Question

the store feature is extremely helpful when you are trying to determine the value of an expression at one or two input values of x. but, if you want to know an expressions value for multiple inputs, then tables are a much better tool.
exercise #4: the expression $x^3 + 2x^2 - 16x - 32$ has an integer zero somewhere on the interval $0 \leq x \leq 10$. use a table to find the zero on this interval. show the table.

table commands can be particularly good at establishing proof that two expressions are equivalent. this is particularly helpful when youve done a number of manipulations and you want to have confidence that youve produced an algebraically equivalent expression.
exercise #5: consider the more complex algebraic expression shown below:
$$(x + 5)(x + 8) - (x + 3)(x - 2)$$
(a) this relatively complex expression simplifies into a linear binomial expression. determine this expression carefully. show your work below.

(b) set up a table using the original expression and the one you found in (a) over the interval $0 \leq x \leq 5$. compare values to determine if you correctly simplified the original expression.

$x$	$y_1 = $	$y_2 = $
1
2
3
4
5

common core algebra ii, unit #1 - essential algebra concepts - lesson #6
emathinstruction, red hook, ny 12571, \textcopyright 2015

Explanation:

Exercise #4

Step1: Define the expression

Let $f(x) = x^3 + 2x^2 -16x -32$

Step2: Evaluate at integer x from 0 to 10

• $f(0) = 0^3 + 2(0)^2 -16(0) -32 = -32$
• $f(1) = 1^3 + 2(1)^2 -16(1) -32 = 1+2-16-32 = -45$
• $f(2) = 2^3 + 2(2)^2 -16(2) -32 = 8+8-32-32 = -48$
• $f(3) = 3^3 + 2(3)^2 -16(3) -32 = 27+18-48-32 = -35$
• $f(4) = 4^3 + 2(4)^2 -16(4) -32 = 64+32-64-32 = 0$

Step1: Expand each product

First product: $(x+5)(x+8) = x^2 +8x +5x +40 = x^2 +13x +40$
Second product: $(x+3)(x-2) = x^2 -2x +3x -6 = x^2 +x -6$

Step2: Subtract the two expansions

$(x^2 +13x +40) - (x^2 +x -6)$

Step3: Simplify the expression

$x^2 +13x +40 -x^2 -x +6 = 12x +46$

Step1: Define $y_1$ and $y_2$

Let $y_1=(x+5)(x+8)-(x+3)(x-2)$, $y_2=12x+46$

Step2: Evaluate for $x=0$ to $5$

• $x=0$: $y_1=(5)(8)-(3)(-2)=40+6=46$; $y_2=12(0)+46=46$
• $x=1$: $y_1=(6)(9)-(4)(-1)=54+4=58$; $y_2=12(1)+46=58$
• $x=2$: $y_1=(7)(10)-(5)(0)=70-0=70$; $y_2=12(2)+46=70$
• $x=3$: $y_1=(8)(11)-(6)(1)=88-6=82$; $y_2=12(3)+46=82$
• $x=4$: $y_1=(9)(12)-(7)(2)=108-14=94$; $y_2=12(4)+46=94$
• $x=5$: $y_1=(10)(13)-(8)(3)=130-24=106$; $y_2=12(5)+46=106$

Answer:

$x$	$f(x)$
1	-45
2	-48
3	-35
4	0
5	53
6	148
7	291
8	488
9	745
10	1068

The integer zero on $0 \leq x \leq 10$ is $x=4$.

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Exercise #5 (a)