Polynomial Models
Definition: A polynomial is an algebraic expression made up of sums and products of
variables and constants. The general form is a xn + b xn-1 +…+ k in this a, b,..., k are the
constants, n is a positive integer
Learn How To:
Add, subtract this is done by combining like terms, with two of more terms. Also,
multiply, divide, solve equations with polynomials, factoring and expanding equations.
Addition:
Polynomials are added in the same basic way as monomials are. The easiest way to be
sure of the proper combination of like terms, is to use communicative property of
addition. If a polynomial contains only one variable, that is raised in decreasing or in
increasing powers.
Ex: 3k2 - 5x + 6x3 - 9
This could be arranged either 6x2 + 3x 2 - 5x - 9 or -9 – 5x + 3 x2 + 6x3 decreasing order
is preferred
The communicative property of addition is used when the polynomials contains more
than one variable, alphabetical arrangement is preferred.
Ex: -5x2 + 3x + 2
rearrange the terms of each polynomial first, in decreasing
2
7x – 12x + 3
powers of x, then line up like terms in a vertical column
2x2 - 9x + 5
Ex: add 3a – 5c + 2b with 6a + 10c with –3b + 2c rearrange 3a + 2b – 5c, 6a + 10c,
-3b + 2c line up the like terms vertically
3a + 2 b – 5c
6a
+ 10c
-3b + 2c
(The sum is) 9a – b + 7c
(When terms are missing the coefficient is zero, they can be left blank)
Questions: Add the following polynomials
1. x + 5y
3x – y
5. 10y – 3x
4y – 7x
-6y + 8x
2. a2 b – r
2a2 b + 6r
6.
3. 2d3 + 7d2 – 5d + 2
-6d 3– 9d2 +7d + 12
3d3 + 4d2 – 2d – 8
4.
4m + n
-m–n
m2 – 5mn + 6n2
4m2 – 8mn – 2n2
Subtracting:
Subtracting polynomials follows the same procedures as in addition of polynomials
Ex: find the difference of 3a2 - 2 + 5a and 6a2 + 4a – 5
Solution
3a2 + 5a – 2
6a2 + 4a – 5
minuend
-3a2 + a + 3
subtracted
Be careful to read problem so that you know which way to subtract
Ex: subtract r2 – 2rs + 5s2 from r2 + 2rs + s2
 First polynomial given is the subtracted
r2 + 2rs + s2
r2 – 2rs + s2
4rs
To check you works simplify compare the minuend with the sum for the subtracted and
difference. ( r2 – 2rs + 5s2) + 4rs = r2 + 2rs + s2 if the answer is yes than your check
proves your work.
- Subtract the lower polynomial form the upper.
Questions:
1. 7x + 8
-3x + 4
2. –x2 + 3xy – 7y2
3. 6a + 2b
-5x2 - 7xy + 3y 2
-8a – b
4. Find the difference of 5x – 3y and 2x + 4y.
5. 13x2- 7xy – 5y2
-8x 2
- y2
6. Find the difference between the sum 4a2 – 3a + 9 and 5a2 + 7a +2, and the
expression 9a2 + 4a – 15.
Multiplication:
To multiply, you use the Distributive Property, Rules for signed numbers, and Laws of
exponents, and combine them together to find the product. The basic form of the
Distributive Property is a(b+c) = ab +ac.
Ex.
5x ( -2x +7) = (5x)(-2x) + (5x)(+7) = -10x2 + 35x
18(a/6 – b/9 + c/2) = 18(a/6) + (-b/9) + 18(c/2) = 3a- 2b + 9c
this is a review of monomial and multinomial, product of two binomials,
Distributive property the multiplier, the first binomial, is multiplied by each term it the
multiplicand.
Ex.
(a+b)(c+d) we would multiply (a+b) by c and d, and get:
Ex.
Ex.
(a+b)c + (a+b)d now we apply the Distributive property and we would get
ac + bc + ad + bd
(2+3)(4+1) would this be equal to 25, (5)(5)
(2+3)4 + (2+3)1
(2)4 + (3)4 + (2)1 + (3)1
8 + 12 +2 +3
20 + 5
25 which was is the same as (5)(5) so it is right
(x +5)(x-2)
(x+5)(x-2)= (x+5)x + (x+5)(-2)= x2 + 5x – 2x – 10 combine like terms
x2 + 3x –10
or you can arrange it vertically and multiply, as if it were a two-digit numbers
x- 2
x+ 5 you would multiply 5 to the –2, then 5 by x
+5x – 10 then you would multiply the x by the –2, then x by x
+x2-2x
now add these together.
2
x +3x-10
You still get the same answer, but you might find the vertical way
easier.
Ex:
(2x-1) (3x2+5x-7)
3x2+5x-7
x
2x-1 multiply terms by 1 first.
2
-3x -5x+7
6x3+10x2-14x
add together
6x2 +7x2-19x+7
** remember to arrange the polynomials in either decreasing or increasing powers, this
makes the process easier.
Questions:
(Multiply as indicated)
1.
.
3.
