Solve 4x^2+11x+5

$$4 { x }^{ 2 } +11x+5$$

Factor

$$4\left(x-\frac{-\sqrt{41}-11}{8}\right)\left(x-\frac{\sqrt{41}-11}{8}\right)$$

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.

$$4x^{2}+11x+5=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 4\times 5}}{2\times 4}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-11±\sqrt{121-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$x=\frac{-11±\sqrt{121-80}}{2\times 4}$$

Add 121 to -80.

$$x=\frac{-11±\sqrt{41}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-11±\sqrt{41}}{8}$$

Now solve the equation x=\frac{-11±\sqrt{41}}{8} when ± is plus. Add -11 to \sqrt{41}.

$$x=\frac{\sqrt{41}-11}{8}$$

Now solve the equation x=\frac{-11±\sqrt{41}}{8} when ± is minus. Subtract \sqrt{41} from -11.

$$x=\frac{-\sqrt{41}-11}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-11+\sqrt{41}}{8} for x_{1} and \frac{-11-\sqrt{41}}{8} for x_{2}.

$$4x^{2}+11x+5=4\left(x-\frac{\sqrt{41}-11}{8}\right)\left(x-\frac{-\sqrt{41}-11}{8}\right)$$

Evaluate

$$4x^{2}+11x+5$$