Solve 12x^2-19x+5

$$12 { x }^{ 2 } -19x+5$$

Factor

$$\left(3x-1\right)\left(4x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-19 ab=12\times 5=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -19.

$$a=-15 b=-4$$

Rewrite 12x^{2}-19x+5 as \left(12x^{2}-15x\right)+\left(-4x+5\right).

$$\left(12x^{2}-15x\right)+\left(-4x+5\right)$$

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

$$3x\left(4x-5\right)-\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(3x-1\right)$$

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.

$$12x^{2}-19x+5=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 12\times 5}}{2\times 12}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 12\times 5}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-19\right)±\sqrt{361-48\times 5}}{2\times 12}$$

Multiply -48 times 5.

$$x=\frac{-\left(-19\right)±\sqrt{361-240}}{2\times 12}$$

Add 361 to -240.

$$x=\frac{-\left(-19\right)±\sqrt{121}}{2\times 12}$$

Take the square root of 121.

$$x=\frac{-\left(-19\right)±11}{2\times 12}$$

The opposite of -19 is 19.

$$x=\frac{19±11}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{19±11}{24}$$

Now solve the equation x=\frac{19±11}{24} when ± is plus. Add 19 to 11.

$$x=\frac{30}{24}$$

Reduce the fraction \frac{30}{24} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{19±11}{24} when ± is minus. Subtract 11 from 19.

$$x=\frac{8}{24}$$

Reduce the fraction \frac{8}{24} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and \frac{1}{3} for x_{2}.

$$12x^{2}-19x+5=12\left(x-\frac{5}{4}\right)\left(x-\frac{1}{3}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-19x+5=12\times \frac{4x-5}{4}\left(x-\frac{1}{3}\right)$$

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

$$12x^{2}-19x+5=12\times \frac{4x-5}{4}\times \frac{3x-1}{3}$$

Multiply \frac{4x-5}{4} times \frac{3x-1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-19x+5=12\times \frac{\left(4x-5\right)\left(3x-1\right)}{4\times 3}$$

Multiply 4 times 3.

$$12x^{2}-19x+5=12\times \frac{\left(4x-5\right)\left(3x-1\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}-19x+5=\left(4x-5\right)\left(3x-1\right)$$

Evaluate

$$\left(3x-1\right)\left(4x-5\right)$$