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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.
Rewrite x^{2}-6x-160 as \left(x^{2}-16x\right)+\left(10x-160\right).
Factor out x in the first and 10 in the second group.
Factor out common term x-16 by using distributive property.
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.
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.
Square -6.
Multiply -4 times -160.
Add 36 to 640.
Take the square root of 676.
The opposite of -6 is 6.
Now solve the equation x=\frac{6±26}{2} when ± is plus. Add 6 to 26.
Divide 32 by 2.
Now solve the equation x=\frac{6±26}{2} when ± is minus. Subtract 26 from 6.
Divide -20 by 2.
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}.
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.