(x+3) (x+4)
(x+y) (x2+xy+y2)
5.
(a+b) (a+b)
2.
(x2-2) (x2-5)
4.
(x2-5) (x2+3)
Dividing:
To divide a polynomial, you can divide any number of terms by a monomial. You would
divide each term of the polynomial by the monomial, the divisor.
DIVIDEND= DIVISOR X QUOTIENT + REMAINDER
You can check your work by comparing the dividend with the product of the divisor and
the quotient. If there is a remainder then it must be added on to the product of the divisor
and the quotient before you compare it with the dividend.
Ex. Divide (12a3 + 27a2) by 3a
Answer:(12a3 + 27a2) / 3a = 12a3 + 27a2 = 4a2 +9a
3a
3a
check: 3a(4a2 +9a) = 12a3 + 27a2
Ex. Divide (6x3 – 18x2 + 12x) by –2x
Answer: (6x3 – 18x2 + 12x) / (-2x)= 6x3 + -18x2 + 12x = -3x2 +9x - 6
-2x -2x
-2x
Check: -2x(-3x2+ 9x –6) = 6x3 – 18x2 + 12x
Questions:
Divide the following
1. 9a3b2 – 15a2b4
-3ab2
2. 12x2 – 3x
3x
3. 4a2- 8a + 3 by 2a
Expanding:
When expanding, you take the number outside of the bracket and multiply it to
each of the terms inside of the bracket.
Ex:
Ex:
3x(2x2-5x-7)
3x(2x2)+3x(-5x)+3x(-7)
6x3-15x2-21x
(apply to all terms)
3(2m+5n)-4(5m-3n)+6(3m-5n)
(3)(2m)+(3)(5n)-(4)(5m)-(-4)(-3n)+(6)(3m)+(6)(-5n)
6m+15n-20m+12n+18m-30n
Questions:
Expand the following
1. 4(3a-2)
2. x(5x + 3y)
3. –3(x-1)
4. m(5m2 – 3m – 4)
5. –y( x-y + 3yz)
Factoring:
For factoring you would find the Greatest Common factor, which applies to all of the
terms. You would then place the factor, on the outside of the bracket and then factor out
the brackets with the greatest common factor.
Ex. 15x2y – 25x3y2
5x2y(3-5xy)
15x2y-25x3y2= 5x2y(3-5xy)
Ex. 3x2y +5y2 this polynomial cannot be factored into a simpler polynomial using this
method.
Questions:
1. x3 – 5x2 + x
2. mn2 – m2n
3. a2b2 – ab
4. 4a2- 6b2
5. 6a(a-b) + 4b(a-b) + 3(a-b)
Polynomial Models:
A polynomial model is an equation that is used to solve situations that may occur in every
day life (but not always). You solve them by using equations that are usually cubic of
higher order functions.
In January of the year 2000, Mike’s dad Harry was eleven times as old as my son. How
old was my son Jake in January of 2000?
Let “H” stand Harry, and let “M” stand for Mike’s age. Then H = 11M in the year 2000.
In the year 2012, Harry and Mike will be twelve years older, so their ages will be H + 12
and M + 12. Also, Harry will be three times as old as Mike so
H + 12 = 3(M + 12)
Okay so now you have two equations, each with two variables
H= 11M
M+12 = 3W+36
Next use the first equation to simplify the second; since H + 11M, plug in “11M” for “H”
in the second equation:
H + 12 = 3M + 36
(11M) + 12 = 3M + 36
11M – 3M = 36 – 12
8W = 24
W=3
Remember that the problem did not ask for the value for variable M; it only asked for the
age of a person. So the answer is that Mike is 3 years old at the beginning of the year
2000.
In three more years, Miguel’s grandfather will be six times as old as Miguel was last
year. When Miguel’s present age is added to his grandfather’s present age, the total is 68.
How old is each one now?
Let Miguel’s present age be “m” and let his grandfather’s age be “g”. Then m + g = 68.
Miguel’s age “last year” was m-1. His grandfathers age in three more years will be g + 3.
The grandfather’s age three years from now is six times Miguel’s age lat year; in other
words:
g + 3 = 6(m-1)
Which gives you two equations with two variables:
m+ g + 68
g+3 = 6(m-1)
Solving the first equation, you get m= 68-g. the problem will work out a bit differently in
the middle, but answer will be the same at the end. So plug this into the second equation.
g + 3 = 6m – 6
g + 3 = 6 (68-g) –6
g + 3 = 408 – 6g – 6
g +3 = 402-6g
g + 6g = 402 – 3
7g = 399
g = 57
Since “g” stands for the grandfather’s current age, then the grandfather is 57 years old.
Since m + g = 68, then m = 11, and Miguel is presently eleven years old.
