Solve 4x^2+10x+8

$$4 { x }^{ 2 } +10x+8$$

Factor

$$2\left(2x^{2}+5x+4\right)$$

Solution Steps

Factor out 2. Polynomial 2x^{2}+5x+4 is not factored since it does not have any rational roots.

$$2\left(2x^{2}+5x+4\right)$$

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.

$$4x^{2}+10x+8=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 4\times 8}}{2\times 4}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-10±\sqrt{100-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-10±\sqrt{100-128}}{2\times 4}$$

Add 100 to -128.

$$x=\frac{-10±\sqrt{-28}}{2\times 4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$4x^{2}+10x+8$$

Evaluate

$$4x^{2}+10x+8$$