Question

Determine whether f(x) = 0 has any repeated real solutions

Answer 1

To determine if the equation f(x) = 0 has repeated real solutions, we need to analyze its graph and see if it touches or crosses the x-axis at multiple points.

Step-by-step explanation:

To determine whether the equation f(x) = 0 has any repeated real solutions, we need to analyze its graph. If the graph of the equation touches the x-axis at a specific point, it means there is a real solution. If the graph crosses the x-axis at a specific point, it means there is a repeated real solution.

For example, if the graph of f(x) = x^2 touches the x-axis at (0, 0), it means x = 0 is a real solution. But if the graph of f(x) = x^2 - 4 crosses the x-axis at (2, 0) and (-2, 0), it means x = 2 and x = -2 are repeated real solutions.

Therefore, if the graph of f(x) = 0 only touches or crosses the x-axis at distinct points, it means there are no repeated real solutions.

Answer 2

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