Solve left(-3+4x^2-4xright)+left(-1-3x^2-4right)

$$\left( -3+4 { x }^{ 2 } -4x \right) + \left( -1-3 { x }^{ 2 } -4 \right)$$

Evaluate

$$x^{2}-4x-8$$

Solution Steps

Subtract 1 from -3 to get -4.

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

Combine 4x^{2} and -3x^{2} to get x^{2}.

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

Subtract 4 from -4 to get -8.

$$-8+x^{2}-4x$$

Factor

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

Steps Using the Quadratic Formula

Subtract 1 from -3 to get -4.

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

Combine 4x^{2} and -3x^{2} to get x^{2}.

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

Subtract 4 from -4 to get -8.

$$factor(-8+x^{2}-4x)$$

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.

$$x^{2}-4x-8=0$$

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.

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

Square -4.

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

Multiply -4 times -8.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2}$$

Add 16 to 32.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{3}}{2}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{2} when ± is plus. Add 4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+4}{2}$$

Divide 4+4\sqrt{3} by 2.

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

Now solve the equation x=\frac{4±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 4.

$$x=\frac{4-4\sqrt{3}}{2}$$

Divide 4-4\sqrt{3} by 2.

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

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

$$x^{2}-4x-8=\left(x-\left(2\sqrt{3}+2\right)\right)\left(x-\left(2-2\sqrt{3}\right)\right)$$