Factor
\left(3z-10\right)\left(3z+2\right)
Evaluate
\left(3z-10\right)\left(3z+2\right)
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a+b=-24 ab=9\left(-20\right)=-180
Factor the expression by grouping. First, the expression needs to be rewritten as 9z^{2}+az+bz-20. To find a and b, set up a system to be solved.
1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15
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 -180.
1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3
Calculate the sum for each pair.
a=-30 b=6
The solution is the pair that gives sum -24.
\left(9z^{2}-30z\right)+\left(6z-20\right)
Rewrite 9z^{2}-24z-20 as \left(9z^{2}-30z\right)+\left(6z-20\right).
3z\left(3z-10\right)+2\left(3z-10\right)
Factor out 3z in the first and 2 in the second group.
\left(3z-10\right)\left(3z+2\right)
Factor out common term 3z-10 by using distributive property.
9z^{2}-24z-20=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(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 9\left(-20\right)}}{2\times 9}
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(-24\right)±\sqrt{576-4\times 9\left(-20\right)}}{2\times 9}
Square -24.
z=\frac{-\left(-24\right)±\sqrt{576-36\left(-20\right)}}{2\times 9}
Multiply -4 times 9.
z=\frac{-\left(-24\right)±\sqrt{576+720}}{2\times 9}
Multiply -36 times -20.
z=\frac{-\left(-24\right)±\sqrt{1296}}{2\times 9}
Add 576 to 720.
z=\frac{-\left(-24\right)±36}{2\times 9}
Take the square root of 1296.
z=\frac{24±36}{2\times 9}
The opposite of -24 is 24.
z=\frac{24±36}{18}
Multiply 2 times 9.
z=\frac{60}{18}
Now solve the equation z=\frac{24±36}{18} when ± is plus. Add 24 to 36.
z=\frac{10}{3}
Reduce the fraction \frac{60}{18} to lowest terms by extracting and canceling out 6.
z=-\frac{12}{18}
Now solve the equation z=\frac{24±36}{18} when ± is minus. Subtract 36 from 24.
z=-\frac{2}{3}
Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.
9z^{2}-24z-20=9\left(z-\frac{10}{3}\right)\left(z-\left(-\frac{2}{3}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{10}{3} for x_{1} and -\frac{2}{3} for x_{2}.
9z^{2}-24z-20=9\left(z-\frac{10}{3}\right)\left(z+\frac{2}{3}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
9z^{2}-24z-20=9\times \frac{3z-10}{3}\left(z+\frac{2}{3}\right)
Subtract \frac{10}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
9z^{2}-24z-20=9\times \frac{3z-10}{3}\times \frac{3z+2}{3}
Add \frac{2}{3} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
9z^{2}-24z-20=9\times \frac{\left(3z-10\right)\left(3z+2\right)}{3\times 3}
Multiply \frac{3z-10}{3} times \frac{3z+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
9z^{2}-24z-20=9\times \frac{\left(3z-10\right)\left(3z+2\right)}{9}
Multiply 3 times 3.
9z^{2}-24z-20=\left(3z-10\right)\left(3z+2\right)
Cancel out 9, the greatest common factor in 9 and 9.
x ^ 2 -\frac{8}{3}x -\frac{20}{9} = 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 9
r + s = \frac{8}{3} rs = -\frac{20}{9}
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{4}{3} - u s = \frac{4}{3} + u
Two numbers r and s sum up to \frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{3} = \frac{4}{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>
(\frac{4}{3} - u) (\frac{4}{3} + u) = -\frac{20}{9}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{9}
\frac{16}{9} - u^2 = -\frac{20}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{20}{9}-\frac{16}{9} = -4
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{4}{3} - 2 = -0.667 s = \frac{4}{3} + 2 = 3.333
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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