# Unit 7 (Chapter 7) Polynomials (P

```Unit 7 (Chapter 7) Polynomials (P. 280 - 327)
Section 7. 1 Adding and Subtracting Polynomials. (p. 285 – 289)
Polynomial – An algebra expression consisting of 1 or more terms
There are three main types that we consider:
(i)
Monomial (1 term)

eg: 2x, 5x3y2
(ii)
Binomial (2 terms)

eg: 2x + 3, 5a2 – 7b, 3x2 + 5x
(iii)
Trinomial (3 terms)

eg: x2 + 5x + 4, 3b4 + 6b2 - 12
Finding the Degree of a Polynomial
1.) When the expression consists of a monomial add the exponents of the variables to
determine the degree
ex:
2x3  degree 1
3x4y1  degree 4 + 1 = 5
3x2yz5  degree 2 + 1 + 5 = 8
2.) When the problem is a polynomial with the same variable , the degree is the largest
exponent.
ex:
3x5 +8x4 – 2x3 - 4x2 +5x – 3
 degree 5 (highest exponent)
3.) When the problem is a polynomial with multiple variables, add the exponents of each
monomial (term). The largest one is the degree
ex:
3x5y4z10 – 2x7y4z8 + 3x10y10 – 5x4yz2
5 + 4 + 10
= 19
7+4+8
= 18
10 + 10
= 20
4+1+2
=9
So the degree is 20
Unit 7 – Polynomials
-1-
Recall that we can only add or subtract terms that are alike:
ex:
2x, 5x, -10x are like terms
Exponents must be the same when
performing + or - !!!
2x2, 5x, -10x3 are unlike terms
EX.1
(a) Simplify the following:
11a + 6b – 5a – 3b + 2a – 1b + 6b – 3a
= 11a – 5a + 2a – 3a + 6b – 3b – 1b + 6b
= 5a + 8b
:regroup like parts then add/subtract as normal
(b) Evaluate the above polynomial for a = 3 and b = -2
5a + 8b
: simply substitute in the number for the appropriate variable
= 5(3) + 8(-2)
: simplify
= 15 – 16
= -1
EX. 2: Simplify and then evaluate for p = -2
3p2 + 2p3 + 6p2 - 4p + 6p + 5p3
= 5p3 + 2p3 + 6p2 + 3p2 – 4p + 6p
= 7p3 + 9p2 + 2p
= 7(-2)2 + 9(-2) + 2(-2)
= 7(4) – 18 – 4
= 28 – 18 – 4
=6
EX.3 Simplify
(a) (2x2 + 3x + 4) + (1x2 – 2x – 5)
= 2x2 + 1x2 + 3x – 2x + 4 – 5
= 3x2 + 1x – 1
: regrouping is not necessary but it certainly
organizes the problem much better!
: always be focused on BEDMAS!
: regroup
: combine like parts
we could also use algebra tiles to solve. (This technique shown represents adding in columns
with one expression on top of another.)


Unit 7 – Polynomials
-2-
2x2 + 3x + 4
+
1x2 – 2x – 5
3x2
+
1x
-
1
3x2 + 1x – 1
(b) (-2y2 + 3y – 5) + ( -y2 – y + 1)
= -2y2 - y2 + 3y – y – 5 + 1
= -3y2 + 2y – 4
= -3y2 + 2y – 4
We need to be extra careful when subtracting tems:
ex: (-3) – (2) = -5
(-3) – (-2) = -3 + 2 = -1
EX.4 Simplify
(a) (2x – 1) – ( -4x + 2)
= 2x – 1 + 4x – 2
= 2x + 4x – 1 – 2
= 6x – 3
: when subtracting a negative it is the same as
: use distributive property to remove brackets first. Remember the
number is understood to be a 1 in front of a variable or bracket if it
is not written in! In this case we are multiplying by a -1.
