Solve 1-x-12x^2

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=-1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

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

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

Factor out 3x in -12x^{2}+3x.

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

Factor out common term -4x+1 by using distributive property.

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

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.

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

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-12\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-12\right)}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-12\right)}$$

Take the square root of 49.

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

The opposite of -1 is 1.

$$x=\frac{1±7}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{1±7}{-24}$$

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

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

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{1±7}{-24} when ± is minus. Subtract 7 from 1.

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

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}.

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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

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

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

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

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.

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

Multiply -3 times -4.

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

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

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

Evaluate

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