Solve 6x^2-13x+6=

$$6 { x }^{ 2 } -13x+6 =$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=6\times 6=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite 6x^{2}-13x+6 as \left(6x^{2}-9x\right)+\left(-4x+6\right).

$$\left(6x^{2}-9x\right)+\left(-4x+6\right)$$

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

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

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

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

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.

$$6x^{2}-13x+6=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\times 6}}{2\times 6}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-13\right)±\sqrt{169-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$x=\frac{-\left(-13\right)±\sqrt{169-144}}{2\times 6}$$

Add 169 to -144.

$$x=\frac{-\left(-13\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-\left(-13\right)±5}{2\times 6}$$

The opposite of -13 is 13.

$$x=\frac{13±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{13±5}{12}$$

Now solve the equation x=\frac{13±5}{12} when ± is plus. Add 13 to 5.

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

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

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

Now solve the equation x=\frac{13±5}{12} when ± is minus. Subtract 5 from 13.

$$x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

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

$$6x^{2}-13x+6=6\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)$$

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

$$6x^{2}-13x+6=6\times \frac{2x-3}{2}\left(x-\frac{2}{3}\right)$$

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

$$6x^{2}-13x+6=6\times \frac{2x-3}{2}\times \frac{3x-2}{3}$$

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

$$6x^{2}-13x+6=6\times \frac{\left(2x-3\right)\left(3x-2\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-13x+6=6\times \frac{\left(2x-3\right)\left(3x-2\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}-13x+6=\left(2x-3\right)\left(3x-2\right)$$

Evaluate

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