(a) Find the rational zeros and then the other zeros of the polynomial function $$ f(x)=3 x^{3}+5 x^{2}+75 x+125 $$, that is, solve $$ f(x)=0 $$. (b) Factor $$ f(x) $$ into linear factors. (a) Select the correct choice below and fill in any answer box(es) within your choice. (Type an exact answer, using radicals and $$ i $$ as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. There is only one rational zero, $$ \square $$, and the other zeros are $$ \square $$. C. There are no rational zeros. The other zeros are Beros and they are

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Alright, I need to solve the polynomial equation $$ f(x) = 3x^3 + 5x^2 + 75x + 125 = 0 $$ by finding all its zeros, both rational and irrational. Let's break this down step by step.

**Step 1: Identify Possible Rational Zeros**

First, I'll use the Rational Root Theorem to find possible rational zeros. The Rational Root Theorem states that any possible rational zero, expressed in lowest terms $$ \frac{p}{q} $$, has $$ p $$ as a factor of the constant term and $$ q $$ as a factor of the leading coefficient.

- **Constant Term:** 125
  - Factors: $$ \pm1, \pm5, \pm25, \pm125 $$
- **Leading Coefficient:** 3
  - Factors: $$ \pm1, \pm3 $$

So, the possible rational zeros are:
$$ \pm1, \pm5, \pm25, \pm125, \pm\frac{1}{3}, \pm\frac{5}{3}, \pm\frac{25}{3}, \pm\frac{125}{3} $$

**Step 2: Test Possible Rational Zeros**

I'll test these candidates by substituting them into $$ f(x) $$ to see if they satisfy $$ f(x) = 0 $$.

1. **Testing $$ x = -5 $$:**
   $$
   f(-5) = 3(-5)^3 + 5(-5)^2 + 75(-5) + 125 = 3(-125) + 5(25) - 375 + 125 = -375 + 125 - 375 + 125 = -500 \neq 0
   $$
   Not a zero.

2. **Testing $$ x = -\frac{5}{3} $$:**
   $$
   f\left(-\frac{5}{3}\right) = 3\left(-\frac{5}{3}\right)^3 + 5\left(-\frac{5}{3}\right)^2 + 75\left(-\frac{5}{3}\right) + 125
   $$
   Calculating each term:
   $$
   3\left(-\frac{125}{27}\right) = -\frac{375}{27} = -\frac{125}{9}
   $$
   $$
   5\left(\frac{25}{9}\right) = \frac{125}{9}
   $$
   $$
   75\left(-\frac{5}{3}\right) = -125
   $$
   Adding them up:
   $$
   -\frac{125}{9} + \frac{125}{9} - 125 + 125 = 0
   $$
   So, $$ x = -\frac{5}{3} $$ is a rational zero.

**Step 3: Factor Out the Rational Zero**

Now that I've found a rational zero, I'll factor $$ f(x) $$ by dividing it by $$ (x + \frac{5}{3}) $$.

Using polynomial long division or synthetic division, I'll divide $$ 3x^3 + 5x^2 + 75x + 125 $$ by $$ x + \frac{5}{3} $$.

After performing the division, I'll obtain a quadratic factor:
$$
f(x) = \left(x + \frac{5}{3}\right)(3x^2 + 0x + 25)
$$
Simplifying the quadratic:
$$
3x^2 + 25
$$

**Step 4: Solve the Quadratic Equation**

Now, I'll solve $$ 3x^2 + 25 = 0 $$ for the remaining zeros.

$$
3x^2 = -25 \
x^2 = -\frac{25}{3} \
x = \pm\sqrt{-\frac{25}{3}} \
x = \pm\frac{5}{\sqrt{3}}i
$$

**Step 5: Summarize the Zeros**

- **Rational Zero:** $$ x = -\frac{5}{3} $$
- **Irrational Zeros:** $$ x = \frac{5}{\sqrt{3}}i $$ and $$ x = -\frac{5}{\sqrt{3}}i $$

**Step 6: Factor $$ f(x) $$ into Linear Factors**

Using the zeros found, the polynomial can be factored as:
$$
f(x) = 3\left(x + \frac{5}{3}\right)\left(x - \frac{5}{\sqrt{3}}i\right)\left(x + \frac{5}{\sqrt{3}}i\right)
$$
Simplifying:
$$
f(x) = 3\left(x + \frac{5}{3}\right)\left(x^2 + \frac{25}{3}\right)
$$

**Conclusion:**

- **Rational Zero:** $$ -\frac{5}{3} $$
- **Other Zeros:** $$ \frac{5}{\sqrt{3}}i $$ and $$ -\frac{5}{\sqrt{3}}i $$

(a) The rational zero is $$ -\frac{5}{3} $$, and the other zeros are $$ \frac{5}{\sqrt{3}}i $$ and $$ -\frac{5}{\sqrt{3}}i $$. (b) The factored form of $$ f(x) $$ is: $$ f(x) = 3\left(x + \frac{5}{3}\right)\left(x^2 + \frac{25}{3}\right) $$

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