Solve x^2+19x-42

$${ x }^{ 2 } +19x-42$$

Factor

$$\left(x-2\right)\left(x+21\right)$$

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.

$$a+b=19 ab=1\left(-42\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 19.

$$a=-2 b=21$$

Rewrite x^{2}+19x-42 as \left(x^{2}-2x\right)+\left(21x-42\right).

$$\left(x^{2}-2x\right)+\left(21x-42\right)$$

Factor out x in the first and 21 in the second group.

$$x\left(x-2\right)+21\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+21\right)$$

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.

$$x^{2}+19x-42=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\left(-42\right)}}{2}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\left(-42\right)}}{2}$$

Multiply -4 times -42.

$$x=\frac{-19±\sqrt{361+168}}{2}$$

Add 361 to 168.

$$x=\frac{-19±\sqrt{529}}{2}$$

Take the square root of 529.

$$x=\frac{-19±23}{2}$$

Now solve the equation x=\frac{-19±23}{2} when ± is plus. Add -19 to 23.

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

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-19±23}{2} when ± is minus. Subtract 23 from -19.

$$x=-\frac{42}{2}$$

Divide -42 by 2.

$$x=-21$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -21 for x_{2}.

$$x^{2}+19x-42=\left(x-2\right)\left(x-\left(-21\right)\right)$$

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

$$x^{2}+19x-42=\left(x-2\right)\left(x+21\right)$$

Evaluate

$$\left(x-2\right)\left(x+21\right)$$