Download Special Products – Blue Level Problems In

Special Products – Blue Level Problems
In Exercises 1 – 4, simplify by multiplying and then adding and subtracting. Write
the result as a polynomial in standard form.
1. (x – 3)2 + (x + 5)2
2. (x – 4)(x + 4) – (x + 2)2
3. (2x – 1)2 - (x + 3)2
4. (2x + 9)(3x – 7) – (5x + 2)2
5. Use the distributive property to show that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
In Exercises 6 – 7, use the result from Exercise 5 to find the product.
6. (x - 3y + 2z)2
7. (x2 + 4x – 5)2
8. Find the product.
a. (x – 5)(x2 + 5x + 25)
b. (x – 2)(x2 + 2x + 4)
c. Use the result of part (a) and (b) to suggest a general product formula involving
(x – a) and another factor. Use multiplication to test your formula.
9. a. Simplify (x – 5)(x3 + 5x2 + 25x + 125).
b. From your answer to part (a) suggest another product formula like the one from
Exercise 8.
10. Transformation Find A, B, and C by rewriting the expression in the left column so
that it appears in the form listed in the right column.
−2(x − 1)2 + 3
C + x(B + Ax)
In Exercises 11 - 13, Rick Claims that if you multiply four consecutive integers
together and add 1, you always get a perfect square.
11. Show that Rick’s statement is true if the smallest of the integers is 2.
12. Show that Rick’s statement is true if the smallest of the integers is 3.
13. Suppose the smallest of the four integers is n – 1. Find the polynomial expression in
terms of the variable n that represents the number Rick is talking about. Give the
expression in standard form.
14. Show that the expression you wrote in Exercise 13 is the square of the trinomial
n2 + n – 1.
15. Alvin's Theorem
Ms. Powers asked her class to look at a list of square numbers and see if they could
find any interesting patterns. After a few minutes, Alvin raised his hand and said,
"Ms. Powers, I think I've found something cool. If I take two consecutive squares
and subtract them, the difference is always the sum of two consecutive integers."
"Show the class what you mean by that, Alvin," said the teacher.
Alvin wrote the following on the board:
Turning to the class, he said, "I call this 'Alvin's Theorem.'"
Ms. Powers smiled and said, "Very good, but if you want to call it a theorem, you
must be able to prove that it's true for every possible pair of consecutive square
numbers. Can you do that using algebra?"
Alvin replied, "Oh yes, I can do that, too. Here's how."
a. What might Alvin have written on the board next?
b. Later, Alvin found a similar pattern for the difference of every other square,
such as 49 - 25 or 64 - 36. What pattern do you think he discovered this time?
Special Products – Blue Level Solutions
1. 2x2 + 4x + 34
2. –4x - 20
3. 3x2 - 10x - 8
4. -19x2 - 7x – 67
5. (a + b + c)2 = (a + b + c) (a + b + c) = a(a + b + c) + b(a + b + c) + c(a + b + c) =
a2 + ab + ac + ab + b2 + bc + ac + bc + c2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
6. x2 + 9y2 +4z2 – 6xy + 4xz – 12yz
7. x4 + 8x3 +6x2 – 40x + 25
8. a. x3 - 125
b. x3 - 8
c. (x – a)(x2 + ax + a2) = x3 – a3
9. a. x4 – 625
b. (x – a)(x3 + ax2 + a2x + a3) = x4 – a4
10. A = -2, B = 4, C = 1
11. (2)(3)(4)(5) + 1 = 121 = 112
12. (3)(4)(5)(6) + 1 = 361 = 192
13. 14. 15. a. For every possible pair of consecutive squares, the difference is ALWAYS the
sum of two consecutive integers. Well, this "Alvin's Theorem" could be
explained with a simple algebraic equation. Let's start.
Alvin said that If he took two consecutive squares and subtracted them, the
difference is always the sum of two consecutive integers. So, let the 1st
number be "n" and the 2nd one (it's consecutive integer) is "n + 1".
The difference of two consecutive integers can be expressed then as
(n + 1)2 - (n)2
From any basic algebra books, we can find this formula:
(a + b)2 = a2 + 2ab + b2
"a" is analogous to "n" and "b" is similar to "1". This will make:
(n + 1)2 = n2 + 2* n * 1 + 12
= n2 + 2 * n + 1
Then, our equation will become:
(n + 1)2 - n2 = (n2 + 2 * n + 1) - n2
=2*n+1
=n+n+1
= n + (n + 1)
= (n + 1) + n
There's another simple way to prove this "Alvin's Theorem". We know the
formula of subtraction for square variables is:
a2 - b2 = (a + b) * (a - b)
Let a = n + 1
b=n
Substitute (n + 1) and n to the formula:
(n + 1)2 - n2 = ((n + 1) + n) * ((n + 1) - n)
= (2 * n + 1) * 1
=2*n+1
=n+n+1
= n + (n + 1)
= (n + 1) + n
Algebraic conclusion: (n + 1)2 - n2 = (n + 1) + n
note:
Strictly you can use "n - 1" as a substitute for "n" and "n" as substitute for
"n + 1". It won't make any changes to the final solution.
"Banyak jalan menuju Roma" meaning: There are many ways to find an answer.
b. The difference of every square is always equal to 4 times the integer lying
between the two integers that have been squared.
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set.
In some cases, sources were not known.
Problems
Bibliography Information
1-9, 11-14
Larson, Ron, Laurie Boswell, Timothy D.
Kanold, and Lee Stiff. Algebra 1
Concepts and Skills. Evanston: McDougal
Littell, 2001. Print.
10
http://balancedassessment.concord.org/
15
The Math Forum @ Drexel
(http://mathforum.org/)