Solve -9x^2+6x-1

$$-9 { x }^{ 2 } +6x-1$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -9x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=6 ab=-9\left(-1\right)=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -9x^{2}+6x-1 as \left(-9x^{2}+3x\right)+\left(3x-1\right).

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

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

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

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

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

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.

$$-9x^{2}+6x-1=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-6±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

Multiply 36 times -1.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-9\right)}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$x=\frac{-6±0}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-6±0}{-18}$$

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{1}{3} for x_{2}.

$$-9x^{2}+6x-1=-9\left(x-\frac{1}{3}\right)\left(x-\frac{1}{3}\right)$$

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

$$-9x^{2}+6x-1=-9\times \frac{-3x+1}{-3}\left(x-\frac{1}{3}\right)$$

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

$$-9x^{2}+6x-1=-9\times \frac{-3x+1}{-3}\times \frac{-3x+1}{-3}$$

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

$$-9x^{2}+6x-1=-9\times \frac{\left(-3x+1\right)\left(-3x+1\right)}{-3\left(-3\right)}$$

Multiply -3 times -3.

$$-9x^{2}+6x-1=-9\times \frac{\left(-3x+1\right)\left(-3x+1\right)}{9}$$

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

$$-9x^{2}+6x-1=-\left(-3x+1\right)\left(-3x+1\right)$$

Evaluate

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