If 5x4-14x³+18x²+40x+16=(x²-4x+8)(ax²+bc+c) find a,b and c and hence find the four solutions of the equation 5x4-14x³+18x²+40x+16

Given that x³-1=(x-&)(ax²+bx+c) find the values of a,b and c and hence find the three roots of the equation x³=1
{These three roots can be referred to as 'the cube roots of unity' and are often written as 1, w, w²}

To find the values of a, b, and c in the equation 5x⁴ - 14x³ + 18x² + 40x + 16 = (x² - 4x + 8)(ax² + bc + c), we can expand the right side of the equation and compare it with the given equation.

Expanding (x² - 4x + 8)(ax² + bc + c):
(ax² + bc + c)(x² - 4x + 8) = ax^4 - 4ax³ + 8ax² + bcx² - 4bcx + 8bc + cx² - 4cx + 8c

Now we can compare the expanded equation with the given equation:

ax^4 - 4ax³ + 8ax² + bcx² - 4bcx + 8bc + cx² - 4cx + 8c = 5x⁴ - 14x³ + 18x² + 40x + 16

Comparing the coefficients of like terms on both sides, we get:

a = 5 (coefficient of x⁴)
-4a + bc - 4b = -14 (coefficient of x³)
8a + c - 4c = 18 (coefficient of x²)
8b - 4c = 40 (coefficient of x)
8c = 16 (constant term)

From the last equation, we can find the value of c:
8c = 16
c = 2

Using the value of c, we can solve the third equation:
8a + c - 4c = 18
8a + 2 - 4(2) = 18
8a - 8 = 18 - 2
8a = 16
a = 2

Now we have the values of a and c. We can substitute them into the second equation:
-4a + bc - 4b = -14
-4(2) + 2b - 4b = -14
-8 + 2b - 4b = -14
-8 -2b = -14
-2b = -14 + 8
-2b = -6
b = 3

Therefore, the values of a, b, and c in the equation (x² - 4x + 8)(ax² + bc + c) are a = 2, b = 3, and c = 2.

To find the four solutions of the equation 5x⁴ - 14x³ + 18x² + 40x + 16, we can set the equation equal to zero:

5x⁴ - 14x³ + 18x² + 40x + 16 = 0

Now we can factor the equation using the values of a, b, and c that we found:

(x² - 4x + 8)(2x² + 3x + 2) = 0

To find the solutions, we can set each factor equal to zero and solve for x:

x² - 4x + 8 = 0
Using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)
x = (4 ± √((-4)² - 4(1)(8))) / (2(1))
x = (4 ± √(16 - 32)) / 2
x = (4 ± √(-16)) / 2
x = (4 ± 4i) / 2
x = 2 ± 2i

Therefore, the first two solutions of the equation are x = 2 + 2i and x = 2 - 2i.

Now let's solve for the second factor:

2x² + 3x + 2 = 0
Using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)
x = (-3 ± √(3² - 4(2)(2))) / (2(2))
x = (-3 ± √(9 - 16)) / 4
x = (-3 ± √(-7)) / 4

Since there is a square root of a negative number, we can conclude that the remaining two solutions of the equation are complex numbers.

In conclusion, the four solutions of the equation 5x⁴ - 14x³ + 18x² + 40x + 16 are:
x = 2 + 2i, x = 2 - 2i, and two complex solutions from (-3 ± √(-7)) / 4.