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2\left(x^{2}-2x+1\right)-8=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
2x^{2}-4x+2-8=0
Use the distributive property to multiply 2 by x^{2}-2x+1.
2x^{2}-4x-6=0
Subtract 8 from 2 to get -6.
x^{2}-2x-3=0
Divide both sides by 2.
a+b=-2 ab=1\left(-3\right)=-3
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
a=-3 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(x^{2}-3x\right)+\left(x-3\right)
Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).
x\left(x-3\right)+x-3
Factor out x in x^{2}-3x.
\left(x-3\right)\left(x+1\right)
Factor out common term x-3 by using distributive property.
x=3 x=-1
To find equation solutions, solve x-3=0 and x+1=0.
2\left(x^{2}-2x+1\right)-8=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
2x^{2}-4x+2-8=0
Use the distributive property to multiply 2 by x^{2}-2x+1.
2x^{2}-4x-6=0
Subtract 8 from 2 to get -6.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}
Multiply -8 times -6.
x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}
Add 16 to 48.
x=\frac{-\left(-4\right)±8}{2\times 2}
Take the square root of 64.
x=\frac{4±8}{2\times 2}
The opposite of -4 is 4.
x=\frac{4±8}{4}
Multiply 2 times 2.
x=\frac{12}{4}
Now solve the equation x=\frac{4±8}{4} when ± is plus. Add 4 to 8.
x=3
Divide 12 by 4.
x=-\frac{4}{4}
Now solve the equation x=\frac{4±8}{4} when ± is minus. Subtract 8 from 4.
x=-1
Divide -4 by 4.
x=3 x=-1
The equation is now solved.
2\left(x^{2}-2x+1\right)-8=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
2x^{2}-4x+2-8=0
Use the distributive property to multiply 2 by x^{2}-2x+1.
2x^{2}-4x-6=0
Subtract 8 from 2 to get -6.
2x^{2}-4x=6
Add 6 to both sides. Anything plus zero gives itself.
\frac{2x^{2}-4x}{2}=\frac{6}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{4}{2}\right)x=\frac{6}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-2x=\frac{6}{2}
Divide -4 by 2.
x^{2}-2x=3
Divide 6 by 2.
x^{2}-2x+1=3+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=4
Add 3 to 1.
\left(x-1\right)^{2}=4
Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-1\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x-1=2 x-1=-2
Simplify.
x=3 x=-1
Add 1 to both sides of the equation.