Solve -15f-10f^2

$$- 15 f - 10 f ^ { 2 }$$

Factor

$$-5f\left(2f+3\right)$$

Solution Steps

Factor out 5.

$$5\left(-3f-2f^{2}\right)$$

Consider -3f-2f^{2}. Factor out f.

$$f\left(-3-2f\right)$$

Rewrite the complete factored expression.

$$5f\left(-2f-3\right)$$

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.

$$-10f^{2}-15f=0$$

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.

$$f=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\left(-10\right)}$$

Take the square root of \left(-15\right)^{2}.

$$f=\frac{-\left(-15\right)±15}{2\left(-10\right)}$$

The opposite of -15 is 15.

$$f=\frac{15±15}{2\left(-10\right)}$$

Multiply 2 times -10.

$$f=\frac{15±15}{-20}$$

Now solve the equation f=\frac{15±15}{-20} when ± is plus. Add 15 to 15.

$$f=\frac{30}{-20}$$

Reduce the fraction \frac{30}{-20} to lowest terms by extracting and canceling out 10.

$$f=-\frac{3}{2}$$

Now solve the equation f=\frac{15±15}{-20} when ± is minus. Subtract 15 from 15.

$$f=\frac{0}{-20}$$

Divide 0 by -20.

$$f=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{2} for x_{1} and 0 for x_{2}.

$$-10f^{2}-15f=-10\left(f-\left(-\frac{3}{2}\right)\right)f$$

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

$$-10f^{2}-15f=-10\left(f+\frac{3}{2}\right)f$$

Add \frac{3}{2} to f by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-10f^{2}-15f=-10\times \frac{-2f-3}{-2}f$$

Cancel out 2, the greatest common factor in -10 and -2.

$$-10f^{2}-15f=5\left(-2f-3\right)f$$

Evaluate

$$-5f\left(2f+3\right)$$