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a+b=-5 ab=3\left(-2\right)=-6
Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
1,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-6 b=1
The solution is the pair that gives sum -5.
Rewrite 3x^{2}-5x-2 as \left(3x^{2}-6x\right)+\left(x-2\right).
Factor out 3x in 3x^{2}-6x.
Factor out common term x-2 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.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\left(-2\right)}}{2\times 3}
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(-5\right)±\sqrt{25-4\times 3\left(-2\right)}}{2\times 3}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-12\left(-2\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 3}
Multiply -12 times -2.
x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 3}
Add 25 to 24.
x=\frac{-\left(-5\right)±7}{2\times 3}
Take the square root of 49.
x=\frac{5±7}{2\times 3}
The opposite of -5 is 5.
Multiply 2 times 3.
Now solve the equation x=\frac{5±7}{6} when ± is plus. Add 5 to 7.
Divide 12 by 6.
Now solve the equation x=\frac{5±7}{6} when ± is minus. Subtract 7 from 5.
Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -\frac{1}{3} for x_{2}.
Simplify all the expressions of the form p-\left(-q\right) to p+q.
3x^{2}-5x-2=3\left(x-2\right)\times \left(\frac{3x+1}{3}\right)
Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
Cancel out 3, the greatest common factor in 3 and 3.
x ^ 2 -\frac{5}{3}x -\frac{2}{3} = 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 3
r + s = \frac{5}{3} rs = -\frac{2}{3}
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{5}{6} - u s = \frac{5}{6} + u
Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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{5}{6} - u) (\frac{5}{6} + u) = -\frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}
\frac{25}{36} - u^2 = -\frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{3}-\frac{25}{36} = -\frac{49}{36}
Simplify the expression by subtracting \frac{25}{36} on both sides
u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{6} - \frac{7}{6} = -0.333 s = \frac{5}{6} + \frac{7}{6} = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.