Question

1. think about the expression ((x - 8)(x + 4)).

(a) for what values of (x) will this expression be equal to zero? show how you arrived at your answer.
(b) write this product as an equivalent trinomial.
(c) show that this trinomial is also equal to zero at the larger value of (x) from part (a).

Explanation:

Part (a)

Step1: Recall zero - product property

The zero - product property states that if \(ab = 0\), then either \(a=0\) or \(b = 0\) (or both). For the expression \((x - 8)(x + 4)=0\), we set each factor equal to zero.
So we have two equations: \(x-8 = 0\) or \(x + 4=0\).

Step2: Solve the first equation

For the equation \(x-8=0\), we add 8 to both sides of the equation.
\(x-8 + 8=0 + 8\), which simplifies to \(x = 8\).

Step3: Solve the second equation

For the equation \(x + 4=0\), we subtract 4 from both sides of the equation.
\(x+4-4=0 - 4\), which simplifies to \(x=-4\).

Step1: Use the distributive property (FOIL method)

To expand \((x - 8)(x + 4)\), we use the FOIL method. FOIL stands for First, Outer, Inner, Last.

• First: Multiply the first terms of each binomial: \(x\times x=x^{2}\)
• Outer: Multiply the outer terms: \(x\times4 = 4x\)
• Inner: Multiply the inner terms: \(-8\times x=-8x\)
• Last: Multiply the last terms: \(-8\times4=-32\)

Step2: Combine like terms

Now we combine the like terms (the terms with \(x\)).
\(x^{2}+4x-8x - 32=x^{2}+(4x-8x)-32\)
\(4x-8x=-4x\), so the expression becomes \(x^{2}-4x - 32\)

Step1: Identify the larger value from part (a)

From part (a), we found that the solutions of \((x - 8)(x + 4)=0\) are \(x = 8\) and \(x=-4\). The larger value of \(x\) is \(x = 8\).

Step2: Substitute \(x = 8\) into the trinomial

We substitute \(x = 8\) into the trinomial \(x^{2}-4x - 32\) (from part (b)).
First, calculate \(x^{2}\) when \(x = 8\): \(8^{2}=64\)
Then, calculate \(-4x\) when \(x = 8\): \(-4\times8=-32\)
Now, substitute these values into the trinomial: \(64-32 - 32\)

Step3: Simplify the expression

\(64-32-32=(64-32)-32=32 - 32=0\)
Since when we substitute \(x = 8\) into the trinomial \(x^{2}-4x - 32\), we get \(0\), the trinomial is equal to zero at \(x = 8\).

Answer:

The values of \(x\) for which the expression \((x - 8)(x + 4)\) is equal to zero are \(x = 8\) and \(x=-4\).

Part (b)