Find all real and imaginary solutions to the equation 4x^3 - 7x^2 - 16x + 28 = 0.

Steps to Solve

1. Set up the equation

We are given the cubic equation:
$$ 4x^3 - 7x^2 - 16x + 28 = 0 $$

2. Find Rational Roots

Use the Rational Root Theorem to identify possible rational roots. The factors of the constant term (28) are:
$$ \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 $$
And the factors of the leading coefficient (4) are:
$$ \pm 1, \pm 2, \pm 4 $$
The possible rational roots are the ratios of these factors.

3. Test possible roots

We can test some possible rational roots. Let's test ( x = 2 ): $$ 4(2)^3 - 7(2)^2 - 16(2) + 28 = 4(8) - 7(4) - 32 + 28 $$ $$ = 32 - 28 - 32 + 28 = 0 $$ Therefore, ( x = 2 ) is a root.

4. Perform synthetic division

Now, use synthetic division to divide the polynomial by ( x - 2 ):

2 |  4  -7  -16  28 
    |      8   2 -28
---------------------
      4   1  -14  0

The quotient is:
$$ 4x^2 + x - 14 $$

5. Factor or use the quadratic formula on the quotient

Next, we need to solve:
$$ 4x^2 + x - 14 = 0 $$
Using the quadratic formula where ( a = 4, b = 1, c = -14 ): $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Calculating the discriminant:
$$ b^2 - 4ac = 1^2 - 4(4)(-14) = 1 + 224 = 225 $$

Now plug this into the formula:
$$ x = \frac{-1 \pm \sqrt{225}}{2(4)} = \frac{-1 \pm 15}{8} $$

6. Calculate the roots

Calculating the two roots:

• For ( +15 ): $$ x = \frac{14}{8} = \frac{7}{4} $$

• For ( -15 ): $$ x = \frac{-16}{8} = -2 $$

7. Summary of the solutions

The solutions to the original equation are:

• Real solutions: ( x = 2, \frac{7}{4}, -2 )
• No imaginary solutions since all roots are real.