Consider $42-R-R^{2}$. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-R^{2}-R+42$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-R^{2}+aR+bR+42$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=-42=-42$$
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 $-42$.
$$1,-42$$ $$2,-21$$ $$3,-14$$ $$6,-7$$
Calculate the sum for each pair.
$$1-42=-41$$ $$2-21=-19$$ $$3-14=-11$$ $$6-7=-1$$
The solution is the pair that gives sum $-1$.
$$a=6$$ $$b=-7$$
Rewrite $-R^{2}-R+42$ as $\left(-R^{2}+6R\right)+\left(-7R+42\right)$.
$$\left(-R^{2}+6R\right)+\left(-7R+42\right)$$
Factor out $R$ in the first and $7$ in the second group.
$$R\left(-R+6\right)+7\left(-R+6\right)$$
Factor out common term $-R+6$ by using distributive property.
$$\left(-R+6\right)\left(R+7\right)$$
Rewrite the complete factored expression.
$$2\left(-R+6\right)\left(R+7\right)$$
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 $ax^{2}+bx+c=0$.
$$-2R^{2}-2R+84=0$$
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.
