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p+q=-7 pq=3\times 2=6
Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa+2. To find p and q, set up a system to be solved.
-1,-6 -2,-3
Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 6.
-1-6=-7 -2-3=-5
Calculate the sum for each pair.
p=-6 q=-1
The solution is the pair that gives sum -7.
\left(3a^{2}-6a\right)+\left(-a+2\right)
Rewrite 3a^{2}-7a+2 as \left(3a^{2}-6a\right)+\left(-a+2\right).
3a\left(a-2\right)-\left(a-2\right)
Factor out 3a in the first and -1 in the second group.
\left(a-2\right)\left(3a-1\right)
Factor out common term a-2 by using distributive property.
3a^{2}-7a+2=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.
a=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 2}}{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.
a=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 2}}{2\times 3}
Square -7.
a=\frac{-\left(-7\right)±\sqrt{49-12\times 2}}{2\times 3}
Multiply -4 times 3.
a=\frac{-\left(-7\right)±\sqrt{49-24}}{2\times 3}
Multiply -12 times 2.
a=\frac{-\left(-7\right)±\sqrt{25}}{2\times 3}
Add 49 to -24.
a=\frac{-\left(-7\right)±5}{2\times 3}
Take the square root of 25.
a=\frac{7±5}{2\times 3}
The opposite of -7 is 7.
a=\frac{7±5}{6}
Multiply 2 times 3.
a=\frac{12}{6}
Now solve the equation a=\frac{7±5}{6} when ± is plus. Add 7 to 5.
a=2
Divide 12 by 6.
a=\frac{2}{6}
Now solve the equation a=\frac{7±5}{6} when ± is minus. Subtract 5 from 7.
a=\frac{1}{3}
Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.
3a^{2}-7a+2=3\left(a-2\right)\left(a-\frac{1}{3}\right)
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}.
3a^{2}-7a+2=3\left(a-2\right)\times \frac{3a-1}{3}
Subtract \frac{1}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
3a^{2}-7a+2=\left(a-2\right)\left(3a-1\right)
Cancel out 3, the greatest common factor in 3 and 3.
x ^ 2 -\frac{7}{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{7}{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{7}{6} - u s = \frac{7}{6} + u
Two numbers r and s sum up to \frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{3} = \frac{7}{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{7}{6} - u) (\frac{7}{6} + u) = \frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}
\frac{49}{36} - u^2 = \frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{3}-\frac{49}{36} = -\frac{25}{36}
Simplify the expression by subtracting \frac{49}{36} on both sides
u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{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{7}{6} - \frac{5}{6} = 0.333 s = \frac{7}{6} + \frac{5}{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.