Consider the following.
f(x) = −4x3 + 15x2 − 8x − 3
(a)
List the possible rational zeros of f. (Enter your answers as a comma-separated list.)
x = 
 
(b)
Sketch the graph of f so that some of the possible zeros in part (a) can be disregarded. Determine all real zeros of f. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.)
x = 
 


### (a) List the possible rational zeros of \( f(x) = -4x^3 + 15x^2 - 8x - 3 \)

To find the **possible rational zeros**, we use the **Rational Root Theorem**, which states that any rational zero of a polynomial function is of the form:

\[
\frac{p}{q}
\]

Where:
- \( p \) is a factor of the constant term (in this case, -3).
- \( q \) is a factor of the leading coefficient (in this case, -4).

The constant term is -3, and its factors are \( \pm 1, \pm 3 \).

The leading coefficient is -4, and its factors are \( \pm 1, \pm 2, \pm 4 \).

Thus, the possible rational zeros are:

\[
\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{3}{4}
\]

Simplifying this list gives:

\[
x = \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}
\]

### (b) Sketch the graph and determine the real zeros

To find the **real zeros**, we will test these possible rational zeros using synthetic division or substitution. Let's try some of the simpler candidates like \( x = 1 \), \( x = -1 \), \( x = 3 \), and \( x = -3 \) first.

1. **Test \( x = 1 \):**

Substituting \( x = 1 \) into \( f(x) \):

\[
f(1) = -4(1)^3 + 15(1)^2 - 8(1) - 3 = -4 + 15 - 8 - 3 = 0
\]

Thus, \( x = 1 \) is a zero.

2. **Use synthetic division to divide \( f(x) \) by \( x - 1 \):**

Now divide \( f(x) \) by \( x - 1 \) using synthetic division:

\[
f(x) = (-4x^3 + 15x^2 - 8x - 3) \div (x - 1)
\]

Performing synthetic division:

\[
\begin{array}{r|rrrr}
1 & -4 & 15 & -8 & -3 \\
  &    & -4 & 11 & 3 \\
\hline
  & -4 & 11 & 3  & 0 \\
\end{array}
\]

The quotient is \( -4x^2 + 11x + 3 \).

3. **Solve \( -4x^2 + 11x + 3 = 0 \):**

We now solve the quadratic equation \( -4x^2 + 11x + 3 = 0 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -4 \), \( b = 11 \), and \( c = 3 \).

\[
x = \frac{-11 \pm \sqrt{11^2 - 4(-4)(3)}}{2(-4)} = \frac{-11 \pm \sqrt{121 + 48}}{-8} = \frac{-11 \pm \sqrt{169}}{-8}
\]

\[
x = \frac{-11 \pm 13}{-8}
\]

Thus, the solutions are:

\[
x = \frac{-11 + 13}{-8} = \frac{2}{-8} = -\frac{1}{4}, \quad x = \frac{-11 - 13}{-8} = \frac{-24}{-8} = 3
\]

### Conclusion: Real Zeros

The real zeros of \( f(x) \) are:

\[
x = 1, x = 3, x = -\frac{1}{4}
\]

So the zeros are \( x = 1, 3, -\frac{1}{4} \).

### Do you want details or have any questions?

### Five related questions:
1. How do you perform synthetic division?
2. What is the Rational Root Theorem and how does it work?
3. Can a polynomial have irrational or complex roots?
4. What does the multiplicity of a zero mean?
5. How do you factor polynomials?

### Tip:
To confirm if a value is a root of a polynomial, substituting it into the polynomial is a quick way to check!

Consider the following. f(x) = −4x^3 + 15x^2 − 8x − 3 (a) List the possible rational zeros of f. (b) Sketch the graph of f so that some of the possible zeros in part (a) can be disregarded. Determine all real zeros of f.

Explore how to list the possible rational zeros of the polynomial f(x) = −4x^3 + 15x^2 − 8x − 3 using the Rational Root Theorem. Learn to solve for the real zeros through synthetic division and apply the quadratic formula. Detailed step-by-step explanation included. Suitable for Grades 10-12 students.

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