1. X⁵-4x⁴-2x³-2x³+4x²+x=0
2. X³-6x²+11x-6=0
3. X⁴+4x³-3x²-14x=8
4. X⁴-2x³-2x²=0
Find the roots for these problem show your work​

So the roots of the original equation are:

x = 0, x = 1 + √3, x = 1 - √3

Let's solve each of these equations and find their roots.

x⁵ - 4x⁴ - 2x³ - 2x³ + 4x² + x = 0:

To factorize this equation, we can factor out an "x" term:

x(x⁴ - 4x³ - 4x² + 4x + 1) = 0

Now, we have two factors:

x = 0

To find the roots of the second factor, x⁴ - 4x³ - 4x² + 4x + 1 = 0, we can use numerical methods or approximation techniques.

Unfortunately, this equation does not have any simple or rational roots. The approximate solutions for this equation are:

x ≈ -1.2385

x ≈ -0.4516

x ≈ 0.2188

x ≈ 3.4714

x³ - 6x² + 11x - 6 = 0:

This equation can be factored using synthetic division or by guessing and checking.

One possible root of this equation is x = 1.

By performing synthetic division, we can obtain the following factorization:

(x - 1)(x² - 5x + 6) = 0

Now, we have two factors:

x - 1 = 0

x = 1

x² - 5x + 6 = 0

To find the roots of the quadratic equation x² - 5x + 6 = 0, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -5, and c = 6.

Substituting these values into the quadratic formula, we get:

x = (5 ± √(25 - 24)) / 2

x = (5 ± √1) / 2

x = (5 ± 1) / 2

So the roots of the quadratic equation are:

x ≈ 2

x ≈ 3

Therefore, the roots of the original equation are:

x = 1, x ≈ 2, x ≈ 3

x⁴ + 4x³ - 3x² - 14x = 8:

To solve this equation, we need to move all the terms to one side to obtain a polynomial equation equal to zero:

x⁴ + 4x³ - 3x² - 14x - 8 = 0

Unfortunately, this equation does not have any simple or rational roots. We can use numerical methods or approximation techniques to find the roots.

Approximate solutions for this equation are:

x ≈ -2.5223

x ≈ -0.4328

x ≈ 1.6789

x ≈ 3.2760

x⁴ - 2x³ - 2x² = 0:

To solve this equation, we can factor out an "x²" term:

x²(x² - 2x - 2) = 0

Now, we have two factors:

x² = 0

x = 0

x² - 2x - 2 = 0

To find the roots of the quadratic equation x² - 2x - 2 = 0, we can again use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -2, and c = -2. Substituting these values into the quadratic formula, we get:

x = (2 ± √(4 - 4(1)(-2))) / (2(1))

x = (2 ± √(4 + 8)) / 2

x = (2 ± √12) / 2

x = (2 ± 2√3) / 2

x = 1 ± √3

So the roots of the original equation are:

x = 0, x = 1 + √3, x = 1 - √3

For similar questions on roots

https://brainly.com/question/14807889

#SPJ8