Solve 12x^2-7x+1

$$12 { x }^{ 2 } -7x+1$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-7 ab=12\times 1=12$$

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 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 12x^{2}-7x+1 as \left(12x^{2}-4x\right)+\left(-3x+1\right).

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

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

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

Factor out common term 3x-1 by using distributive property.

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

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.

$$12x^{2}-7x+1=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2\times 12}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 12}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2\times 12}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2\times 12}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{7±1}{24}$$

Now solve the equation x=\frac{7±1}{24} when ± is plus. Add 7 to 1.

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

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

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

Now solve the equation x=\frac{7±1}{24} when ± is minus. Subtract 1 from 7.

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

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 \frac{1}{3} for x_{1} and \frac{1}{4} for x_{2}.

$$12x^{2}-7x+1=12\left(x-\frac{1}{3}\right)\left(x-\frac{1}{4}\right)$$

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

$$12x^{2}-7x+1=12\times \frac{3x-1}{3}\left(x-\frac{1}{4}\right)$$

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

$$12x^{2}-7x+1=12\times \frac{3x-1}{3}\times \frac{4x-1}{4}$$

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

$$12x^{2}-7x+1=12\times \frac{\left(3x-1\right)\left(4x-1\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}-7x+1=12\times \frac{\left(3x-1\right)\left(4x-1\right)}{12}$$

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

$$12x^{2}-7x+1=\left(3x-1\right)\left(4x-1\right)$$

Evaluate

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