: regroup like terms. Note the sign changes NOTE:
The regrouping step is
: combine like terms
not necessary. It is to
keep you more
organized and aware
of signs. Hence less
careless mistakes!
(b) (-2y2 – 4y + 1) – ( -4y2 – 3y + 2)
= -2y2 – 4y + 1 + 4y2 + 3y - 2
= -2y2 + 4y2 – 4y + 3y + 1 – 2
= 2y2 – y - 1
= 2y2 – 1y - 1
EX.5 Simplify and evaluate for x = 1
(5x2 + 8x – 3) – (2x2 + 2x – 2)
= 5x2 + 8x – 3 – 2x2 - 2x + 2
= 5x2 – 2x2 + 8x – 2x – 3 + 2
= 3x2 + 6x – 1
for x = 1
=3x2 + 6x - 1
3(1)2 + 6(1) - 1
= 3(1) + 6 – 1
=3+6 -1
=8
Unit 7 – Polynomials
-3-
Seatwork/homework: p. 288 – 289 #5acegi, 11aceg, 12, 14ab, 16, 17ab(sketch/use algebra
tiles)
Section 7.2 Multiplying and Dividing Monomials (p. 290 – 294)
Multiplying
EX.1 State the product
(a) 2x(3x) = 6x2
(b) -3a(4a3b) = -12a4b
(c) 2x2(-3x2y) = - 6x4y
(d) (-5y)(3y)(-2z) = 30y2z
(e) (2abc)(-2abc)(3abc) = -12a3b3c3
EX.2 Simplify completely:
(a) (x2y3)2 = (x2)2 (y3)2 = x4y6
We need to remember
all the laws of
exponents we learned
in Unit 2 - chapter 5!!!
: recall power to a power property
(b) (x2)3 (y3)3 (z2)4 = x6 y9 z8
(c) (2 x2 y2 z4)3 = 23x6 y6 z12 = 8 x6 y6 z12
(d) (-2 a2 b3) 2 = (-2)2 a4 b6 = 4 a4 b6
(e) (-2a2 b3) 3 = (-2)3 a6 b9
(f) -(2x2 y3) 4 = -(2)4 x8 y12 = - 16 x8 y12
Seatwork/homework: p 292 #2, 5aceg, 6acegi, 7abcd, 9ace, 13aegh
Dividing
EX. 1
(a)
 15a
= -5a
3
(e)
 18ab
(b)
= -3b
6a
(c)
(d)
5n 5
3n 2
=
 5x 5 y 7
 3x
(f) -15
5n 3
3
 15 x 6 y 8
 9x 4 y3
5 2
y4
= 3 x9 y
y

3y3
=
(g) 26 p3  14 p =
 15 y 4
3y3
= -5y
26 p 3 13 p 2
=
14 p
7
= 3xy
Unit 7 – Polynomials
-4-
Seatwork/homework: P. 293 #3, 8abcd, 11ace, 14aceij
EX.2 Evaluate for x = -1
(a) – x 3
= - (-1) 3
= - ( -1)
=1
(b) (2x 2)(x 3)
= 2x 5
= 2(-1) 5
= 2(-1)
= -2
: use exponent laws to simplify expression
: substitute x = -1 into expression
: evaluate
Seatwork/homework: p. 293 #18
Section 7.3 Multiplying a Polynomial by a Monomial. (p. 295 – 299)
EX.1 Expand
(a) 2x(3x + 6)
= 2x(3x) + 2x(6)
:use distributive property
: It is not necessary to show this step. Just remember the
operation used and the appropriate law of exponents you perform.
WATCH signs!
= 6x2 + 12x
(b) 3y(-2y2 – 3y + 2)
= -6y3 – 9y2 +6y
EX.2 Simplify
(a) x(x – 1) – x(x + 1)
= x2 – 1x – x2 – 1x
= - 2x
: using distributive property twice …WATCH signs!
