Polynomials in Several Variables

I. Definitions
A. Two or more terms in an ex pression that have the same variable(s) and the same exponent(s)
on those variables are called like terms .

B. Examples of Like Terms
1. 3x + x
2. 14xyz + 7xyz – 8xyz
3. 12x² – 15x²
4. 72x²y + 83x²y

C. Examples of Unlike Terms
1. 3x – y
2. 17xy + 8yz – 5
3. 13x³ + 45x² – 10x + 6
4. –80x²y + 102xy²

II. Combining Like Terms
A. Procedure
1. Determine which are the like terms.
2. Add/ Subtract the coefficients .
3. Leave the exp onents alone .

B. Examples – Simplify by combining like terms
1. 9x – x = (9 – 1)x
Answer: 8x

2. 7a – a + 2a = (7 – 1 + 2)a
Answer: 8a

3. Now you try one: 12b – 6b + b
Answer: 7b

4. –9x – 1 + x – 4 = (–9x + x) + (–1 – 4)
Answer: -8x – 5

5. 5(y – 3) – (y – 4) = 5y – 15 – y + 4 = (5y – y) + (–15 + 4)
Answer: 4y – 11

6. –3(3x – 2) – 4(2x + 3) = -9x + 6 – 8x – 12 = (–9x – 8x) + (6 – 12)
Answer: -17x – 6

7. Now you try one: –4(3x – 2) – 6
Answer: –12x + 2

III. Evaluating Expressions
A. Procedure
1. Substitute the given value for each variable .
2. Use the order of ope rations to simplify the expression.

B. Examples - Evaluate each polynomial for x = 2 and y = –3.
1. x³y – xy + 2
We first substitute for x & y.
(2)³(–3) − (2)( –3) + 2 Exponents first.
(8)( –3) − (2)( –3) + 2 Multiply left to right.
−24 + 6 + 2 Add.
Answer: −16

2. Now you try one: 2x²y − 5y + 3
Answer: −6

IV. Multiplying
A. Remember to use the Distributive Property .
1. Multiply coefficients.
2. Add exponents (when multiplying like bases).
3. Combine like terms (but don't add the exponents at this step , only the coefficients!).

B. Examples - Find each product.
1. (4x²y)(5xy)

On this one, the Distributive Property does not apply, so we multiply the coefficients
and add the exponents on the like bases.

(4)(5)(x²⁺¹)(y¹⁺¹)
Answer: 20x³y²

2. −b(a³ − ab + b³)
Here we will distribute "−b" to all the terms in the parentheses:

(−b)(a³) − (−b)(ab) + (−b)(b³) Recall the rules for the signs when multiplying ; add
exponents of like bases:
−a³b + a(b¹⁺¹) − (b¹⁺³) Note that we put our answer in alphabetical order.

Answer: −a³b + ab² − b⁴

3. Now you try one: 3xy²(6x² − 2y)

Answer: 18x³y² − 6xy³

4. (xy − 5)²
his is a special product. If you remember the pattern, great! If not, remember what it
means to square something.

(xy − 5)² = (xy)² − (2)(xy)(5) + (5)² Simplify each term.

Answer: (xy − 5)² = x²y² − 10xy + 25

5. (x² + y)(x² − y)
Again, we have a special product, the product of conjugates .

(x² + y)(x² − y) = (x²)(x²) − (y)(y) = x⁴ − y²

6. Now you try one: (x + y)(x² + 3xy + y²)

Answer: x³ + 4x²y + 4xy² + y³

V. Applications
A. Substitute the values that you are given into each formula.

B. Examples
1. An object that is falling or vertically projected into the air has its height, in feet, above
the ground given by

s = −16t² + v₀t + s₀

where s is the height, in feet, v₀ is the original velocity of the object in feet per second,
t is the time the object is in motion, in seconds, and s₀ is the height, in feet, from which
the object is dropped or projected. The figure (bottom of left-hand column, page 356)
shows that a ball is thrown straight up from a rooftop at an original velocity of 80 feet
per second from a height of 96 feet. The ball misses the rooftop on its way down and
eventually strikes the ground. Use the formula and this information to solve Exercises
89 - 91.

How high above the ground will the ball be 4 seconds after being thrown? (#90)

First, we need to determine what we are given and substitute these values into the
formula.

The initial velocity is 80 feet per second. The ball is thrown from a height of 96 feet. It
is flight for 4 seconds. So matching these values up with the variables they represent,
the formula becomes:
s = −16(4)² + 80(4) + 96 Use PEMDAS to simplify the right-hand side.
s = −16(16) + 80(4) + 96
s = −256 + 320 + 96
s = 160 Answer the question.

Answer: The object is at a height of 160 feet after 4 seconds.

2. Now you try one:
The graph (top of right-hand column, page 356) visually displays the information about
the thrown ball described in Exercises 89-91. The horizontal axis represents the ball's
time in motion, in seconds. The vertical axis represents the ball's height above the
ground, in feet. Use the graph to solve Exercises 92-97.

After how many seconds does the ball reach its maximum height above the ground?
What is a reasonable estimate of this maximum height? (Page 356, #97)

Answer: The ball reaches its maximum height after about 2.5 seconds. The
maximum height is about 200 feet.