Question

exercise #5: the product of three binomials, just like the product of two, can be found with repeated applications of the distributive property.
(a) find the product: ((x - 2)(x + 4)(x - 7)). use area arrays to help keep track of the product.
(b) for what three values of (x) will the cubic polynomial that you found in part (a) have a value of zero? what famous law is this known as?
(c) test one of the three values you found in (b) to verify that it is a zero of the cubic polynomial.

Explanation:

Part (a)

Step 1: Multiply the first two binomials

First, multiply \((x - 2)(x + 4)\) using the distributive property (FOIL method).
\[

$$\begin{align*} (x - 2)(x + 4)&=x\cdot x+x\cdot 4-2\cdot x - 2\cdot 4\\ &=x^{2}+4x-2x - 8\\ &=x^{2}+2x - 8 \end{align*}$$

\]

Step 2: Multiply the result by the third binomial

Now, multiply \((x^{2}+2x - 8)\) with \((x - 7)\)
\[

$$\begin{align*} &(x^{2}+2x - 8)(x - 7)\\ =&x^{2}\cdot x+x^{2}\cdot(-7)+2x\cdot x+2x\cdot(-7)-8\cdot x-8\cdot(-7)\\ =&x^{3}-7x^{2}+2x^{2}-14x - 8x + 56\\ =&x^{3}+(-7x^{2}+2x^{2})+(-14x-8x)+56\\ =&x^{3}-5x^{2}-22x + 56 \end{align*}$$

\]

To find the values of \(x\) for which the cubic polynomial \(y=x^{3}-5x^{2}-22x + 56\) is zero, we set \(y = 0\), i.e., \(x^{3}-5x^{2}-22x + 56=0\). We can use the fact that if \((x - a)\) is a factor of the polynomial, then \(x=a\) is a root. From the original binomials \((x - 2)(x + 4)(x - 7)\), we know that by the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), if \((x - 2)(x + 4)(x - 7)=0\), then \(x-2=0\) or \(x + 4=0\) or \(x - 7=0\). Solving these equations:

• For \(x-2=0\), we get \(x = 2\)
• For \(x + 4=0\), we get \(x=-4\)
• For \(x - 7=0\), we get \(x = 7\)

This law is known as the zero - product property.

Step 1: Substitute \(x = 2\) into the polynomial

We have the polynomial \(P(x)=x^{3}-5x^{2}-22x + 56\)
Substitute \(x = 2\) into \(P(x)\):
\[

$$\begin{align*} P(2)&=2^{3}-5\times2^{2}-22\times2+56\\ &=8-5\times4-44 + 56\\ &=8 - 20-44 + 56 \end{align*}$$

\]

Step 2: Simplify the expression

\[

$$\begin{align*} P(2)&=(8 + 56)-(20 + 44)\\ &=64-64\\ &=0 \end{align*}$$

\]
Since \(P(2) = 0\), \(x = 2\) is a zero of the cubic polynomial.

Answer:

The product is \(x^{3}-5x^{2}-22x + 56\)

Part (b)