Solve -3x^2-18x-24

$$-3 { x }^{ 2 } -18x-24$$

Factor

$$-3\left(x+2\right)\left(x+4\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}-6x-8\right)$$

Consider -x^{2}-6x-8. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-2 b=-4$$

Rewrite -x^{2}-6x-8 as \left(-x^{2}-2x\right)+\left(-4x-8\right).

$$\left(-x^{2}-2x\right)+\left(-4x-8\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(-x-2\right)+4\left(-x-2\right)$$

Factor out common term -x-2 by using distributive property.

$$\left(-x-2\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$3\left(-x-2\right)\left(x+4\right)$$

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.

$$-3x^{2}-18x-24=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-18\right)±\sqrt{324+12\left(-24\right)}}{2\left(-3\right)}$$

Multiply 12 times -24.

$$x=\frac{-\left(-18\right)±\sqrt{324-288}}{2\left(-3\right)}$$

Add 324 to -288.

$$x=\frac{-\left(-18\right)±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{-\left(-18\right)±6}{2\left(-3\right)}$$

The opposite of -18 is 18.

$$x=\frac{18±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{18±6}{-6}$$

Now solve the equation x=\frac{18±6}{-6} when ± is plus. Add 18 to 6.

$$x=\frac{24}{-6}$$

Divide 24 by -6.

$$x=-4$$

Now solve the equation x=\frac{18±6}{-6} when ± is minus. Subtract 6 from 18.

$$x=\frac{12}{-6}$$

Divide 12 by -6.

$$x=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and -2 for x_{2}.

$$-3x^{2}-18x-24=-3\left(x-\left(-4\right)\right)\left(x-\left(-2\right)\right)$$

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

$$-3x^{2}-18x-24=-3\left(x+4\right)\left(x+2\right)$$

Evaluate

$$-3\left(x+2\right)\left(x+4\right)$$