Solve 1-x^2-2x+1

$$1 - x ^ { 2 } - 2 x + 1$$

Factor

$$-\left(x-\left(-\sqrt{3}-1\right)\right)\left(x-\left(\sqrt{3}-1\right)\right)$$

Steps Using the Quadratic Formula

Add 1 and 1 to get 2.

$$factor(2-x^{2}-2x)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-x^{2}-2x+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2\left(-1\right)}$$

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{3}}{-2}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{-2} when ± is plus. Add 2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{-2}$$

Divide 2+2\sqrt{3} by -2.

$$x=-\left(\sqrt{3}+1\right)$$

Now solve the equation x=\frac{2±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$x=\frac{2-2\sqrt{3}}{-2}$$

Divide 2-2\sqrt{3} by -2.

$$x=\sqrt{3}-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\left(1+\sqrt{3}\right) for x_{1} and -1+\sqrt{3} for x_{2}.

$$-x^{2}-2x+2=-\left(x-\left(-\left(\sqrt{3}+1\right)\right)\right)\left(x-\left(\sqrt{3}-1\right)\right)$$

Evaluate

$$2-2x-x^{2}$$

Solution Steps

Add 1 and 1 to get 2.

$$2-x^{2}-2x$$