Solve 9x^2-8x-1

$$9 { x }^{ 2 } -8x-1$$

Factor

$$\left(x-1\right)\left(9x+1\right)$$

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.

$$a+b=-8 ab=9\left(-1\right)=-9$$

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 -9.

$$1,-9 3,-3$$

Calculate the sum for each pair.

$$1-9=-8 3-3=0$$

The solution is the pair that gives sum -8.

$$a=-9 b=1$$

Rewrite 9x^{2}-8x-1 as \left(9x^{2}-9x\right)+\left(x-1\right).

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

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

$$9x\left(x-1\right)+x-1$$

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

$$\left(x-1\right)\left(9x+1\right)$$

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.

$$9x^{2}-8x-1=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 9\left(-1\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-8\right)±\sqrt{64-36\left(-1\right)}}{2\times 9}$$

Multiply -36 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64+36}}{2\times 9}$$

Add 64 to 36.

$$x=\frac{-\left(-8\right)±\sqrt{100}}{2\times 9}$$

Take the square root of 100.

$$x=\frac{-\left(-8\right)±10}{2\times 9}$$

The opposite of -8 is 8.

$$x=\frac{8±10}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{8±10}{18}$$

Now solve the equation x=\frac{8±10}{18} when ± is plus. Add 8 to 10.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{8±10}{18} when ± is minus. Subtract 10 from 8.

$$x=-\frac{2}{18}$$

Reduce the fraction \frac{-2}{18} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -\frac{1}{9} for x_{2}.

$$9x^{2}-8x-1=9\left(x-1\right)\left(x-\left(-\frac{1}{9}\right)\right)$$

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

$$9x^{2}-8x-1=9\left(x-1\right)\left(x+\frac{1}{9}\right)$$

Add \frac{1}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-8x-1=9\left(x-1\right)\times \frac{9x+1}{9}$$

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

$$9x^{2}-8x-1=\left(x-1\right)\left(9x+1\right)$$

Evaluate

$$\left(x-1\right)\left(9x+1\right)$$