Factor x^2-4x-4?

Question

This is for a partial decomposition problem I'm working on.
Im factoring the denominator and im left with
(x+3)(x^2-4x-4)
I cant seem to factor the quadratic trinomial.
My calculator tells me values that I don't get by hand.
Someone help me out and show the steps.

Show me how to arrive to this
x-2(sqrt2+1)
x+2(sqrt2-1)
When expanded these are equal to
x^2-4x-4

Answer 1

Use the quadratic equation to factor x^2 - 4x - 4.

You get: 4 +/- squareroot (16-4*(-4))
-----------------------------------------
2

= 4+/- squareroot 32
---------------------------
2

= 2+/- 2*squareroot 2

Therefore, x^2-4x-4 factors into (x - 2 + 2*squareroot 2) and (x - 2 -2*squareroot 2)

Answer 2

X 2 4x 4

Answer 3

Factor x^2-4x-4
Let us try to make this a perfect square, as we do with
A -2-degree EU will have 2 equal roots if b^2 = 4ac

ax^2+bx+c=0 what we do that we add and subtract c' = b^2/4a in the given quadratic. then take + c' along into square. (x - b/2a)^2 - (h)^2 form will come:

c' = 16/4 =4

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

(x - 2)^2 - [2 sqrt(2)]^2 >>>>> (A^2 - B^2) .. (A+B) (A-B)

[x - 2 + 2 sqrt(2)] [x -2 - 2 sqrt(2)] ====(1)

Test: output is A^2 - B^2 = (x-2)^ - 8 = x^2 - 4x -4 simple

roots
x = 2 [1- sqrt(2)]
x = 2 [1+ sqrt(2)]

now you can use in putting these into fraction of polynomial

Answer 4

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RE:
Factor x^2-4x-4?
This is for a partial decomposition problem I'm working on.
Im factoring the denominator and im left with
(x+3)(x^2-4x-4)
I cant seem to factor the quadratic trinomial.
My calculator tells me values that I don't get by hand.
Someone help me out and show the steps.

Answer 5

It's been a while since I had Algebra, but I don't think it's (x-2)(x-2). Wouldn't that be x^2-4x+4?

Answer 6

Both of their answers are wrong unless you mistyped the sign for the 4 at the end of the equation. You are not going to get a whole number to factor it down to. So that's is as far as it can go unless you use the quadratic formula.

Answer 7

when you can't find the numbers, use the quadratic formula:

a=1
b=-4
c=-4

so you get x=[4+-sqrt(16+16)]/2 =2+-2sqrt(2)
so your problem is now:

[x-2+2sqrt(2)][x-2-2sqrt(2)] ( you change the sign of the solution because you are putting it in the polynomial)
Or,
(x-0.82842712474619)(x-4.82842712474619)

Answer 8

It's not factorable.

Answer 9

(x-2)(x-2) doesnt work for this equation since the product of -2 and -2 is 4 not -4..there has to be some mistake with the question itself when u derived the equation..

Answer 10

You guys just dont know how to factor it factors out