Solve 6-x-x^2=

$$6 - x - x ^ { 2 } =$$

Factor

$$\left(2-x\right)\left(x+3\right)$$

Solution Steps

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

$$-x^{2}-x+6$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

Rewrite -x^{2}-x+6 as \left(-x^{2}+2x\right)+\left(-3x+6\right).

$$\left(-x^{2}+2x\right)+\left(-3x+6\right)$$

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

$$x\left(-x+2\right)+3\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+3\right)$$

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.

$$-x^{2}-x+6=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-1\right)}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±5}{-2}$$

Now solve the equation x=\frac{1±5}{-2} when ± is plus. Add 1 to 5.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{1±5}{-2} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

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

$$-x^{2}-x+6=-\left(x-\left(-3\right)\right)\left(x-2\right)$$

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

$$-x^{2}-x+6=-\left(x+3\right)\left(x-2\right)$$

Evaluate

$$\left(2-x\right)\left(x+3\right)$$