How do I factor this?

Question

How do I factor this:

1x^3 - 4x^2 + 8x - 8 = 0

So far, I am only able to get:

(1x^3 - 4x^2) (+ 8x - 8) = 0
1x^2 (1x - 4) + 8(1x - 1) = 0

The two sets of parenthesis don't match which is where I am stuck. Anyone please help!! Thanks.

Answer 1

Let p(x) = x^3 - 4x^2 + 8x - 8

To factor this, what you want to test are factors of the numeric term (the one with no variables), -8. That is, you want to test:
-1, 1, -2, 2, -4, 4, -8, 8

What you're going to do is plug these values into the function to see if it gives you 0. Once you get some value r, then (x - r) will be a factor, and you will use synthetic long division. Let's start testing.

p(-1) = (-1)^3 - 4(-1)^2 + 8(-1) - 8 = -1 - 4 - 8 - 8 = -21
p(1) = 1^3 - 4(1)^2 + 8(1) - 8 = 1 - 4 + 8 - 8 = -3
p(-2) = (-2)^3 - 4(-2)^2 + 8(-2) - 8 = -8 - 16 - 16 - 8 = [nonzero]
p(2) = 2^3 - 4(2)^2 + 8(2) - 8 = 8 - 1 + 16 - 8 = 0

Since 2 is a root, (x - 2) is a factor.

Use synethetic long division; x - 2 INTO x^3 - 4x^2 + 8x - 8.
I'm not going to show you the details (since long division is extremely difficult to do on here), but you should get a quotient of
x^2 - 2x + 4 and a remainder of 0.

This means
x^3 - 4x^2 + 8x - 8 = (x - 2) (x^2 - 2x + 4)

Equating this to 0,

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

And now you just solve for each factor. The first one is easy (x = 2); the second one requires the quadratic formula, but I'll just complete the square.

x^2 - 2x + 4 = 0
x^2 - 2x + 1 + 3 = 0
(x - 1)^2 + 3 = 0
(x - 1)^2 = -3
x -1 = +/- sqrt(3)i
x = 1 +/- sqrt(3)i

Therefore, your three roots are:
x = {2, 1 + sqrt(3)i, 1 - sqrt(3)i}

Answer 2

Ooops! ( x^3 - ....) x (8x -...) = 8x^4 +-.... you first solution line is messed up big time.

you're trying to factor it? not simplify it. Don't put a 1 in front of the x because 1x = x last I heard... I guess it might be good for a while if you're new to it....Oh well never mind.

you factor it by coming up with factors.

x-4 or x-1 are NOT factors. Review what a factor is.

A factor is a number which when **multiplied** by other numbers gives the result, equivalently for polynomials. A polynomial is factored by discovering the simplest factors which when multiplied together make the polynomiial.

x^2+4x+4 = (x+2)(x+2)

in your above case, once you have enough practice, you will be able to factor it pretty easily by inspection. I am so far out of practice that I had to use trial and error.
I tried X+1, and x-1 but they didnt go into the polynomial evenly.
(you divide using long division but if you don't know what I'm taling about,I can't go into more detail here except to say guessing at x-1 as a divisor for x^3-4x^2+8x-8 the first factor would be X^2 so x-1*(x^2)= x^3-x^2 and subtracting that give a remainder of -3x^2+8x-8 making the next part of the factor -3x
so choosing x-1 and using "long" division you build up the other "factor" x^2-3x+.... term by term.
Anyway I got to x-2 and voila! it factored it evenly, with no remainder. Then I looked at the polynomial and realized that the first and last terms were perfect cubes (one of x, the other of 2)
asking you to try the (easy) result (x-2)^3. but the "long division" method will work if you're out of all other choices, too.

So (x-2)(x-2)(x-2) are the factors (or is it "is the factor"?).

Answer 3

1x^3 - 4x^2 + 8x - 8 = 0 1 root is x=2 s0 (x-2) is a factor

x^2 - 2x + 4
(x-2\x^3-4x^2+8x-8
-(x^2-2x^2)
-2x^2+8x
-(-2x^2+4x)
4x - 8
-(4x-8)
0

(x-2)( x^2 - 2x + 4)
there are no more real factors. You can use the quadratic equation to find the complex factors.

Answer 4

Ok well it's been two years since I've done this, but if I remember correctly then this is what I've come up with, depending on how far you have to go.
1x^3-4x^2+8x^8=0
x^2(x-4+8x^6)=0
I hope that helps.

Answer 5

It factors down to (x-2)(x^2-2x+4). You can't factor it any further using only real numbers.

A quick way to determine if (x-n) is a factor is to use the Root Theorem (I don't remember its actual name), which states that if (x-n) is a factor of a polynomial p(x), then n is a root of p(x)=0 and vice versa. Trying a few values for n, I discovered that 2 works.

Answer 6

f(x) = 1x^3 - 4x^2 + 8x - 8
f(2) = 0
f(x) = (x-2)(x^2-2x+4)

x = 2, 1±√3 i

Answer 7

1x^3 - 4x^2 + 8x - 8 = 0
(x-2)(Ax²+Bx+C)=1x^3 - 4x^2 + 8x - 8
Ax³+Bx²+Cx-2Ax²-2Bx-2c=1x^3 - 4x^2 + 8x - 8

By comparing coeff. of x³:
A=1

By comparing coeff. of x²:
B-2A=-4
B-2=-4
B=-2

-2C=-8
C=4

Substitute A,B,C back into (x-2)(Ax²+Bx+C)=1x^3 - 4x^2 + 8x - 8
(x-2)(x²-2x+4)=1x^3 - 4x^2 + 8x - 8
Since,
B²-4AC=-12
(x²-2x+4) can't be factorise any longer.Hence its
(x-2)(x²-2x+4).

Answer 8

I'm probably wrong but I think
x^2(x - 4) + 8(x-1) =0

Answer 9

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Answer 10

use a grapher