Solve 9x^2-15x+6

$$9 x ^ { 2 } - 15 x + 6$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(3x^{2}-5x+2\right)$$

Consider 3x^{2}-5x+2. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=3\times 2=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite 3x^{2}-5x+2 as \left(3x^{2}-3x\right)+\left(-2x+2\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$9x^{2}-15x+6=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 9\times 6}}{2\times 9}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 9\times 6}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-15\right)±\sqrt{225-36\times 6}}{2\times 9}$$

Multiply -36 times 6.

$$x=\frac{-\left(-15\right)±\sqrt{225-216}}{2\times 9}$$

Add 225 to -216.

$$x=\frac{-\left(-15\right)±\sqrt{9}}{2\times 9}$$

Take the square root of 9.

$$x=\frac{-\left(-15\right)±3}{2\times 9}$$

The opposite of -15 is 15.

$$x=\frac{15±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{15±3}{18}$$

Now solve the equation x=\frac{15±3}{18} when ± is plus. Add 15 to 3.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{15±3}{18} when ± is minus. Subtract 3 from 15.

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

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 1 for x_{1} and \frac{2}{3} for x_{2}.

$$9x^{2}-15x+6=9\left(x-1\right)\left(x-\frac{2}{3}\right)$$

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

$$9x^{2}-15x+6=9\left(x-1\right)\times \frac{3x-2}{3}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9x^{2}-15x+6=3\left(x-1\right)\left(3x-2\right)$$

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

$$x ^ 2 -\frac{5}{3}x +\frac{2}{3} = 0$$

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

$$r + s = \frac{5}{3} rs = \frac{2}{3}$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = \frac{2}{3}$$

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

$$\frac{25}{36} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = \frac{2}{3}-\frac{25}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =\frac{5}{6} - \frac{1}{6} = 0.667 s = \frac{5}{6} + \frac{1}{6} = 1.000$$

Evaluate

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