Saturday, October 08, 2016

August 2016 Common Core Algebra II Regents Part 3

I don't usually post the answers to the Algebra II exam, but I've received enough requests that I've decided to post some of the open-ended questions.
Some questions were posted earlier here, here and here.

August 2016, Algebra II Part 3

33. Find algebraically the zeros for p(x) = x3 + x2 - 4x - 4.

On the set of axes below, graph y = p(x).

Group the first two terms and the last two terms of the expression. Note that this causes a sign change in the second group:

p(x) = (x3 + x2) - (4x + 4)

Factor each pair:
p(x) = x2(x + 1) - 4(x + 1)

Do you see that each pair has (x + 1) in it? That can be factored, and put up front:
p(x) = (x + 1)(x2 - 4)

Almost done. the x2 expression is the Difference of Squares and can be factored further:
p(x) = (x + 1)(x + 2)(x - 2)

Set p(x) = 0
0 = (x + 1)(x + 2)(x - 2)

(x + 1) = 0 or (x + 2) = 0 or (x - 2) = 0

x = -1 or x = -2 or x = 2

That's your answer.

Put the equation in your calculator to get a table of values. From part 1, you know you have these points:
the roots (-1, 0), (-2, 0) and (2, 0) as well as (0, -4), the y-intercept.
You need x = 1. p(1) = (1)3 + (1)2 - 4(1) - 4 = 1 + 1 - 4 - 4 = -6. So (1, -6) is a point on the curve.
Important: You need to see the curve, so use the calculator!

34. One of the medical uses of Iodine–131 (I–131), a radioactive isotope of iodine, is to enhance x-ray images. The half-life of I–131 is approximately 8.02 days. A patient is injected with 20 milligrams of I–131. Determine, to the nearest day, the amount of time needed before the amount of I–131 in the patient’s body is approximately 7 milligrams.

Given the fractions, exponents and subscripts in this problem, I'm going to use an image file to show my work.

Basically, you need to use the formula for exponential decay: y = a(b)x
In this case, y = 7, your target amount, a = 20, the initial amount, b = .5, which is half, and x = t/8.02 days because you are finding the number of 8.02 day increments.
Finding the exponent means using log.5 7/20 = 1.5145731728...
Multiply by 8.02 and round down. The answer is 12 days.

Any questions?

Thoughts, comments, concerns?

More to come.


CCSSIMath said...

p(x) = x³ + x² - 4x - 4

We all saw this first: switch the 2nd and 3rd terms.

p(x) = x³ - 4x + x² - 4
= x(x² - 4) + (x² - 4)

It's a non-obvious ploy, but it obviates the need to change signs, a common cause of errors. We think students should be exposed to factoring "tricks" like this. Sometimes such rearrangements are actually the only way that works.

What percentage of students factored this correctly? Are there stats?

(x, why?) said...

Thanks for the comment.

I wouldn't call it non-obvious if only because I actually taught grouping that way first in my Algebra 1 class this past spring. But the switching confused several of them, so I tried the other way. This was met with "Why didn't we do it this way in the first place!" So I kept doing this.

That said, the distributed minus sign threw a few of them. Some figured it out when -- using this problem as an example -- they got a -4 in one term and a +4 in the other.

If there are any stats, it would likely only show how many points were scored on the entire problem, not just the factoring. I haven't seen them.