Factor

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

$(x−16)(x+10)$

Evaluate

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

$(x−16)(x+10)$

Graph

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

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.

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

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

Calculate the sum for each pair.

a=-16 b=10

The solution is the pair that gives sum -6.

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

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

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

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

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

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

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

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

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{36-4\left(-160\right)}}{2}

Square -6.

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

Multiply -4 times -160.

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

Add 36 to 640.

x=\frac{-\left(-6\right)±26}{2}

Take the square root of 676.

x=\frac{6±26}{2}

The opposite of -6 is 6.

x=\frac{32}{2}

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

x=16

Divide 32 by 2.

x=\frac{-20}{2}

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

x=-10

Divide -20 by 2.

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

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+10\right)

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

x ^ 2 -6x -160 = 0

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.

r + s = 6 rs = -160

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 = 3 - u s = 3 + u

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>

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

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

9 - u^2 = -160

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

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

Simplify the expression by subtracting 9 on both sides

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

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

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

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