Solve 16x^2+8x+4

$$16 { x }^{ 2 } +8x+4$$

Factor

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

Solution Steps

Factor out 4. Polynomial 4x^{2}+2x+1 is not factored since it does not have any rational roots.

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

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.

$$16x^{2}+8x+4=0$$

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.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16\times 4}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64\times 4}}{2\times 16}$$

Multiply -64 times 4.

$$x=\frac{-8±\sqrt{64-256}}{2\times 16}$$

Add 64 to -256.

$$x=\frac{-8±\sqrt{-192}}{2\times 16}$$

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

$$16x^{2}+8x+4$$

Evaluate

$$16x^{2}+8x+4$$