Finding the Zeros of a Polynomial Function

Oct 13, 2025 by talynt.org 119 views

Finding the Zeros of a Polynomial Function

To find the zeros of a polynomial function, we need to solve the equation d(x) = 0. In this case, the function is d(x) = -12x^3 + 33x^2 + 9x. We can start by factoring out the greatest common factor (GCF) of the terms, which is x.

Factoring Out the GCF

The GCF of the terms is x, so we can factor it out as follows:

d(x) = x(-12x^2 + 33x + 9)

Finding the Zeros of the Quadratic Function

Now we need to find the zeros of the quadratic function -12x^2 + 33x + 9. We can use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -12, b = 33, and c = 9. Plugging these values into the formula, we get:

x = (-(33) ± √((33)^2 - 4(-12)(9))) / 2(-12) x = (-33 ± √(1089 + 432)) / -24 x = (-33 ± √1521) / -24 x = (-33 ± 39) / -24

Solving for x

Now we need to solve for x by simplifying the two possible solutions:

x = (-33 + 39) / -24 x = 6 / -24 x = -1/4

x = (-33 - 39) / -24 x = -72 / -24 x = 3

Finding the Multiplicities of the Zeros

To find the multiplicities of the zeros, we need to look at the original function d(x) = -12x^3 + 33x^2 + 9x. We can see that the factor (x + 1/4) appears only once, so the multiplicity of the zero x = -1/4 is 1.

Similarly, the factor (x - 3) appears only once, so the multiplicity of the zero x = 3 is 1.

Conclusion

In conclusion, the zeros of the function d(x) = -12x^3 + 33x^2 + 9x are x = -1/4 and x = 3, with multiplicities of 1.

Final Answer

The final answer is: $\boxed{-\frac{1}{4}, 3}$

Finding the Zeros of a Polynomial Function

To find the zeros of a polynomial function, we need to solve the equation d(x) = 0. In this case, the function is d(x) = -12x^3 + 33x^2 + 9x. We can start by factoring out the greatest common factor (GCF) of the terms, which is x.

Factoring Out the GCF

The GCF of the terms is x, so we can factor it out as follows:

d(x) = x(-12x^2 + 33x + 9)

Finding the Zeros of the Quadratic Function

Now we need to find the zeros of the quadratic function -12x^2 + 33x + 9. We can use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -12, b = 33, and c = 9. Plugging these values into the formula, we get:

x = (-(33) ± √((33)^2 - 4(-12)(9))) / 2(-12) x = (-33 ± √(1089 + 432)) / -24 x = (-33 ± √1521) / -24 x = (-33 ± 39) / -24

Solving for x

Now we need to solve for x by simplifying the two possible solutions:

x = (-33 + 39) / -24 x = 6 / -24 x = -1/4

x = (-33 - 39) / -24 x = -72 / -24 x = 3

Finding the Multiplicities of the Zeros

To find the multiplicities of the zeros, we need to look at the original function d(x) = -12x^3 + 33x^2 + 9x. We can see that the factor (x + 1/4) appears only once, so the multiplicity of the zero x = -1/4 is 1.

Similarly, the factor (x - 3) appears only once, so the multiplicity of the zero x = 3 is 1.

Conclusion

In conclusion, the zeros of the function d(x) = -12x^3 + 33x^2 + 9x are x = -1/4 and x = 3, with multiplicities of 1.

Q&A

Q: What is the greatest common factor (GCF) of the terms in the function d(x) = -12x^3 + 33x^2 + 9x?

A: The GCF of the terms is x.

Q: How do we find the zeros of the quadratic function -12x^2 + 33x + 9?

A: We can use the quadratic formula to solve for x.

Q: What is the quadratic formula?

A: The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a.

Q: What are the values of a, b, and c in the quadratic formula?

A: In this case, a = -12, b = 33, and c = 9.

Q: How do we simplify the two possible solutions for x?

A: We can simplify the solutions by dividing the numerator and denominator by their greatest common factor.

Q: What is the multiplicity of the zero x = -1/4?

A: The multiplicity of the zero x = -1/4 is 1.

Q: What is the multiplicity of the zero x = 3?

A: The multiplicity of the zero x = 3 is 1.

Q: What are the zeros of the function d(x) = -12x^3 + 33x^2 + 9x?

A: The zeros of the function are x = -1/4 and x = 3.

Final Answer

The final answer is: $\boxed{-\frac{1}{4}, 3}$