We have the trinomial factored as;
[tex]2z^2+19z-21\text{ = (2z+21)(z-1)}[/tex]Here, we want to factorize the given quadratic equation
To do this, we will have to rewrite the trinomial
To rewrite, we need to change the middle term to a sum
We need to find two entities of the term z, which when added, will give +19z and when multiplied, they have a product that is equal to (product of the first and the last term);
[tex]2z^2\text{ }\times(-21)=-42z^2[/tex]These terms are -2z and 21z
Rewriting the polynomial, we have;
[tex]\begin{gathered} 2z^2-2z+21z-21 \\ =\text{ 2z(z-1) + 21(z-1)} \\ =\text{ (2z+21)(z-1)} \end{gathered}[/tex]To get the proper factors, the first thing we have to do is multiply the first and last terms
the product is -42z^2
Now, we know that the factors will include z, so we need not bother about that
We list out the factors of -42 (consider the negative and positive numbers since the factor itself is negative)
We have these factors as;
1, -1 , 2 , -2 , 3 , -3 , 6, -6 , 7 , -7 , 14 , -14 , 21, -21 and 42, -42
Now, which of these two can we add that will give +19?
As we can see; -2 and +21 fits this situation perfectly
All we need to do is to add the z after them
Thus, we simply will have -2z and 21z