Solve x^2+4x-252

$${ x }^{ 2 } +4x-252$$

Factor

$$\left(x-14\right)\left(x+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-252. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-252\right)=-252$$

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 -252.

$$-1,252 -2,126 -3,84 -4,63 -6,42 -7,36 -9,28 -12,21 -14,18$$

Calculate the sum for each pair.

$$-1+252=251 -2+126=124 -3+84=81 -4+63=59 -6+42=36 -7+36=29 -9+28=19 -12+21=9 -14+18=4$$

The solution is the pair that gives sum 4.

$$a=-14 b=18$$

Rewrite x^{2}+4x-252 as \left(x^{2}-14x\right)+\left(18x-252\right).

$$\left(x^{2}-14x\right)+\left(18x-252\right)$$

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

$$x\left(x-14\right)+18\left(x-14\right)$$

Factor out common term x-14 by using distributive property.

$$\left(x-14\right)\left(x+18\right)$$

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.

$$x^{2}+4x-252=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-252\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-252\right)}}{2}$$

Multiply -4 times -252.

$$x=\frac{-4±\sqrt{16+1008}}{2}$$

Add 16 to 1008.

$$x=\frac{-4±\sqrt{1024}}{2}$$

Take the square root of 1024.

$$x=\frac{-4±32}{2}$$

Now solve the equation x=\frac{-4±32}{2} when ± is plus. Add -4 to 32.

$$x=\frac{28}{2}$$

Divide 28 by 2.

$$x=14$$

Now solve the equation x=\frac{-4±32}{2} when ± is minus. Subtract 32 from -4.

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

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 14 for x_{1} and -18 for x_{2}.

$$x^{2}+4x-252=\left(x-14\right)\left(x-\left(-18\right)\right)$$

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

$$x^{2}+4x-252=\left(x-14\right)\left(x+18\right)$$

Evaluate

$$\left(x-14\right)\left(x+18\right)$$