1. X^4 - 2X - 8 = 0

2. 16X^4 - 1 = 0

3 -3X^3 - 18X^2 + 27X + 162 =

plz help would appreciate it

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- Mar 23rd 2007, 11:26 AMinobd22a little bit confused on these 3
1. X^4 - 2X - 8 = 0

2. 16X^4 - 1 = 0

3 -3X^3 - 18X^2 + 27X + 162 =

plz help would appreciate it - Mar 23rd 2007, 11:48 AMecMathGeek
**1.**This is not factorable. It also cannot be put into the quadratic equation. And even if you knew how to use synthetic division, on this problem it wouldn't work. Pretty much it cannot be solved without a computer. In other words, dont' worry about trying to solve number 1.

**2.**This is a difference of squares:

16x^4 = (4x^2)^2

1 = (1)^1

We can use the formula: a^2 - b^2 = (a - b)(a + b)

16x^4 - 1 = (4x^2 - 1)(4x^2 + 1)

4x^2 - 1 is also a difference of squares:

4x^2 = (2x)^2

1 = (1)^2

So using the formula to factor this, we get:

4x^2 - 1 = (2x - 1)(2x + 1) ... Now we need to put all the factors together:

16x^4 - 1 = 0

(4x^2 - 1)(4x^2 + 1) = 0

(2x - 1)(2x + 1)(4x^2 + 1) = 0

Set each factor equal to 0:

2x - 1 = 0

2x = 1

**x = 1/2**

2x + 1 = 0

2x = -1

**x = -1/2**

4x^2 + 1 = 0

4x^2 = -1

x^2 = -1/4

x = sqrt(-1/4) ... but this is the square root of a negative number, so it's undefined (unless you know how to work with the imaginary number i).

**3.**We can factor this by grouping. First, divide everything by -3:

x^3 + 6x^2 - 9x - 54 = 0

Now, group the first two terms together and the last two terms together:

(x^3 + 6x^2) + (-9x - 54) = 0

Factor out the greatest common factor in each:

x^2(x + 6) - 9(x + 6) = 0

Since (x + 6) is common in both terms, we can factor this out to get:

(x + 6)(x^2 - 9) = 0

x^2 - 9 is a difference of squares:

x^2 = (x)^2

9 = (3)^2

So,

x^2 - 9 = (x - 3)(x + 3)

Regrouping our factors we get:

x^3 + 6x^2 - 9x - 54 = 0

(x + 6)(x^2 - 9) = 0

(x + 6)(x - 3)(x + 3) = 0

Set each factor equal to 0:

x + 6 = 0

**x = -6**

x - 3 = 0

**x = 3**

x + 3 = 0

**x = -3**