Factor the expression by grouping. First, the expression needs to be rewritten as $z^{2}+az+bz+28$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-11$$ $$ab=1\times 28=28$$
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 $28$.
$$-1,-28$$ $$-2,-14$$ $$-4,-7$$
Calculate the sum for each pair.
$$-1-28=-29$$ $$-2-14=-16$$ $$-4-7=-11$$
The solution is the pair that gives sum $-11$.
$$a=-7$$ $$b=-4$$
Rewrite $z^{2}-11z+28$ as $\left(z^{2}-7z\right)+\left(-4z+28\right)$.
$$\left(z^{2}-7z\right)+\left(-4z+28\right)$$
Factor out $z$ in the first and $-4$ in the second group.
$$z\left(z-7\right)-4\left(z-7\right)$$
Factor out common term $z-7$ by using distributive property.
$$\left(z-7\right)\left(z-4\right)$$
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 $ax^{2}+bx+c=0$.
$$z^{2}-11z+28=0$$
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.
