Solve 165-3x^2-18x

$$165 - 3 x ^ { 2 } - 18 x$$

Factor

$$3\left(5-x\right)\left(x+11\right)$$

Solution Steps

Factor out 3.

$$3\left(55-x^{2}-6x\right)$$

Consider 55-x^{2}-6x. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-6x+55$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+55. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-55=-55$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -55.

$$1,-55 5,-11$$

Calculate the sum for each pair.

$$1-55=-54 5-11=-6$$

The solution is the pair that gives sum -6.

$$a=5 b=-11$$

Rewrite -x^{2}-6x+55 as \left(-x^{2}+5x\right)+\left(-11x+55\right).

$$\left(-x^{2}+5x\right)+\left(-11x+55\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(-x+5\right)+11\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(x+11\right)$$

Rewrite the complete factored expression.

$$3\left(-x+5\right)\left(x+11\right)$$

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.

$$-3x^{2}-18x+165=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-3\right)\times 165}}{2\left(-3\right)}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\left(-3\right)\times 165}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-18\right)±\sqrt{324+12\times 165}}{2\left(-3\right)}$$

Multiply 12 times 165.

$$x=\frac{-\left(-18\right)±\sqrt{324+1980}}{2\left(-3\right)}$$

Add 324 to 1980.

$$x=\frac{-\left(-18\right)±\sqrt{2304}}{2\left(-3\right)}$$

Take the square root of 2304.

$$x=\frac{-\left(-18\right)±48}{2\left(-3\right)}$$

The opposite of -18 is 18.

$$x=\frac{18±48}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{18±48}{-6}$$

Now solve the equation x=\frac{18±48}{-6} when ± is plus. Add 18 to 48.

$$x=\frac{66}{-6}$$

Divide 66 by -6.

$$x=-11$$

Now solve the equation x=\frac{18±48}{-6} when ± is minus. Subtract 48 from 18.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -11 for x_{1} and 5 for x_{2}.

$$-3x^{2}-18x+165=-3\left(x-\left(-11\right)\right)\left(x-5\right)$$

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

$$-3x^{2}-18x+165=-3\left(x+11\right)\left(x-5\right)$$

Evaluate

$$3\left(5-x\right)\left(x+11\right)$$