Factor the expression by grouping. First, the expression needs to be rewritten as $-3x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=-3=-3$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
$$a=3$$ $$b=-1$$
Rewrite $-3x^{2}+2x+1$ as $\left(-3x^{2}+3x\right)+\left(-x+1\right)$.
$$\left(-3x^{2}+3x\right)+\left(-x+1\right)$$
Factor out $3x$ in $-3x^{2}+3x$.
$$3x\left(-x+1\right)-x+1$$
Factor out common term $-x+1$ by using distributive property.
$$\left(-x+1\right)\left(3x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$-3x^{2}+2x+1=0$$
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.
Now solve the equation $x=\frac{-2±4}{-6}$ when $±$ is plus. Add $-2$ to $4$.
$$x=\frac{2}{-6}$$
Reduce the fraction $\frac{2}{-6}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{1}{3}$$
Now solve the equation $x=\frac{-2±4}{-6}$ when $±$ is minus. Subtract $4$ from $-2$.
$$x=-\frac{6}{-6}$$
Divide $-6$ by $-6$.
$$x=1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{1}{3}$ for $x_{1}$ and $1$ for $x_{2}$.
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 $x^2+Bx+C=0$.
$$x ^ 2 -\frac{2}{3}x -\frac{1}{3} = 0$$
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 $rs = C$
$$r + s = \frac{2}{3} $$ $$ rs = -\frac{1}{3}$$
Two numbers $r$ and $s$ sum up to $\frac{2}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{2}{3} = \frac{1}{3}$. 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 $u$.
$$r = \frac{1}{3} - u$$ $$s = \frac{1}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{1}{3}$
