touches the x-axis and “bounces” back up. This indicates an even multiplicity.
3. At the zero $x = 0$: The graph crosses the x-axis. This indicates an odd multiplicity. The simplest odd multiplicity is 1.

Now, let’s evaluate the options:

• "-3 with even multiplicity.": FALSE.
• "-3 with odd multiplicity.": TRUE.
• “0 with even multiplicity.”: FALSE.
• “0 with 1 multiplicity.”: TRUE (since 1 is an odd number, and the graph crosses here).
• "-1 with even multiplicity.": TRUE.
• "-1 with odd multiplicity.": FALSE.

Final Answer: The correct statements are:

• -3 with odd multiplicity.
• 0 with 1 multiplicity.
• -1 with even multiplicity.

Question 4: Degree from Zeros and Multiplicities (from file image_d10bf1.png)

The Question: Let $p(x)$ be a polynomial and 7, 9, 2 are the only zeros of the polynomial with multiplicities 3, 1, 4 respectively. Then the degree of the polynomial can be at least __________.

Core Concept: The minimum possible degree of a polynomial is the sum of the multiplicities of all its distinct zeros.

Detailed Solution:

1. List the zeros and their multiplicities:
   • Zero at $x=7$ has multiplicity 3.
   • Zero at $x=9$ has multiplicity 1.
   • Zero at $x=2$ has multiplicity 4.
2. Sum the multiplicities:
   • Minimum Degree = $3 + 1 + 4 = 8$.
3. This means the polynomial, when factored, would look something like $p(x) = c(x-7)^3(x-9)^1(x-2)^4$, where $c$ is a constant. The highest power of $x$ would be $x^3 \cdot x^1 \cdot x^4 = x^8$.

Final Answer: The degree of the polynomial can be at least 8.

Question 5: Zeros of a Factored Polynomial (from file image_d10b5e.png)

The Question: Suppose $p(x) = (x+4)^5(2x^2 + 4)$, choose the set of correct options.

Core Concept: To find the real zeros of a factored polynomial, set each factor equal to zero and solve for $x$. The multiplicity of a zero is the exponent on its corresponding factor.

Detailed Solution:

To find the zeros, we set $p(x) = 0$:

This means either the first factor is zero, or the second factor is zero.

1. Analyze the first factor:

   • $(x+4)^5 = 0$
   • $x+4 = 0$
   • $x = -4$.
   • This gives us a real zero at -4. The multiplicity is the exponent, so the multiplicity is 5.

2. Analyze the second factor:

   • $2x^2 + 4 = 0$
   • $2x^2 = -4$
   • $x^2 = -2$
   • There is no real number whose square is negative. Therefore, this factor gives no real roots.

3. Combine the findings and evaluate the options:

   • The only real zero is -4.
   • The multiplicity of this zero is 5.
   • The total number of real roots is 1.

Let’s check the statements:

• “4 is the zero of the polynomial”: FALSE.
• "-4 is the zero of the polynomial": TRUE.
• “Only one root is of multiplicity 5”: TRUE. (The root -4 is the only one, and its multiplicity is 5).
• “Two roots are of multiplicity 5”: FALSE.
• “There is no real root”: FALSE.
• “The number of real roots is 2”: FALSE.
• “The number of real roots is 1”: TRUE.

Final Answer: The correct options are:

• -4 is the zero of the polynomial
• Only one root is of multiplicity 5
• The number of real roots is 1