Solve 4x^2+2x-1

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

Factor

$$4\left(x-\frac{-\sqrt{5}-1}{4}\right)\left(x-\frac{\sqrt{5}-1}{4}\right)$$

Steps Using the Quadratic Formula

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.

$$4x^{2}+2x-1=0$$

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.

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

Square 2.

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

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\left(-1\right)}}{2\times 4}$$

Multiply -16 times -1.

$$x=\frac{-2±\sqrt{4+16}}{2\times 4}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2\times 4}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2\sqrt{5}}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{8} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{8}$$

Divide -2+2\sqrt{5} by 8.

$$x=\frac{\sqrt{5}-1}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{8} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{8}$$

Divide -2-2\sqrt{5} by 8.

$$x=\frac{-\sqrt{5}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{5}}{4} for x_{1} and \frac{-1-\sqrt{5}}{4} for x_{2}.

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

Evaluate

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