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2\left(15z^{2}-4z-4\right)
Factor out 2.
a+b=-4 ab=15\left(-4\right)=-60
Consider 15z^{2}-4z-4. Factor the expression by grouping. First, the expression needs to be rewritten as 15z^{2}+az+bz-4. To find a and b, set up a system to be solved.
1,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-10 b=6
The solution is the pair that gives sum -4.
\left(15z^{2}-10z\right)+\left(6z-4\right)
Rewrite 15z^{2}-4z-4 as \left(15z^{2}-10z\right)+\left(6z-4\right).
5z\left(3z-2\right)+2\left(3z-2\right)
Factor out 5z in the first and 2 in the second group.
\left(3z-2\right)\left(5z+2\right)
Factor out common term 3z-2 by using distributive property.
2\left(3z-2\right)\left(5z+2\right)
Rewrite the complete factored expression.
30z^{2}-8z-8=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.
z=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 30\left(-8\right)}}{2\times 30}
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.
z=\frac{-\left(-8\right)±\sqrt{64-4\times 30\left(-8\right)}}{2\times 30}
Square -8.
z=\frac{-\left(-8\right)±\sqrt{64-120\left(-8\right)}}{2\times 30}
Multiply -4 times 30.
z=\frac{-\left(-8\right)±\sqrt{64+960}}{2\times 30}
Multiply -120 times -8.
z=\frac{-\left(-8\right)±\sqrt{1024}}{2\times 30}
Add 64 to 960.
z=\frac{-\left(-8\right)±32}{2\times 30}
Take the square root of 1024.
z=\frac{8±32}{2\times 30}
The opposite of -8 is 8.
z=\frac{8±32}{60}
Multiply 2 times 30.
z=\frac{40}{60}
Now solve the equation z=\frac{8±32}{60} when ± is plus. Add 8 to 32.
z=\frac{2}{3}
Reduce the fraction \frac{40}{60} to lowest terms by extracting and canceling out 20.
z=-\frac{24}{60}
Now solve the equation z=\frac{8±32}{60} when ± is minus. Subtract 32 from 8.
z=-\frac{2}{5}
Reduce the fraction \frac{-24}{60} to lowest terms by extracting and canceling out 12.
30z^{2}-8z-8=30\left(z-\frac{2}{3}\right)\left(z-\left(-\frac{2}{5}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and -\frac{2}{5} for x_{2}.
30z^{2}-8z-8=30\left(z-\frac{2}{3}\right)\left(z+\frac{2}{5}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
30z^{2}-8z-8=30\times \frac{3z-2}{3}\left(z+\frac{2}{5}\right)
Subtract \frac{2}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
30z^{2}-8z-8=30\times \frac{3z-2}{3}\times \frac{5z+2}{5}
Add \frac{2}{5} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
30z^{2}-8z-8=30\times \frac{\left(3z-2\right)\left(5z+2\right)}{3\times 5}
Multiply \frac{3z-2}{3} times \frac{5z+2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
30z^{2}-8z-8=30\times \frac{\left(3z-2\right)\left(5z+2\right)}{15}
Multiply 3 times 5.
30z^{2}-8z-8=2\left(3z-2\right)\left(5z+2\right)
Cancel out 15, the greatest common factor in 30 and 15.
x ^ 2 -\frac{4}{15}x -\frac{4}{15} = 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.This is achieved by dividing both sides of the equation by 30
r + s = \frac{4}{15} rs = -\frac{4}{15}
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 = \frac{2}{15} - u s = \frac{2}{15} + u
Two numbers r and s sum up to \frac{4}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{15} = \frac{2}{15}. 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>
(\frac{2}{15} - u) (\frac{2}{15} + u) = -\frac{4}{15}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{15}
\frac{4}{225} - u^2 = -\frac{4}{15}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{4}{15}-\frac{4}{225} = -\frac{64}{225}
Simplify the expression by subtracting \frac{4}{225} on both sides
u^2 = \frac{64}{225} u = \pm\sqrt{\frac{64}{225}} = \pm \frac{8}{15}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{2}{15} - \frac{8}{15} = -0.400 s = \frac{2}{15} + \frac{8}{15} = 0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.