Help with math (Algebra II) factoring.?

Question

Solve the following equations...

^=powerof

1. 8x^5-6x^2+12x^3-9=0

2. x^3=9x

3. 2x^4+5x^3+4x^2+5x=-2


If you don't want to do all, can you do one and show a relatively detailed process of how you figured it out. Thanks! I really need to figure these out soon.

Answer 1

PROBLEM 1:

Start by factoring a common 2x² out of the first two terms:
2x²(4x^3 - 3) + 12x^3 - 9 = 0

Now factor a common 3 out of the last two terms:
2x²(4x^3 - 3) + 3(4x^3 - 3) = 0

Ah, now you can factor out the common 4x^3 - 3:
(2x² + 3)(4x^3 - 3) = 0

Now you have two equations:
2x² + 3 = 0
and
4x^3 - 3 = 0

Solving them individually:
2x² = -3
Hmm... that will give you x² = -3/2 and will lead to a pair of imaginary roots.

4x^3 - 3 = 0
x^3 = 3/4
x = cuberoot(6/8)
x = cuberoot(6) / cuberoot (8)
x = cuberoot(6) / 2

PROBLEM 2:

Let's put everything on one side:

x^3 - 9x = 0

Factor the common x:
x(x² - 9) = 0

Now you have a difference of squares:
x(x - 3)(x + 3) = 0

x = 0, x = 3, x = -3

PROBLEM 3:

Here you need to try to find some possible values first. Again write it with everything on one side:
2x^4 + 5x^3 + 4x^2 + 5x + 2 = 0

Possible answers will be the factors of the last coefficient (2), divided by factors of the first coefficient (2 in front of x^4). This includes positive and negative factors:

Choices: -2, -1, 1, 2.
Possible roots:
(-2, -1, 1, 2, -1/2, 1/2).

Try these until you find a couple of roots. I found x = -2 and x = -1/2 gave a zero root, so (x + 2) and (x + 1/2) can be factored out.

(x + 2)(x + 1/2) = x² + 2½x + 1

I like to double this to get rid of fractions:
2x² + 5x + 2

Now use synthetic division to divide the original equation by this:
................................................. x² ........ + 1
2x² + 5x + 2 ) 2x^4 + 5x^3 + 4x² + 5x + 2
...................... 2x^4 + 5x^3 + 2x²
..................... ......................... 2x² + 5x + 2
..................... ......................... 2x² + 5x + 2

So you get x² + 1 as a final factor (more imaginary roots).
(x + 2)(2x + 1)(x² + 1) = 0

Solutions:
x = -2, x = -1/2, x = +/-i

Answer 2

the respond is 4. You attain that answer with the help of: x^2 - 4x = 0; upload 4x to the suited component providing you with x^2 = 4x; subtract an x from the two component and you get x=4. Plugging it back into the equation, 4^2 - 4(4) = 0

Answer 3

1. 8x^5+12x^3-6x^2-9=0
(8x^5+12x^3)+(-6x^2-9)=0
4x^3(2x^2+3)-3x(2x^2+3)=0
(4x^3-3x)(2x^2+3)=0
x(4x^2-3)(2x^2+3)=0
x=0 4x^2-3=0 2x^2+3=0
x=SR 3/4 x= SR -3/2

2. x^3-9x=0
x(x^2-9)=0
x(x-3)(x+3)=0
x=0 x=3 x= -3

3. 2x^4+5x^3+4x^2+5x+2=0
(2x^4+4x^2+2)+(5x^3+5x)=0
(2x^2+2)(x^2+1) + 5x(x^2+1)=0
2x^2+2=0 x^2+1=0 5x=0 x^2+1=0
x=SR -1 x=SR -1 x=0 x=SR -1

Answer 4

i can technically give you answers for numbers 1 and 3, but they're very ugly looking and are probably not the right answers. as for number two, that ones easy...

-take the "x" on the "9x" and divide both sides by that "x"
-this leaves you with (x^3/x)=9
-the (x^3/x) also means (x^3-x^1) which becomes (x^2)
-so now you've got (x^2=9)
-if you take the square root of both sides it becomes (x=+3 and x=-3)

i believe that's right.

Answer 5

2. x^3=9x
x^3 - 9x = 0
x( x^2 - 9) = 0
x (x - 3) (x + 3) = 0