Solve 6x^2-11x+3

$$6 { x }^{ 2 } -11x+3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-11 ab=6\times 3=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite 6x^{2}-11x+3 as \left(6x^{2}-9x\right)+\left(-2x+3\right).

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

Factor out 3x in the first and -1 in the second group.

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

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

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

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.

$$6x^{2}-11x+3=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\times 3}}{2\times 6}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 6\times 3}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-24\times 3}}{2\times 6}$$

Multiply -24 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-72}}{2\times 6}$$

Add 121 to -72.

$$x=\frac{-\left(-11\right)±\sqrt{49}}{2\times 6}$$

Take the square root of 49.

$$x=\frac{-\left(-11\right)±7}{2\times 6}$$

The opposite of -11 is 11.

$$x=\frac{11±7}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{11±7}{12}$$

Now solve the equation x=\frac{11±7}{12} when ± is plus. Add 11 to 7.

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

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{11±7}{12} when ± is minus. Subtract 7 from 11.

$$x=\frac{4}{12}$$

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

$$x=\frac{1}{3}$$

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

$$6x^{2}-11x+3=6\left(x-\frac{3}{2}\right)\left(x-\frac{1}{3}\right)$$

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

$$6x^{2}-11x+3=6\times \frac{2x-3}{2}\left(x-\frac{1}{3}\right)$$

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

$$6x^{2}-11x+3=6\times \frac{2x-3}{2}\times \frac{3x-1}{3}$$

Multiply \frac{2x-3}{2} times \frac{3x-1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x+3=6\times \frac{\left(2x-3\right)\left(3x-1\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-11x+3=6\times \frac{\left(2x-3\right)\left(3x-1\right)}{6}$$

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

$$6x^{2}-11x+3=\left(2x-3\right)\left(3x-1\right)$$

Evaluate

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