:combine like terms
(b) -3x(x + 2) + 2x(2x – 1)
= -3x2 – 6x + 4x2 – 2x
Unit 7 – Polynomials
-5-
= 1x2 – 8x
EX.3 Use algebra tiles to simplify
(a) 2(x + 4)
remember that 2 means 2 unit tiles
and x + 4 is the x-tile
plus 4 unit tiles
When we set up the tiles to perform multiplying these factors we sketch a rectangle that is 2
units wide and x + 4 units long:
x+4
2
We now fill in the rectangle with tiles:
x+4
2
We see what is inside the box are 2 x-tiles and 8 unit tiles to represent the expression 2x + 8
2(x + 4) = 2x + 8
(b) Try
2x(x + 4)
We see 2 x2 tiles and 8 x–tiles
2x2 + 8x
Unit 7 – Polynomials
-6-
Seatwork/homework: p. 298 - 299 #1, 3aceg, 4, 5aceg, 6aceg, 8aceg
Section 7.5 Dividing a Polynomial by a Monomial. (P. 306 – 308)
Again we need to consider our laws of exponents for this section as well.
EX.1 Simplify each of the following:
4 x 2  8x
2x
(a)
=
: divide the denominator into each term of the numerator
4 x 2 8x

2x 2x
: use simple division on the numbers (i.e. reduce fractions) and
remember to subtract the exponents (quotient Rule).
= 2x + 4
(b)
5 x 2  10 x
5
5 x 2 10 x
=

5
5
2
=x –2
: separate expressions
: reduce fractions and apply quotient rule for exponents
3b 3  6b 2 9b
(c)
6b
3
3b
6b 2 9b
=


6b
6b 6b
1
3
= b2  b 
2
2
(d)
: separate expressions
: reduce fractions and apply quotient rule for exponents
10 x 2 y  5 xy 2  15 xy
5 xy
: the expression can have more than one variable just remember
to group and work the like parts together.
2
2
10 x y 5 xy
15 xy


5 xy
5 xy
5 xy
= 2x + 1y – 3
=
(e)
16a 5b 4  20a 4b 4  24a 6b 3
4 a 3b 2
=
16a 5b 4
-
20a 4 b 4
+
4a 3b 2
4a 3b 2
= 4a2b2 – 5ab2 + 6a3b
Unit 7 – Polynomials
246a 6b 3
4a 3 b 2
-7-
Seatwork/homework: p. 307 #1ac, 2ac, 5ac, 6ac, 9
Unit 7 – Polynomials
-8-
Section 7.4 Factoring Polynomials (p. 300 - 305)
Certain definitions and skills from past courses need to be understood before we learn to
factors polynomials:
Factor – A whole number that exactly divides into another whole number
ex: 3 is a factor of 12 because 12  3 = 4
GCF - The largest possible whole number that can divide into 2 or more quantities:
ex: consider 12 and 48 - the largest number (factor) that can divide into these is 12
Prime Number – A whole number that is only divisible by one and itself (only has two factors)
ex: 2, 3, 5, 7, 11, 13, 17,19, … are a few examples of prime numbers
Composite number – A whole number that has more than two factors. Every composite
number can be written as the product of prime factors. This product is called its prime
factorization. For instance, the prime factorization of 12 is 2 · 2 · 3.
The prime factorization of a whole number is the number written as the product of its prime
factors. There is exactly one prime factorization for any composite number.
EX.1 Express as a product of prime numbers. (Hint: it helps to make a factor tree!)
(a)
36
(b)
72
6 ∙ 6
9
2 ∙ 3 ∙ 2∙ 3
8
3 3 2 4
 36 = 2  2  3  3
3
3 2 2 2
72 = 3 × 3 × 2 × 2 × 2
We can use prime factorization (or factors trees) to determine the GCF of 2 or more numbers.
EX.2 Determine the GCF of 24 and 36.
