Solve 2f^2+5f

$$2 f ^ { 2 } + 5 f$$

Factor

$$f\left(2f+5\right)$$

Solution Steps

Factor out f.

$$f\left(2f+5\right)$$

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.

$$2f^{2}+5f=0$$

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.

$$f=\frac{-5±\sqrt{5^{2}}}{2\times 2}$$

Take the square root of 5^{2}.

$$f=\frac{-5±5}{2\times 2}$$

Multiply 2 times 2.

$$f=\frac{-5±5}{4}$$

Now solve the equation f=\frac{-5±5}{4} when ± is plus. Add -5 to 5.

$$f=\frac{0}{4}$$

Divide 0 by 4.

$$f=0$$

Now solve the equation f=\frac{-5±5}{4} when ± is minus. Subtract 5 from -5.

$$f=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$f=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -\frac{5}{2} for x_{2}.

$$2f^{2}+5f=2f\left(f-\left(-\frac{5}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$2f^{2}+5f=2f\left(f+\frac{5}{2}\right)$$

Add \frac{5}{2} to f by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2f^{2}+5f=2f\times \frac{2f+5}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2f^{2}+5f=f\left(2f+5\right)$$

Evaluate

$$f\left(2f+5\right)$$