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Let's start with part (a) of the problem, which asks:

### Expression: \((x - 8)(x + 4)\)

#### (a) For what values of \(x\) will this expression be equal to zero?

To solve this, set the expression equal to zero and solve for \(x\):
\[
(x - 8)(x + 4) = 0
\]

Using the **zero product property**, we know that if the product of two factors is zero, then at least one of the factors must be zero. So, set each factor equal to zero:

1. \(x - 8 = 0 \implies x = 8\)
2. \(x + 4 = 0 \implies x = -4\)

Thus, the expression is equal to zero when \(x = 8\) or \(x = -4\).

#### (c) Show that this trinomial is also equal to zero at the same values of \(x\).

To answer part (c), we first expand the given expression \((x - 8)(x + 4)\) to a trinomial:

\[
(x - 8)(x + 4) = x(x + 4) - 8(x + 4)
\]
\[
= x^2 + 4x - 8x - 32
\]
\[
= x^2 - 4x - 32
\]

Now, we have the trinomial \(x^2 - 4x - 32\). To show that this trinomial is zero at \(x = 8\) and \(x = -4\), substitute these values into the trinomial.

1. For \(x = 8\):
   \[
   8^2 - 4(8) - 32 = 64 - 32 - 32 = 0
   \]
   
2. For \(x = -4\):
   \[
   (-4)^2 - 4(-4) - 32 = 16 + 16 - 32 = 0
   \]

Thus, the trinomial \(x^2 - 4x - 32\) is indeed zero at \(x = 

Think about the expression (x - 8)(x + 4). For what values of x will this expression be equal to zero? Show how you arrived at your answer. Show that this trinomial is also equal to zero at the same values of x.

This lesson explains how to solve the quadratic equation (x - 8)(x + 4) by factoring and applying the Zero Product Property. It also covers expanding binomials to form a trinomial and verifying solutions by substitution. Suitable for Grades 9-10.

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