Solve x^2+10x-56

$${ x }^{ 2 } +10x-56$$

Factor

$$\left(x-4\right)\left(x+14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-56\right)=-56$$

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 -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 10.

$$a=-4 b=14$$

Rewrite x^{2}+10x-56 as \left(x^{2}-4x\right)+\left(14x-56\right).

$$\left(x^{2}-4x\right)+\left(14x-56\right)$$

Factor out x in the first and 14 in the second group.

$$x\left(x-4\right)+14\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+14\right)$$

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.

$$x^{2}+10x-56=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-56\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-56\right)}}{2}$$

Multiply -4 times -56.

$$x=\frac{-10±\sqrt{100+224}}{2}$$

Add 100 to 224.

$$x=\frac{-10±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-10±18}{2}$$

Now solve the equation x=\frac{-10±18}{2} when ± is plus. Add -10 to 18.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-10±18}{2} when ± is minus. Subtract 18 from -10.

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

Divide -28 by 2.

$$x=-14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -14 for x_{2}.

$$x^{2}+10x-56=\left(x-4\right)\left(x-\left(-14\right)\right)$$

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

$$x^{2}+10x-56=\left(x-4\right)\left(x+14\right)$$

Evaluate

$$\left(x-4\right)\left(x+14\right)$$