Factor out $5$. Polynomial $6a^{2}-8a+3$ is not factored since it does not have any rational roots.
$$5\left(6a^{2}-8a+3\right)$$
Steps Using the Quadratic Formula
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 $ax^{2}+bx+c=0$.
$$30a^{2}-40a+15=0$$
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.
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
$$30a^{2}-40a+15$$
Steps Using Direct Factoring Method
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 $30$
$$x ^ 2 -\frac{4}{3}x +\frac{1}{2} = 0$$
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 $rs = C$
$$r + s = \frac{4}{3} $$ $$ rs = \frac{1}{2}$$
Two numbers $r$ and $s$ sum up to $\frac{4}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{4}{3} = \frac{2}{3}$. 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 $u$.
$$r = \frac{2}{3} - u$$ $$s = \frac{2}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{1}{2}$
