Solve -3x^2-39x-36

$$-3 { x }^{ 2 } -39x-36$$

Factor

$$-3\left(x+1\right)\left(x+12\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}-13x-12\right)$$

Consider -x^{2}-13x-12. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-13 ab=-\left(-12\right)=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-1 b=-12$$

Rewrite -x^{2}-13x-12 as \left(-x^{2}-x\right)+\left(-12x-12\right).

$$\left(-x^{2}-x\right)+\left(-12x-12\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(-x-1\right)+12\left(-x-1\right)$$

Factor out common term -x-1 by using distributive property.

$$\left(-x-1\right)\left(x+12\right)$$

Rewrite the complete factored expression.

$$3\left(-x-1\right)\left(x+12\right)$$

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.

$$-3x^{2}-39x-36=0$$

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.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\left(-3\right)\left(-36\right)}}{2\left(-3\right)}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\left(-3\right)\left(-36\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-39\right)±\sqrt{1521+12\left(-36\right)}}{2\left(-3\right)}$$

Multiply 12 times -36.

$$x=\frac{-\left(-39\right)±\sqrt{1521-432}}{2\left(-3\right)}$$

Add 1521 to -432.

$$x=\frac{-\left(-39\right)±\sqrt{1089}}{2\left(-3\right)}$$

Take the square root of 1089.

$$x=\frac{-\left(-39\right)±33}{2\left(-3\right)}$$

The opposite of -39 is 39.

$$x=\frac{39±33}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{39±33}{-6}$$

Now solve the equation x=\frac{39±33}{-6} when ± is plus. Add 39 to 33.

$$x=\frac{72}{-6}$$

Divide 72 by -6.

$$x=-12$$

Now solve the equation x=\frac{39±33}{-6} when ± is minus. Subtract 33 from 39.

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

Divide 6 by -6.

$$x=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -12 for x_{1} and -1 for x_{2}.

$$-3x^{2}-39x-36=-3\left(x-\left(-12\right)\right)\left(x-\left(-1\right)\right)$$

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

$$-3x^{2}-39x-36=-3\left(x+12\right)\left(x+1\right)$$

Evaluate

$$-3\left(x+1\right)\left(x+12\right)$$