Solve 6x^2-11x+4

$$6 { x }^{ 2 } -11x+4$$

Factor

$$\left(2x-1\right)\left(3x-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-11 ab=6\times 4=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite 6x^{2}-11x+4 as \left(6x^{2}-8x\right)+\left(-3x+4\right).

$$\left(6x^{2}-8x\right)+\left(-3x+4\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(3x-4\right)-\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(2x-1\right)$$

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.

$$6x^{2}-11x+4=0$$

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.

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

Square -11.

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

Multiply -4 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-24\times 4}}{2\times 6}$$

Multiply -24 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 6}$$

Add 121 to -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-\left(-11\right)±5}{2\times 6}$$

The opposite of -11 is 11.

$$x=\frac{11±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{11±5}{12}$$

Now solve the equation x=\frac{11±5}{12} when ± is plus. Add 11 to 5.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{11±5}{12} when ± is minus. Subtract 5 from 11.

$$x=\frac{6}{12}$$

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

$$x=\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and \frac{1}{2} for x_{2}.

$$6x^{2}-11x+4=6\left(x-\frac{4}{3}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x+4=6\times \frac{3x-4}{3}\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x+4=6\times \frac{3x-4}{3}\times \frac{2x-1}{2}$$

Multiply \frac{3x-4}{3} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x+4=6\times \frac{\left(3x-4\right)\left(2x-1\right)}{3\times 2}$$

Multiply 3 times 2.

$$6x^{2}-11x+4=6\times \frac{\left(3x-4\right)\left(2x-1\right)}{6}$$

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

$$6x^{2}-11x+4=\left(3x-4\right)\left(2x-1\right)$$

Evaluate

$$\left(2x-1\right)\left(3x-4\right)$$