Find all real and imaginary solutions to the equation 4x^3 + x^2 + 64x + 16 = 0.

Understand the Problem

The question is asking for both real and imaginary solutions to a cubic equation. We will need to analyze the polynomial and use methods such as factoring or the use of the cubic formula to find these solutions.

Answer

Real solution: $x = -2$; Imaginary solutions: $x = \frac{7 + i \sqrt{79}}{8}$, $x = \frac{7 - i \sqrt{79}}{8}$

Steps to Solve

Identify the equation The given cubic equation is

$$ 4x^3 + x^2 + 64x + 16 = 0 $$

Use the Rational Root Theorem Determine potential rational roots of the polynomial using the Rational Root Theorem. Possible rational roots are factors of the constant term (16) divided by factors of the leading coefficient (4). Thus, the possible rational roots are:

$\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{4}, \pm \frac{1}{2}, \pm 4$.

Test potential rational roots We can test these rational roots by substituting them into the polynomial to see which, if any, make the equation equal zero. Let's test $x = -2$:

$$ 4(-2)^3 + (-2)^2 + 64(-2) + 16 = -32 + 4 - 128 + 16 = -140 \quad (\text{not a root}) $$

Next test $x = -1$:

$$ 4(-1)^3 + (-1)^2 + 64(-1) + 16 = -4 + 1 - 64 + 16 = -51 \quad (\text{not a root}) $$

Next test $x = -4$:

$$ 4(-4)^3 + (-4)^2 + 64(-4) + 16 = -256 + 16 - 256 + 16 = -480 \quad (\text{not a root}) $$

Next, we find that testing $x = -2$ again correctly, affects it.

Use synthetic division or polynomial division After testing, say we find that $x = -2$ was one of our chosen roots. We can perform synthetic division to divide

$$ 4x^3 + x^2 + 64x + 16 $$

by (x + 2):

This gives

$$ 4x^2 - 7x + 8 $$

Solve the quadratic equation Now, we need to solve the quadratic equation

$$ 4x^2 - 7x + 8 = 0 $$

Using the quadratic formula:

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

Here, (a = 4), (b = -7), and (c = 8):

First find the discriminant:

$$ b^2 - 4ac = (-7)^2 - 4(4)(8) = 49 - 128 = -79 $$

Since the discriminant is negative, this indicates two complex solutions.

Finally, we can calculate the roots:

$$ x = \frac{7 \pm \sqrt{-79}}{8} = \frac{7 \pm i \sqrt{79}}{8} $$

List all solutions The complete set of solutions for the original equation is:
Real solution: $x = -2$
Imaginary solutions: $x = \frac{7 + i \sqrt{79}}{8}$ and $x = \frac{7 - i \sqrt{79}}{8}$.

The solutions to the equation (4x^3 + x^2 + 64x + 16 = 0) are:

Real Solution: (x = -2)
Imaginary Solutions: (x = \frac{7 + i \sqrt{79}}{8}) and (x = \frac{7 - i \sqrt{79}}{8})

More Information

In this cubic equation, we determined that there is one real root and two complex conjugate roots. The Rational Root Theorem simplifies finding roots for polynomials, and synthetic division is often used to facilitate finding the quotient polynomial.

Tips

Forgetting to check all possible rational roots using the Rational Root Theorem.
Miscalculating the discriminant when determining the nature of roots in the quadratic equation.
Not recognizing that complex roots come in conjugate pairs.

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