(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:

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!