Solve x^2-x+1-3x-3

$$x ^ { 2 } - x + 1 - 3 x - 3$$

Evaluate

$$x^{2}-4x-2$$

Solution Steps

Combine -x and -3x to get -4x.

$$x^{2}-4x+1-3$$

Subtract 3 from 1 to get -2.

$$x^{2}-4x-2$$

Factor

$$\left(x-\left(2-\sqrt{6}\right)\right)\left(x-\left(\sqrt{6}+2\right)\right)$$

Steps Using the Quadratic Formula

Combine -x and -3x to get -4x.

$$factor(x^{2}-4x+1-3)$$

Subtract 3 from 1 to get -2.

$$factor(x^{2}-4x-2)$$

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.

$$x^{2}-4x-2=0$$

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.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8}}{2}$$

Add 16 to 8.

$$x=\frac{-\left(-4\right)±\sqrt{24}}{2}$$

Take the square root of 24.

$$x=\frac{-\left(-4\right)±2\sqrt{6}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{4±2\sqrt{6}}{2} when ± is plus. Add 4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+4}{2}$$

Divide 4+2\sqrt{6} by 2.

$$x=\sqrt{6}+2$$

Now solve the equation x=\frac{4±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from 4.

$$x=\frac{4-2\sqrt{6}}{2}$$

Divide 4-2\sqrt{6} by 2.

$$x=2-\sqrt{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2+\sqrt{6} for x_{1} and 2-\sqrt{6} for x_{2}.

$$x^{2}-4x-2=\left(x-\left(\sqrt{6}+2\right)\right)\left(x-\left(2-\sqrt{6}\right)\right)$$