Solve x^2-6x-160

$$x^2-6x-160$$

Factor

$$\left(x-16\right)\left(x+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-160. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-160\right)=-160$$

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 -160.

$$1,-160 2,-80 4,-40 5,-32 8,-20 10,-16$$

Calculate the sum for each pair.

$$1-160=-159 2-80=-78 4-40=-36 5-32=-27 8-20=-12 10-16=-6$$

The solution is the pair that gives sum -6.

$$a=-16 b=10$$

Rewrite x^{2}-6x-160 as \left(x^{2}-16x\right)+\left(10x-160\right).

$$\left(x^{2}-16x\right)+\left(10x-160\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-16\right)+10\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+10\right)$$

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.

$$x^{2}-6x-160=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-160\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-160\right)}}{2}$$

Multiply -4 times -160.

$$x=\frac{-\left(-6\right)±\sqrt{36+640}}{2}$$

Add 36 to 640.

$$x=\frac{-\left(-6\right)±\sqrt{676}}{2}$$

Take the square root of 676.

$$x=\frac{-\left(-6\right)±26}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±26}{2}$$

Now solve the equation x=\frac{6±26}{2} when ± is plus. Add 6 to 26.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{6±26}{2} when ± is minus. Subtract 26 from 6.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 16 for x_{1} and -10 for x_{2}.

$$x^{2}-6x-160=\left(x-16\right)\left(x-\left(-10\right)\right)$$

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

$$x^{2}-6x-160=\left(x-16\right)\left(x+10\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -6x -160 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 6 rs = -160$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -160

$$(3 - u) (3 + u) = -160$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$9 - u^2 = -160$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -160-9 = -169$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 169 u = \pm\sqrt{169} = \pm 13$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =3 - 13 = -10 s = 3 + 13 = 16$$

Evaluate

$$\left(x-16\right)\left(x+10\right)$$