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

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

Evaluate

$$5-x-2x^{2}$$

Solution Steps

Combine -x and x to get 0.

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

Add -1 and 3 to get 2.

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

Combine -2x^{2} and x^{2} to get -x^{2}.

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

Add 2 and 3 to get 5.

$$-x^{2}-x^{2}+5-x$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$-2x^{2}+5-x$$

Factor

$$-2\left(x-\frac{-\sqrt{41}-1}{4}\right)\left(x-\frac{\sqrt{41}-1}{4}\right)$$

Steps Using the Quadratic Formula

Combine -x and x to get 0.

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

Add -1 and 3 to get 2.

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

Combine -2x^{2} and x^{2} to get -x^{2}.

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

Add 2 and 3 to get 5.

$$factor(-x^{2}-x^{2}+5-x)$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$factor(-2x^{2}+5-x)$$

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.

$$-2x^{2}-x+5=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\times 5}}{2\left(-2\right)}$$

Multiply 8 times 5.

$$x=\frac{-\left(-1\right)±\sqrt{1+40}}{2\left(-2\right)}$$

Add 1 to 40.

$$x=\frac{-\left(-1\right)±\sqrt{41}}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{41}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±\sqrt{41}}{-4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{-4} when ± is plus. Add 1 to \sqrt{41}.

$$x=\frac{\sqrt{41}+1}{-4}$$

Divide 1+\sqrt{41} by -4.

$$x=\frac{-\sqrt{41}-1}{4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{-4} when ± is minus. Subtract \sqrt{41} from 1.

$$x=\frac{1-\sqrt{41}}{-4}$$

Divide 1-\sqrt{41} by -4.

$$x=\frac{\sqrt{41}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1-\sqrt{41}}{4} for x_{1} and \frac{-1+\sqrt{41}}{4} for x_{2}.

$$-2x^{2}-x+5=-2\left(x-\frac{-\sqrt{41}-1}{4}\right)\left(x-\frac{\sqrt{41}-1}{4}\right)$$