Solve =2-3x-9x^2

$$=2-3x-9 { x }^{ 2 }$$

Factor

$$\left(1-3x\right)\left(3x+2\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-9x^{2}-3x+2$$

Factor the expression by grouping. First, the expression needs to be rewritten as -9x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-9\times 2=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=3 b=-6$$

Rewrite -9x^{2}-3x+2 as \left(-9x^{2}+3x\right)+\left(-6x+2\right).

$$\left(-9x^{2}+3x\right)+\left(-6x+2\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(-3x+1\right)+2\left(-3x+1\right)$$

Factor out common term -3x+1 by using distributive property.

$$\left(-3x+1\right)\left(3x+2\right)$$

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.

$$-9x^{2}-3x+2=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-9\right)\times 2}}{2\left(-9\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-9\right)\times 2}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-3\right)±\sqrt{9+36\times 2}}{2\left(-9\right)}$$

Multiply 36 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\left(-9\right)}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\left(-9\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2\left(-9\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{3±9}{-18}$$

Now solve the equation x=\frac{3±9}{-18} when ± is plus. Add 3 to 9.

$$x=\frac{12}{-18}$$

Reduce the fraction \frac{12}{-18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{3±9}{-18} when ± is minus. Subtract 9 from 3.

$$x=-\frac{6}{-18}$$

Reduce the fraction \frac{-6}{-18} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{3}$$

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

$$-9x^{2}-3x+2=-9\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\frac{1}{3}\right)$$

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

$$-9x^{2}-3x+2=-9\left(x+\frac{2}{3}\right)\left(x-\frac{1}{3}\right)$$

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

$$-9x^{2}-3x+2=-9\times \frac{-3x-2}{-3}\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-9x^{2}-3x+2=-9\times \frac{-3x-2}{-3}\times \frac{-3x+1}{-3}$$

Multiply \frac{-3x-2}{-3} times \frac{-3x+1}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-9x^{2}-3x+2=-9\times \frac{\left(-3x-2\right)\left(-3x+1\right)}{-3\left(-3\right)}$$

Multiply -3 times -3.

$$-9x^{2}-3x+2=-9\times \frac{\left(-3x-2\right)\left(-3x+1\right)}{9}$$

Cancel out 9, the greatest common factor in -9 and 9.

$$-9x^{2}-3x+2=-\left(-3x-2\right)\left(-3x+1\right)$$

Evaluate

$$\left(1-3x\right)\left(3x+2\right)$$