Solve (12-x-x^2)

$$( 12 - x - x ^ { 2 } )$$

Factor

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

Solution Steps

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

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

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+12. 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 -x^{2}-x+12 as \left(-x^{2}+3x\right)+\left(-4x+12\right).

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

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

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

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

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

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.

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

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.

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

Multiply -4 times -1.

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

Multiply 4 times 12.

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

Add 1 to 48.

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

Take the square root of 49.

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

The opposite of -1 is 1.

$$x=\frac{1±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±7}{-2}$$

Now solve the equation x=\frac{1±7}{-2} when ± is plus. Add 1 to 7.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{1±7}{-2} when ± is minus. Subtract 7 from 1.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

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

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

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

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

Evaluate

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