To solve $f(x) = 4x^3 - x^2 + 20x - 5$ and find all roots using the Rational Root Theorem, we proceed step by step.

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Step 1: Use the Rational Root Theorem

The Rational Root Theorem states that any rational root of $f(x)$ is of the form:

$\frac{p}{q}$

where:

• $p$ is a factor of the constant term (-5).
• $q$ is a factor of the leading coefficient (4).

Factors of $p = -5$: $\pm 1, \pm 5$

Factors of $q = 4$: $\pm 1, \pm 2, \pm 4$

Possible rational roots are:

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}$

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Step 2: Test the Rational Roots

We substitute these values into $f(x) = 4x^3 - x^2 + 20x - 5$ to find a root.

Testing $x = 1$:

$f(1) = 4(1)^3 - (1)^2 + 20(1) - 5 = 4 - 1 + 20 - 5 = 18 \quad \text{(Not a root)}$

Testing $x = -1$:

$f(-1) = 4(-1)^3 - (-1)^2 + 20(-1) - 5 = -4 - 1 - 20 - 5 = -30 \quad \text{(Not a root)}$

Testing $x = \frac{1}{2}$:

= 4\left(\frac{1}{8}\right) - \frac{1}{4} + 10 - 5 = \frac{1}{2} - \frac{1}{4} + 5 - 5 = 0 \quad \text{(Root found!)}$$ Thus, $$x = \frac{1}{2}$$ is a root. --- ### Step 3: Perform Synthetic Division We now divide $$f(x)$$ by $$x - \frac{1}{2}$$ using synthetic division to reduce the cubic polynomial to a quadratic. #### Setup: Divide $$4x^3 - x^2 + 20x - 5$$ by $$x - \frac{1}{2}$$. #### Synthetic Division Process: 1. Write coefficients: $$[4, -1, 20, -5]$$. 2. Divide by $$x - \frac{1}{2}$$, so use $$\frac{1}{2}$$ as the divisor. 3. Perform the steps: $$\begin{array}{r|rrrr} \frac{1}{2} & 4 & -1 & 20 & -5 \\ & & 2 & 0 & 10 \\ \hline & 4 & 1 & 20 & 5 \\ \end{array}$$ The quotient is $$4x^2 + x + 20$$, and the remainder is 0, confirming $$x = \frac{1}{2}$$ is a root. --- ### Step 4: Solve the Quadratic $$4x^2 + x + 20 = 0$$ Now solve $$4x^2 + x + 20 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 4$$, $$b = 1$$, $$c = 20$$. $$x = \frac{-1 \pm \sqrt{1^2 - 4(4)(20)}}{2(4)}$$ $$x = \frac{-1 \pm \sqrt{1 - 320}}{8}$$ $$x = \frac{-1 \pm \sqrt{-319}}{8}$$ $$x = \frac{-1 \pm i\sqrt{319}}{8}$$ The two complex roots are: $$x = \frac{-1 + i\sqrt{319}}{8}, \quad x = \frac{-1 - i\sqrt{319}}{8}$$ --- ### Final Answer: The roots of $$f(x) = 4x^3 - x^2 + 20x - 5$$ are: 1. $$x = \frac{1}{2}$$ (rational root) 2. $$x = \frac{-1 + i\sqrt{319}}{8}$$ (complex root) 3. $$x = \frac{-1 - i\sqrt{319}}{8}$$ (complex root) --- Would you like a step-by-step explanation of any part? ### Related Questions: 1. How is the Rational Root Theorem derived? 2. Can you demonstrate long division instead of synthetic division for the cubic polynomial? 3. What is the significance of complex roots in polynomials? 4. How do you interpret the discriminant in the quadratic formula? 5. Can we graph $$f(x)$$ to verify the roots visually? ### Tip: Always check the discriminant ($$b^2 - 4ac$$) before solving a quadratic equation to quickly identify the nature of the roots (real or complex).