Solve -x^2-x+42

$$- { x }^{ 2 } -x+42$$

Factor

$$\left(6-x\right)\left(x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+42. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-42=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=6 b=-7$$

Rewrite -x^{2}-x+42 as \left(-x^{2}+6x\right)+\left(-7x+42\right).

$$\left(-x^{2}+6x\right)+\left(-7x+42\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(-x+6\right)+7\left(-x+6\right)$$

Factor out common term -x+6 by using distributive property.

$$\left(-x+6\right)\left(x+7\right)$$

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.

$$-x^{2}-x+42=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 42}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 42}}{2\left(-1\right)}$$

Multiply 4 times 42.

$$x=\frac{-\left(-1\right)±\sqrt{1+168}}{2\left(-1\right)}$$

Add 1 to 168.

$$x=\frac{-\left(-1\right)±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-1\right)±13}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±13}{-2}$$

Now solve the equation x=\frac{1±13}{-2} when ± is plus. Add 1 to 13.

$$x=\frac{14}{-2}$$

Divide 14 by -2.

$$x=-7$$

Now solve the equation x=\frac{1±13}{-2} when ± is minus. Subtract 13 from 1.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -7 for x_{1} and 6 for x_{2}.

$$-x^{2}-x+42=-\left(x-\left(-7\right)\right)\left(x-6\right)$$

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

$$-x^{2}-x+42=-\left(x+7\right)\left(x-6\right)$$

Evaluate

$$\left(6-x\right)\left(x+7\right)$$