Solve 9c^2-10c+1

$$9 c ^ { 2 } - 10 c + 1$$

Factor

$$\left(c-1\right)\left(9c-1\right)$$

Solution Steps

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

$$a+b=-10 ab=9\times 1=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 -10.

$$a=-9 b=-1$$

Rewrite 9c^{2}-10c+1 as \left(9c^{2}-9c\right)+\left(-c+1\right).

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

Factor out 9c in the first and -1 in the second group.

$$9c\left(c-1\right)-\left(c-1\right)$$

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

$$\left(c-1\right)\left(9c-1\right)$$

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.

$$9c^{2}-10c+1=0$$

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.

$$c=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 9}}{2\times 9}$$

Square -10.

$$c=\frac{-\left(-10\right)±\sqrt{100-4\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$c=\frac{-\left(-10\right)±\sqrt{100-36}}{2\times 9}$$

Add 100 to -36.

$$c=\frac{-\left(-10\right)±\sqrt{64}}{2\times 9}$$

Take the square root of 64.

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

The opposite of -10 is 10.

$$c=\frac{10±8}{2\times 9}$$

Multiply 2 times 9.

$$c=\frac{10±8}{18}$$

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

$$c=\frac{18}{18}$$

Divide 18 by 18.

$$c=1$$

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

$$c=\frac{2}{18}$$

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

$$c=\frac{1}{9}$$

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}.

$$9c^{2}-10c+1=9\left(c-1\right)\left(c-\frac{1}{9}\right)$$

Subtract \frac{1}{9} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9c^{2}-10c+1=9\left(c-1\right)\times \frac{9c-1}{9}$$

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

$$9c^{2}-10c+1=\left(c-1\right)\left(9c-1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 9

$$x ^ 2 -\frac{10}{9}x +\frac{1}{9} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{10}{9} rs = \frac{1}{9}$$

Two numbers r and s sum up to \frac{10}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{9} = \frac{5}{9}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{9} - u s = \frac{5}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{9}

$$(\frac{5}{9} - u) (\frac{5}{9} + u) = \frac{1}{9}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{81} - u^2 = \frac{1}{9}$$

Simplify the expression by subtracting \frac{25}{81} on both sides

$$-u^2 = \frac{1}{9}-\frac{25}{81} = -\frac{16}{81}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{16}{81} u = \pm\sqrt{\frac{16}{81}} = \pm \frac{4}{9}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{5}{9} - \frac{4}{9} = 0.111 s = \frac{5}{9} + \frac{4}{9} = 1$$

Evaluate

$$\left(c-1\right)\left(9c-1\right)$$