Quiz
1. Add the following (remember to rearrange terms in the preferred order)
(a) a2b-r
2a2b + 6r
(b) 2r2 –5r +7
-6r2 + 3r – 2
5r +6r – 8
(d) 11a – 7b + 4c – 3d
-12a + 8b – 5c + 4d
(c) m2 + n + 1; 2m-1; 3m2 –4m +2
(e) 6x2- 3x +2
-9x2 –4x + 6
(f) 3a +4a
-a + 3b
6a – 7b
2. Subtract the following
(a) 4a – 2b + 7c
-2a-2b + 8c
(e) 3x2
-2y2
-2x2 – xy + y2
(b) 2n+4
4n+3
(c) –2a2+ 5a =9
7a2 – a +17
(d) –4.6m + 3.5n
-6.1m – 2.8n
3. Multiply the following
(a) (a+1)(a-2)
(b) (m+3)(m-3)
(c) (x+2)(x-3)
(d) (n+2m)(n+mn-m)
4. Divide the following
(a) 6x3 + 2x2 – 4x + 7
2
(b) 5a-25b+20c
-5
5. Factor the following (if possible)
(a) m8 + 2m4 – 3m
(b) x4 + 4ax3+ 4a2x2 (c) 12a2 – 4ab – 3ax2 + bx2
6. Expand the following
(a) –3(x2-2x –5)
(b) 3(2b+3) –5(3b-5) (c) 18(a-b) + 45( a+b)
Answer the following polynomial model question:
Your uncle walks in, jingling the loose change in his pocket. He tells you that if you can
figure out how many of every coin there are in his pocket.
He tells you that there are 26 coins in his pockets and they are dollars and quarters. And it
adds up to seventeen dollars. How much of each coin does have?
Appendix:
Addition
1.
2.
3.
4.
5.
6.
4x + 4y
3a2 b + 5r
11d3 + 20d2 – 14d – 18
3m
8y – 2x
5m2 – 13mn + 4n2
Subtraction
1. 4x+4
2. –6x2 +10xy-10y2
3. –2a + 3b
4. 3x-7y
5. 5x2 –7xy- 4y2
6. 10a2 –14a+22
Multiplying
1. = x2+3x+4x+12
= x2+7x+12
2. = x4+2x2-5x2+10
= x4-3x2+10
3. = x3+x2y+xy2+yx2+xy2+y3
= x3+x4y2+x2y4+y3
4. = x4+3x2-5x2-15
= x4-2x2-15
5. = a2+2ab+b2
Dividing:
1. Answer: 9a3b2 + -15a2b4 = -3a2 + 5ab2
-3ab2 -3ab2
Check: -3ab2(-3a2 + 5ab2) = 9a3b2 – 15a2b4
2. 4x-1 = 12x –3x
Check: 3x(4x-1)= 12x2 – 3x
3. 2a – 4 = 3
2a
Check: 2a(2a- 4) +3 = 4a2- 8a + 3
Expanding:
1. = 4(3a) + 4(-2)
= 12a-8
2. = x (5x) + (3y)
= 5x2 + 3xy
3. = -3(x) + -3(-1)
= -3x + 3
4. = m(5m2) + m(-3m) + m(-4)
= 5m3 – 3m2 – 4m
5. = -y(x) + -y(-y) + -y(3yz)
= -xy + y2 – 3y2z
Factoring
1. = x(x2 – 5x + 1)
2. = mn(n-m)
3. = ab(ab –1)
4. = 2(2a2 –3b2)
5. = (a-b) (6a + 4b +3)
Quiz:
1. a)3a2b + 5r
b) r2+ 4r – 3
c) 4m2-im+ 3
d)-1a + 1b - 1c + 1d
e) –3x2- 7x + 8
f) 8a
2.a)6a- 1c
b) –2n+ 1
c) -9a2 + 6a –8
d) 1.5m + 6.3n
e) 5x2 –xy – 3y
3.a) a2+ 2a- 1a – 2
a2+1a -2
b) m- 9
c) x2 – 1x -6
d) n2 + mn2+ mn+ 2mn2- 2m2
4. a) 3x3 + x2 – 2x + 3.5
b) –1a+ 5b- 4c
5.a) m(m2+ 2m2-3)
b) x2 (x2+ 4ax+4a2)
c) not possible
6. a) –3x2 +6x + 15
b) –9b+34
c) 63a + 27b
7.
There are twenty six coins in total some of them are dollars; let “d” stand for the
number of dollar coins. The rest for the coins are quarters let “q” stand for the number of
quarters. Then d + q = 26
If your uncle has only one quarter, then 25*1 = 25 cents comes from quarters. If he has
two quarters, then 25*2= 50 cents comes from quarters. Since he has q quarters, then
25*q = 25q cents comes from quarters.
Now, for the dollar part to work, you’ll have too convert to cents. That is, one dollar is
one hundred cents. Since he had 100d cents from the dollar coins.
He has seventeen dollars in total of 1700 cents, part of which is from quarters and part of
which is from dollars.
If you know how to solve systems of liner equations, you can see that you have the
following systems:
D+ q = 26
100d + 25q = 1700
And if you do what you were shown before you would get d = 14
So there would be 14 coins dollars and 12 quarters
Bibliography:
www.sosmath.com
www.alltell.net
www.math.usask.ca
Making Mathematics 10 math book