Factor the expression by grouping. First, the expression needs to be rewritten as $3z^{2}+az+bz+30$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-19$$ $$ab=3\times 30=90$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $90$.
Rewrite $3z^{2}-19z+30$ as $\left(3z^{2}-10z\right)+\left(-9z+30\right)$.
$$\left(3z^{2}-10z\right)+\left(-9z+30\right)$$
Factor out $z$ in the first and $-3$ in the second group.
$$z\left(3z-10\right)-3\left(3z-10\right)$$
Factor out common term $3z-10$ by using distributive property.
$$\left(3z-10\right)\left(z-3\right)$$
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 $ax^{2}+bx+c=0$.
$$3z^{2}-19z+30=0$$
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 $z=\frac{19±1}{6}$ when $±$ is plus. Add $19$ to $1$.
$$z=\frac{20}{6}$$
Reduce the fraction $\frac{20}{6}$ to lowest terms by extracting and canceling out $2$.
$$z=\frac{10}{3}$$
Now solve the equation $z=\frac{19±1}{6}$ when $±$ is minus. Subtract $1$ from $19$.
$$z=\frac{18}{6}$$
Divide $18$ by $6$.
$$z=3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{10}{3}$ for $x_{1}$ and $3$ 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 $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $3$
$$x ^ 2 -\frac{19}{3}x +10 = 0$$
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 $rs = C$
$$r + s = \frac{19}{3} $$ $$ rs = 10$$
Two numbers $r$ and $s$ sum up to $\frac{19}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{19}{3} = \frac{19}{6}$. 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 $u$.
$$r = \frac{19}{6} - u$$ $$s = \frac{19}{6} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 10$
$$(\frac{19}{6} - u) (\frac{19}{6} + u) = 10$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{361}{36} - u^2 = 10$$
Simplify the expression by subtracting $\frac{361}{36}$ on both sides
$$-u^2 = 10-\frac{361}{36} = -\frac{1}{36}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
