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Factor the expression by grouping. First, the expression needs to be rewritten as . To find and , set up a system to be solved.
Since is negative, and have the opposite signs. Since is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product .
Calculate the sum for each pair.
The solution is the pair that gives sum .
Rewrite as .
Factor out in the first and in the second group.
Factor out common term by using distributive property.
Evaluate
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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.
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