Solve 12x^2-15x+3

$$12 { x }^{ 2 } -15x+3$$

Factor

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

Solution Steps

Factor out 3.

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

Consider 4x^{2}-5x+1. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=4\times 1=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite 4x^{2}-5x+1 as \left(4x^{2}-4x\right)+\left(-x+1\right).

$$\left(4x^{2}-4x\right)+\left(-x+1\right)$$

Factor out 4x in the first and -1 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$12x^{2}-15x+3=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 12\times 3}}{2\times 12}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 12\times 3}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-15\right)±\sqrt{225-48\times 3}}{2\times 12}$$

Multiply -48 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 12}$$

Add 225 to -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 12}$$

Take the square root of 81.

$$x=\frac{-\left(-15\right)±9}{2\times 12}$$

The opposite of -15 is 15.

$$x=\frac{15±9}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{15±9}{24}$$

Now solve the equation x=\frac{15±9}{24} when ± is plus. Add 15 to 9.

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

Divide 24 by 24.

$$x=1$$

Now solve the equation x=\frac{15±9}{24} when ± is minus. Subtract 9 from 15.

$$x=\frac{6}{24}$$

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

$$x=\frac{1}{4}$$

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{1}{4} for x_{2}.

$$12x^{2}-15x+3=12\left(x-1\right)\left(x-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-15x+3=12\left(x-1\right)\times \frac{4x-1}{4}$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12x^{2}-15x+3=3\left(x-1\right)\left(4x-1\right)$$

Evaluate

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