I REALLY need help with these problems.. its factoring polynomial equations.. and i kinda understand how to do it BUT these problems are just way over my head would love some help plz here they are
1. 8X^3 - 27 = 0
2. X^3 - 3X^2 = 16x - 48
3. X^4 - 2X - 8 = 0
4. 16X^4 - 1 = 0
5. -3X^3 - 18X^2 + 27X + 162 = 0
These are the 5 problems that i cant seem to solve... plz help would appreciate it alot
1. 1. 8X^3 - 27 = 0Originally Posted by inobd22
Notice that 8, x^3, and 27 are perfect cube terms. We can use the difference of cubes to solve this:
(a^3 - b^3) = (a - b)(a^2 + ab + b^2)
Notice that a is the cubed root of a^3 and b is the cubed root of b^3, so to solve this problem, we need the cubed roots of the two terms.
8x^3 --> 2x
27 --> 3
Therefore,
8x^3 - 27 = 0
(2x - 3)[(2x)^2 + (2x)(3) + (3)^2] = 0
(2x - 3)(4x^2 + 6x + 9) = 0
2. X^3 - 3X^2 = 16x - 48
First, we need to 'set this equal to 0' by moving everything to the left.
x^3 - 3x^2 - 16x + 48 = 0
Now, I'm going to try to factor this by grouping:
(x^3 - 3x^2) + (-16x + 48) = 0
x^2(x - 3) - 16(x - 3) = 0 ... Notice that (x - 3) is common in both terms, I can factor it to get:
(x - 3)(x^2 - 16) = 0 ... Notice that x^2 - 16 is a difference of squares:
(x - 3)(x + 4)(x - 4) = 0
3. X^4 - 2X - 8 = 0
Check to make sure you didn't write this one wrong. I doubt it should have an x^4 term.
I have to go... hopefully someone else can help solve the rest.
These are both going to involve the "Rational Root" theorem:
Given a polynomial equation
px^n + ax^{n-1} + ... + bx + q = 0
any rational roots of this equation (if rational roots exist) will be of the form:
x = (+/-){factor of q}/{factor of p}
So for 3):
x^4 - 2x - 8 = 0
The possible rational roots will be:
x = (+/-)1, 2, 4, 8
Unfortunately none of these work. So a numerical solution is going to be your best bet. I get two real roots:
x = -1.49595 and x = 1.84942
Now you can do some polynomial division and use the quadratic formula to find the two remaining complex solutions:
x = -0.1769734 (+/-)1.69126*I (where I = sqrt(-1), with due apologies to TPH!)
For 5):
-3x^3 - 18x^2 + 27x + 162 = 0
The possible rational roots are:
x = (+/-)1, 2, 3, 6, 27, 54, 81, 162
We find that x = -6, (+/-)3. Since there are 3 solutions listed here, we are done.
-Dan
PS Heh. I always do things the hard way. You can divide the equation in 5 by a -3 to get:
x^3 + 6x^2 - 9x - 54 = 0
which is slightly easier to solve.
-Dan
This is the difference of two squares, and thus factors:
16x^4 - 1 = 0
(4x^2)^2 - 1^2 = 0
(4x^2 - 1)(4x^2 + 1) = 0
Set each factor to 0:
a) 4x^2 - 1 = 0
b) 4x^2 + 1 = 0
Let's do a).
4x^2 - 1 = 0 <-- Again the difference of two squares.
(2x)^2 - 1^2 = 0
(2x - 1)(2x + 1) = 0
Set each factor to 0:
2x - 1 = 0 ==> x = 1/2
2x + 1 = 0 ==> x = -1/2
Now do b).
4x^2 + 1 = 0
x^2 = -1/4
x = (+/-)I/2 (where I = sqrt(-1).)
So the 4 solutions are
x = (+/-)1/2, (+/-)I/2
-Dan
Factor by grouping this one as well:
(x^3 + 6x^2) + (-9x - 54) = 0
x^2(x + 6) - 9(x + 6) = 0
(x + 6)(x^2 - 9) = 0
(x + 6)(x - 3)(x + 3) = 0
By the way, thanks for helping in the last problems, topsquark.
P.S. I really think problem 3 was a typo because I doubt that inobd22 knows about polynomial devision. Even then, the answer came out irrational - which makes it a 'slightly' harder problem than the other 4.
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