36
4
24
9
2 2 3 3
4 6
2 2 2 3
The GCF = multiply all those prime
factors that are common in both
numbers.
GCF = 2 × 2 × 3 = 12
or you may find it easier to consider the prime factorization form of these two numbers:
36 = 2 × 2 × 3 × 3
24 = 2 × 2 × 2 × 3
Unit 7 – Polynomials
So we see that the common factors are 2, 2, and 3. If
we multiply theses together we get the greatest
common factor, GCF = 2 × 2 × 3 = 12
-9-
EX. 3 (A) Determine the GCF of 3x2 and 12 x
3x2
12x
The common factors are 3 and x
3 x2
12 x
so the GCF = 3 ∙ x = 3x
3 x x
4 3 x
2 2 3 x
(B) Find the GCF of 21x3y2 and 14x2y 3
The prime factorization of each of these are:
21 x3y2 = 3 ∙ 7 ∙ x ∙ x ∙ x ∙ y ∙ y
14x2y 3 = 2 ∙ 7 ∙ x ∙ x ∙ y ∙ y ∙ y
: identify common factors
 the GCF = 7 ∙ x ∙ x ∙ y ∙ y = 7x 2 y 2
: note in the answer we have the smallest
exponent of the variables given initially
(C) Find the GCF of 20a3b4c5 and 15a6b2c3
20a3b4c5 = 2 ∙ 2 ∙ 5 ∙ a ∙a ∙ a ∙ b ∙ b ∙ b ∙ b ∙ c ∙ c ∙ c ∙ c ∙ c
15a6b2c3 = 3 ∙ 5 ∙ a ∙ a ∙a ∙ a ∙ a ∙ b ∙ b ∙ c ∙ c ∙ c
 GCF = 5 ∙ a ∙ a ∙ a ∙ b ∙ b ∙ c ∙ c ∙ c = 5a2b2c3
Try these!
(D) Find the GCF of 9x16y14z12 and 3x10y19z14
(E) Find the GCF of 10x4y5, 5x5y4, and 15x3y6
Seatwork/Homework: p. 304 #1, 3
Unit 7 – Polynomials
- 10 -
Factoring Expression with Common Factors
EX. 1 Completely factor each of the following:
(a) 5x + 25
= 5∙x + 5∙ 5
: using prime factorization we can determine the GCF of the terms and let
that be one factor
: note the common factor of 5 is pulled outside as a factor
= 5(x + 5)
: whatever if “left over” becomes the other factor
To check if this is correct we can multiply the factors together and we should get the original
expression.
5(x + 5)
: use distributive property
= 5(x) + 5(5)
= 5x + 25
: so we know we are right!
(b)
(C)
9x2y2 – 6xy2
= 3∙3∙x∙x∙y∙y – 2∙3∙x∙y∙y
: GCF = 3∙x∙y∙y = 3xy2
= 3xy2 (3x – 2)
: this is answer in factored form
8a2b6 + 4a2bc
It is not necessary to show prime factorization to determine
the GCF!
= 4a2b (2b5 + c)
Simply consider each individual “piece” of the terms:
In this case, we know that 4 is the largest factor that can be
divided into 8 and 4.
Also, we have also shown above that we remove the lowest
exponent of the common variables given to find the GCF.
So, we see that for all common variables terms the lowest
exponent on a is 2 the lowest exponent on b is 1.
We do not include c since it is not a common variable.
Therefore, the GCF for this expression is 4 a2b.
This is written in front of the other factor (the remaining terms
combine to give us the other factor).
Don’t forget to multiply back (use distribute property) to check
to see if answer is correct!
