Solve 3x^2+2x^2-3x^2+8x-4

$$3 x ^ { 2 } + 2 x ^ { 2 } - 3 x ^ { 2 } + 8 x - 4$$

Evaluate

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

Solution Steps

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$5x^{2}-3x^{2}+8x-4$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

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

Factor

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

Steps Using the Quadratic Formula

Combine 3x^{2} and 2x^{2} to get 5x^{2}.

$$factor(5x^{2}-3x^{2}+8x-4)$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

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

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.

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

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.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\left(-4\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-8±\sqrt{64+32}}{2\times 2}$$

Add 64 to 32.

$$x=\frac{-8±\sqrt{96}}{2\times 2}$$

Take the square root of 96.

$$x=\frac{-8±4\sqrt{6}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±4\sqrt{6}}{4}$$

Now solve the equation x=\frac{-8±4\sqrt{6}}{4} when ± is plus. Add -8 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}-8}{4}$$

Divide -8+4\sqrt{6} by 4.

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

Now solve the equation x=\frac{-8±4\sqrt{6}}{4} when ± is minus. Subtract 4\sqrt{6} from -8.

$$x=\frac{-4\sqrt{6}-8}{4}$$

Divide -8-4\sqrt{6} by 4.

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

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}.

$$2x^{2}+8x-4=2\left(x-\left(\sqrt{6}-2\right)\right)\left(x-\left(-\sqrt{6}-2\right)\right)$$