Solve -9x^2+9x-2

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

Factor

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

Solution Steps

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.

$$a+b=9 ab=-9\left(-2\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

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

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=6 b=3$$

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

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

Factor out -3x in -9x^{2}+6x.

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

Factor out common term 3x-2 by using distributive property.

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

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.

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

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.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-9\right)\left(-2\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-9±\sqrt{81+36\left(-2\right)}}{2\left(-9\right)}$$

Multiply 36 times -2.

$$x=\frac{-9±\sqrt{81-72}}{2\left(-9\right)}$$

Add 81 to -72.

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

Take the square root of 9.

$$x=\frac{-9±3}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-9±3}{-18}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply -3 times -3.

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

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

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

Evaluate

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