Solve f(x)=-8x^2-3x

$$f ( x ) = - 8 x ^ { 2 } - 3 x$$

Factor

$$-x\left(8x+3\right)$$

Solution Steps

Factor out x.

$$x\left(-8x-3\right)$$

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.

$$-8x^{2}-3x=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\left(-8\right)}$$

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\left(-8\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{3±3}{-16}$$

Now solve the equation x=\frac{3±3}{-16} when ± is plus. Add 3 to 3.

$$x=\frac{6}{-16}$$

Reduce the fraction \frac{6}{-16} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{8}$$

Now solve the equation x=\frac{3±3}{-16} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{-16}$$

Divide 0 by -16.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{8} for x_{1} and 0 for x_{2}.

$$-8x^{2}-3x=-8\left(x-\left(-\frac{3}{8}\right)\right)x$$

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

$$-8x^{2}-3x=-8\left(x+\frac{3}{8}\right)x$$

Add \frac{3}{8} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-8x^{2}-3x=-8\times \frac{-8x-3}{-8}x$$

Cancel out 8, the greatest common factor in -8 and -8.

$$-8x^{2}-3x=\left(-8x-3\right)x$$

Evaluate

$$-x\left(8x+3\right)$$