Solve x^2-3x-378

$${ x }^{ 2 } -3x-378$$

Factor

$$\left(x-21\right)\left(x+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-378. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-378\right)=-378$$

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 -378.

$$1,-378 2,-189 3,-126 6,-63 7,-54 9,-42 14,-27 18,-21$$

Calculate the sum for each pair.

$$1-378=-377 2-189=-187 3-126=-123 6-63=-57 7-54=-47 9-42=-33 14-27=-13 18-21=-3$$

The solution is the pair that gives sum -3.

$$a=-21 b=18$$

Rewrite x^{2}-3x-378 as \left(x^{2}-21x\right)+\left(18x-378\right).

$$\left(x^{2}-21x\right)+\left(18x-378\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-21\right)+18\left(x-21\right)$$

Factor out common term x-21 by using distributive property.

$$\left(x-21\right)\left(x+18\right)$$

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.

$$x^{2}-3x-378=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-378\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-378\right)}}{2}$$

Multiply -4 times -378.

$$x=\frac{-\left(-3\right)±\sqrt{9+1512}}{2}$$

Add 9 to 1512.

$$x=\frac{-\left(-3\right)±\sqrt{1521}}{2}$$

Take the square root of 1521.

$$x=\frac{-\left(-3\right)±39}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±39}{2}$$

Now solve the equation x=\frac{3±39}{2} when ± is plus. Add 3 to 39.

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

Divide 42 by 2.

$$x=21$$

Now solve the equation x=\frac{3±39}{2} when ± is minus. Subtract 39 from 3.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 21 for x_{1} and -18 for x_{2}.

$$x^{2}-3x-378=\left(x-21\right)\left(x-\left(-18\right)\right)$$

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

$$x^{2}-3x-378=\left(x-21\right)\left(x+18\right)$$

Evaluate

$$\left(x-21\right)\left(x+18\right)$$