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3\left(-x^{2}-9x-18\right)
Factor out 3.
a+b=-9 ab=-\left(-18\right)=18
Consider -x^{2}-9x-18. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-18. To find a and b, set up a system to be solved.
-1,-18 -2,-9 -3,-6
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.
-1-18=-19 -2-9=-11 -3-6=-9
Calculate the sum for each pair.
a=-3 b=-6
The solution is the pair that gives sum -9.
\left(-x^{2}-3x\right)+\left(-6x-18\right)
Rewrite -x^{2}-9x-18 as \left(-x^{2}-3x\right)+\left(-6x-18\right).
x\left(-x-3\right)+6\left(-x-3\right)
Factor out x in the first and 6 in the second group.
\left(-x-3\right)\left(x+6\right)
Factor out common term -x-3 by using distributive property.
3\left(-x-3\right)\left(x+6\right)
Rewrite the complete factored expression.
-3x^{2}-27x-54=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(-27\right)±\sqrt{\left(-27\right)^{2}-4\left(-3\right)\left(-54\right)}}{2\left(-3\right)}
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(-27\right)±\sqrt{729-4\left(-3\right)\left(-54\right)}}{2\left(-3\right)}
Square -27.
x=\frac{-\left(-27\right)±\sqrt{729+12\left(-54\right)}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-\left(-27\right)±\sqrt{729-648}}{2\left(-3\right)}
Multiply 12 times -54.
x=\frac{-\left(-27\right)±\sqrt{81}}{2\left(-3\right)}
Add 729 to -648.
x=\frac{-\left(-27\right)±9}{2\left(-3\right)}
Take the square root of 81.
x=\frac{27±9}{2\left(-3\right)}
The opposite of -27 is 27.
x=\frac{27±9}{-6}
Multiply 2 times -3.
x=\frac{36}{-6}
Now solve the equation x=\frac{27±9}{-6} when ± is plus. Add 27 to 9.
x=-6
Divide 36 by -6.
x=\frac{18}{-6}
Now solve the equation x=\frac{27±9}{-6} when ± is minus. Subtract 9 from 27.
x=-3
Divide 18 by -6.
-3x^{2}-27x-54=-3\left(x-\left(-6\right)\right)\left(x-\left(-3\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6 for x_{1} and -3 for x_{2}.
-3x^{2}-27x-54=-3\left(x+6\right)\left(x+3\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +9x +18 = 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 = -9 rs = 18
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{9}{2} - u s = -\frac{9}{2} + u
Two numbers r and s sum up to -9 exactly when the average of the two numbers is \frac{1}{2}*-9 = -\frac{9}{2}. 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{9}{2} - u) (-\frac{9}{2} + u) = 18
To solve for unknown quantity u, substitute these in the product equation rs = 18
\frac{81}{4} - u^2 = 18
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 18-\frac{81}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{81}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{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{9}{2} - \frac{3}{2} = -6 s = -\frac{9}{2} + \frac{3}{2} = -3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.