Factor the expression by grouping. First, the expression needs to be rewritten as $9b^{2}+pb+qb-8$. To find $p$ and $q$, set up a system to be solved.
$$p+q=1$$ $$pq=9\left(-8\right)=-72$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-72$.
Rewrite $9b^{2}+b-8$ as $\left(9b^{2}-8b\right)+\left(9b-8\right)$.
$$\left(9b^{2}-8b\right)+\left(9b-8\right)$$
Factor out $b$ in $9b^{2}-8b$.
$$b\left(9b-8\right)+9b-8$$
Factor out common term $9b-8$ by using distributive property.
$$\left(9b-8\right)\left(b+1\right)$$
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 $ax^{2}+bx+c=0$.
$$9b^{2}+b-8=0$$
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 $b=\frac{-1±17}{18}$ when $±$ is plus. Add $-1$ to $17$.
$$b=\frac{16}{18}$$
Reduce the fraction $\frac{16}{18}$ to lowest terms by extracting and canceling out $2$.
$$b=\frac{8}{9}$$
Now solve the equation $b=\frac{-1±17}{18}$ when $±$ is minus. Subtract $17$ from $-1$.
$$b=-\frac{18}{18}$$
Divide $-18$ by $18$.
$$b=-1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{8}{9}$ for $x_{1}$ and $-1$ for $x_{2}$.
Cancel out $9$, the greatest common factor in $9$ and $9$.
$$9b^{2}+b-8=\left(9b-8\right)\left(b+1\right)$$
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 $9$
$$x ^ 2 +\frac{1}{9}x -\frac{8}{9} = 0$$
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 $rs = C$
$$r + s = -\frac{1}{9} $$ $$ rs = -\frac{8}{9}$$
Two numbers $r$ and $s$ sum up to $-\frac{1}{9}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{1}{9} = -\frac{1}{18}$. 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 $u$.
