# `4x^2- 8x + 4 = 0` Solve by factoring

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### 6 Answers

You should write the constant term 4 as 8 - 4 such that:

`4x^2 - 8x + 8 - 4 = 0`

You need to form 2 groups such that:

`(4x^2 - 4) - (8x - 8) = 0`

You need to factor out 4 in the first group and 8 in the second group such that:

`4(x^2 - 1) - 8(x - 1) = 0`

You should write the difference of squares `x^2 - 1` as a product such that:

4(x-1)(x+1) - 8(x-1) = 0

You need to factor out (x-1) such that:

`(x-1)[4(x+1) - 8] = 0`

`(x-1)(4x + 4 - 8) = 0`

`(x-1)(4x - 4) = 0`

`4(x-1)(x-1) = 0 =gt 4(x-1)^2 = 0`

You need to find the zeroes of the equation `4(x-1)^2 = 0 =gt x-1=0 =gt x_(1,2) = 1`

**Hence, the solutions to the quadratic are `x_(1,2) = 1` .**

Solve `4x^2-8x+4=0` :

`4(x^2-2x+1)=0` factor out the common 4

`4(x-1)^2=0` recognize `x^2-2x+1` as a perfect square trinomial

`(x-1)^2=0 ==> x-1=0 ==> x=1`

From the zero-product property we know that either 4=0 or `(x-1)^2=0` . Since the former is impossible, the result follows.

**The solution is x=1.**

** If you did not recognize `x^2-2x+1` as a perfect square trinomial, you can still factor -- find two numbers p and q such that pq=1 and p+q=-2. The only possibility is for p=q=-1 so you have `(x^2-2x+1)=(x-1)(x-1)=(x-1)^2`

**Sources:**

4x^2 - 8x + 4 = 0

divide by 4

x^2 - 2x + 1 = 0

difference of 2 squares:

( x - 1)(x-1)

Set equal to 0 and solve.

x - 1 = 0

x = 1

4x^2 - 8x + 4 = 0

divide everything out by 4

x^2 - 2x + 1 = 0

Its a perfect square which equals ( x - 1)^2

Set that equal to 0 and solve.

(x - 1)^2 = 0

x - 1 = 0

x = 1

`a=4 ` `b=-8 ` `c=4 ` fist multiply axxc

`4x4=16` now find facts of 16 that add up to -8 that would be -4 and -4 plug those numbers in

`4x^2-4x-4x+4` group the terms

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

`4x(x-1) -4(x-1)`

`(x-1)(4x-4) ` set them equal to 0

`x-1=0`

`x=1`

`4x-4=0`

`4x=4`

`(4x)/4 = 4/4`

`x=1`

**x is equal to 1**