# solve by factoring 4x^2-8x+4=0must show work

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### 2 Answers

4x^2-8x+4=0 by factoring

Factor out the GCF of 4 from each term in the polynomial.

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

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

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

Combine the two common factors of (x-1) by adding the exponents.

4 (x-1)^2 =0

Divide both sides of the equation by 4 which leaves

(x-1)^2 = 0

Set each factor equal to 0

x-1 = 0

Add 1 to each side to solve for x

**x=1**

** **

The equation 4x^2-8x+4=0 has to be solved.

4x^2-8x+4=0

Factor the number 4 that is common to all the terms.

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

Now write -2 as a sum of two numbers t1 and t2 such that their sum t1 + t2 = -2 and the product t1*t2 = 1*1 = 1

-2 can be written as the sum of -1 and -1. This gives:

4*(x^2 - x - x + 1) = 0

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

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

x - 1 = 0

x = 1

The solution of the equation 4x^2-8x+4=0 is x = 1