(d) 8a7b6 + a2bc
= a2b(8a5b5 + c)
Unit 7 – Polynomials
(e) 20x3y4z5+ 15x6y5z4 – 25x5y3z
= 5x3y3z(4yz4 + 3x3y2z3 – 5x2)
- 11 -
(g) -6x2y + 3xy2 – 3x2y3
= -3xy(2x – 1y + xy2)
(f) -3x2y + 9x3y2
= -3x2y(1 – 3xy)
Seatwork/homework: p. 304 – 305 # 4acegi, 8acegi, 12acegi, 13acegi
Factoring using Algebra Tiles
Factor 4x + 8
We see this as
The objective is to make these form a rectangle because length × width = area and the l × w
will then be the factors of this expression.
x+2
4
Thus, 4x + 8 = 4(x + 2)
We could also create a rectangle like:
Which would give us 2(2x + 4) as a factored expression. But since we see that the factor
(2x + 4) can be factored further to 2(x + 4) we are not finished and there will be another step.
We see this more clearly algebraically: 4x +8
= 2(2x +4)
= 2 ∙ 2(x + 2)
= 4(x + 2)
Seatwork/Homework: p. 304 #5
Unit 7 – Polynomials
- 12 -
Section 7. 6 Multiplying Two Binomials. (p. 309 – 313)
FOIL Method: is a method of multiplying two binomial expressions. It is essentially using the
distributive property of multiplication twice. The letters of the acronym tells what order and
which terms will be multiplied.
F–
O–
I–
L–
the first terms in each expression (positions a and c)
the outside terms of the expression (positions a and d)
the inside (middle) terms of the expression (positions b and c)
the last two terms of each expression (positions b and d)
Outside
First
(a + b) (c + d)
Inside
Last
EX. 1 Expand (multiply) each of the following:
(A)
(x + 3) (x + 2)
: apply FOIL method
= x(x) + x(2) + 3(x) + 3(2)
= x2 + 2x + 3x + 6
= x2 + 5x + 6
: group like terms
(b) (2x + 4)(3x – 1)
= 6x2 – 2x + 12x – 4
= 6x2 + 10x – 4
(c) (3x – 2)(5x – 4)
= 15x2 – 12x – 10x + 8
= 15x2 – 22x + 8
Unit 7 – Polynomials
: watch signs when multiplying!
- 13 -
Multiplying Binomials using Algebra Tiles
EX.2 Use algebra tiles to expand (multiply) the following:
(a) (x + 2) (x + 3)
(x + 3)
(x + 2)
= x2+ 5x + 6
(b) (2x + 2)(2x – 1)
(2x - 1)
= 4x2+ 2x - 2
(2x + 2)
(c) (3x + 1) (2x + 1)
(2x + 1)
= 6x2+ 5x +1
(3x + 1)
Seatwork/homework: p. 311 – 313 # 8, 9, 10, 11, 13,15, 17
Unit 7 – Polynomials
- 14 -
Section 7.7 Factoring Trinomials. (p. 314 – 320)
Recall, a polynomial with three terms is called a trinomial.
ex: x2 + 5x + 6,
2x2 – 7x + 12
When we multiply two binomials together we often get a trinomial
ex: (x + 2)( x + 3) = x2 + 5x + 6
(2x + 3) ( x – 4) = 2x2 – 5x – 12
Factoring is the process of reversing this multiplication (FOIL) process.
ex: factoring x2 + 5x + 6 will give us (x + 2)(x + 3)
There are several ways to factor trinomials. We will look at factoring in two ways: (1) using a
method called product and sum (sometimes called trial and error or guess and check) and (2)
using algebra tiles.
EX.1 Factor each of the following
(a) x2 + 5x + 6
=(
)(
)
We need to determine two binomials so that if we foiled we get the
original equation.
If we consider the foiling process we know that if we multiply the first terms together we get
the first term in the trinomial. Hence, we know that the first two terms are x and x  x ∙ x = x2
So we have:
= (x
) (x
)
We also know that from the foiling process the last terms multiplied give us the last term in the
trinomial.
Think: What two numbers multiply to be 6? 1 and 6, -1 and -6, 2 and 3, or -2 and -3
So, how do we know what we need to choose?
