Solve 5*x^2+20x+25

$$5 \cdot x ^ { 2 } + 20 x + 25$$

Factor

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

Solution Steps

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

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

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.

$$5x^{2}+20x+25=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 5\times 25}}{2\times 5}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 5\times 25}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-20±\sqrt{400-20\times 25}}{2\times 5}$$

Multiply -20 times 25.

$$x=\frac{-20±\sqrt{400-500}}{2\times 5}$$

Add 400 to -500.

$$x=\frac{-20±\sqrt{-100}}{2\times 5}$$

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

$$5x^{2}+20x+25$$

Evaluate

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