How do you find all the zeros of #f(x) = x^3 + 4x^2 + 14x + 20#?

Question

How do you find all the zeros of #f(x) =  x^3 + 4x^2 + 14x + 20#?

Lincoln Anderson · Accepted Answer

The roots of the equation are: -2, -1 + 3i, and -1 - 3i.

Cora Alexander · Answer

undefined To find all the zeros of the function $ f(x) = x^3 + 4x^2 + 14x + 20 $, you can use various methods such as the Rational Root Theorem, synthetic division, or graphing techniques.

One approach is to start by checking for rational roots using the Rational Root Theorem. According to this theorem, any rational root of the polynomial $ f(x) $ must be of the form $ \frac{p}{q} $, where $ p $ is a factor of the constant term (in this case, 20) and $ q $ is a factor of the leading coefficient (in this case, 1).

After finding the possible rational roots, you can use synthetic division or polynomial long division to test each potential root until you find the actual roots.

Once you find one root, you can use polynomial division to divide $ f(x) $ by $ (x - \text{root}) $ to obtain a quadratic equation. You can then use the quadratic formula to find the remaining roots.

Repeat this process until you have found all the roots of the polynomial $ f(x) $.

HIX Tutor · Answer

undefined To find all the zeros of $ f(x) = x^3 + 4x^2 + 14x + 20 $, follow these steps:

1. **Look for Rational Roots**: Use the Rational Root Theorem. It says to test possible factors of the constant term (20) over factors of the leading coefficient (1). Possible roots are $ \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 $.

2. **Test Possible Roots**: Plug in each value into $ f(x) $ to see if it equals 0.
   - Test $ x = -2 $:

$$
   f(-2) = (-2)^3 + 4(-2)^2 + 14(-2) + 20
   $$
   $$
   = -8 + 16 - 28 + 20 = 0 
   $$

So, $ x = -2 $ is a root!

3. **Divide the Polynomial**: Now, divide $ f(x) $ by $ (x + 2) $.
   - You can use synthetic division or polynomial long division. Here’s synthetic division:

Set up:

$$
   -2 |  1 \quad 4 \quad 14 \quad 20
       |      -2 \quad -4 \quad -20
       ----------------------
         1 \quad 2 \quad 10 \quad 0
   $$

This gives $ x^2 + 2x + 10 $.

4. **Solve the Quadratic**: Now, we have $ x^2 + 2x + 10 = 0 $. Use the quadratic formula:

$$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$

Here, $ a = 1, b = 2, c = 10 $:

$$
   x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}
   $$
   $$
   = \frac{-2 \pm \sqrt{4 - 40}}{2}
   $$
   $$
   = \frac{-2 \pm \sqrt{-36}}{2} 
   $$
   $$
   = \frac{-2 \pm 6i}{2}
   $$
   $$
   = -1 \pm 3i
   $$

5. **Write the Zeros**: The zeros of the function are:

- $ x = -2 $
   - $ x = -1 + 3i $
   - $ x = -1 - 3i $

So, all the zeros are $ -2, -1 + 3i, -1 - 3i $.