To find the values of $x$ that make the expression {(𝑥2+5𝑥+6)(𝑥2−7𝑥+6)}/(𝑥2−4𝑥−21) equal to zero, we need to find the values of $x$ that make the numerator equal to zero because division by zero is undefined.

First, let’s factor the numerator and denominator:

Numerator: $(x_{2}+5x+6)(x_{2}−7x+6)$

$(x_{2}+5x+6)$ factors as $(x+2)(x+3)$.

$(x_{2}−7x+6)$ factors as $(x−1)(x−6)$.

So, the numerator becomes $(x+2)(x+3)(x−1)(x−6)$.

Denominator: $x_{2}−4x−21$

This quadratic expression factors as $(x−7)(x+3)$.

So, the expression becomes:

{(𝑥+2)(𝑥+3)(𝑥−1)(𝑥−6)}/{(𝑥−7)(𝑥+3)}

To make the expression equal to zero, the numerator must be zero, while the denominator should not be zero. So, the values of $x$ that make the expression equal to zero are the values that make the numerator zero:

$x=−2,x=−3,x=1,x=6$

However, we need to exclude any values that would make the denominator zero, which are $x=−3$ and $x=7$. Therefore, the possible values of $x$ are:

$x=−2,x=1,x=6$

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