We have also seen from the multiplying the outside and inside terms that they end up being
grouped together to give the middle term of the trinomial.
Think: Which of these two factors will give us 5 if we add them?
2 and 3
Therefore we can write the above trinomial as a product of two factors:
= (x + 2) (x + 3)
(b) x2 + 8x + 15
= (x + 5)(x +3)
Unit 7 – Polynomials
: think: What two number multiply to be 15 and add to be 8?
5 and 3
: check by foiling. If you get the original equations then you are
correct.
- 15 -
(c) x2 – 1x – 6
= (x – 3)(x + 2)
:what two numbers × to be -6 and + to be -1?
-3 and 2
(d) x2 + 3x – 28
= (x – 7)(x + 4)
: what two values × to be -28 and + to be 3?
-7 and 4
(e) x2 – 7x + 10
= (x – 2)(x – 5)
: what two numbers × to be 10 and + to be -7?
-2 and -5
In previous examples we see the leading coefficient as 1. There are times when this will be
another number other than 1. It then becomes necessary to factor out the GCF before we use
product and sum method.
EX.2 Factor completely.
(a) 5x2 + 10x + 40
= 5(x2 + 2x + 8)
: determine GCF of 5,10, 40
GCF = 5 (same as leading coefficient)
: now determine the two values whose product = 8 and sum = 2
Note: we can not determine these two factors so we are done.
(b) 4x2 – 16x – 20
= 4(x2 – 4x – 5)
= 4(x – 5)(x + 1)
GCF = 4
: determine two numbers that × to be -5 and + to be -4
(c) 4x2 + 8x – 12
= 4(x2 + 2x – 3)
= 4(x + 3)(x – 1)
:factor the GCF first
: think of two #’s that × = -3 and + = 2? -1 and 3
: Note: it does not matter which order the factors are written in.
-5, 1
TRY THESE!
(d) 6x2 - 42x + 60
(e) 2x3 + 14x2 + 20
Seatwork/ homework: p. 319 # 11acegi, 12acegi, 13acegi, 14acegi, 19
Unit 7 – Polynomials
- 16 -
Factoring using Algebra Tiles
Once again we can use algebra tiles to rearrange into rectangles. Remember as we showed
earlier the l × w gives us the factors.
EX.1 Factor using algebra tiles:
(a) x2 + 6x + 8
We are using 1 x2 tile , 6 x-tiles and 8 unit tiles to form a rectangle
Rearrange in a rectangular form and determine the length and the width which will represent
the factors of the trinomial.
x+2
x+4
Hence x2 + 6x + 8 = (x + 2)(x + 4)
(b) x2 + 4x + 4
(c) x2 + 7x + 6
x+1
x+2
x+6
x+2
 x2 + 4x + 4 = (x + 2)(x + 2)
Unit 7 – Polynomials
 x2 + 7x + 6 = (x + 1)(x + 6)
- 17 -
Seatwork/homework: p. 318 #4, 5
Binomial Products – Special Cases
ex:
(x + 3)(x – 3)
= x2 – 3x + 3x – 9
= x2 - 9
Difference of squares
Notes:
 The factors are the same except for the sign.
 The middle terms will cancel
 Left with two perfect squares that are subtracted (this is why it is called
difference of squares).
EX.1 Expand each of the following:
(a) (x + 9)(x – 9)
= x2 – 9x + 9x – 81
= x2 – 81
(b) (a + 5)(a – 5)
: since middle terms will cancel it is only necessary to complete
F and L of the FOIL method.
= a2 – 25
(c) (10x + 3)(10x – 3)
= 100x2 – 9
(d) (3x – 2y)(3x + 2y)
= 9x2 – 4y2
Seatwork/Homework: p. 313 #15, 16
Perfect Squares
EX.2 Expand each of the following:
(a) (x + 3)2
= (x + 3)(x + 3)
= x2 + 3x + 3x + 9
= x2 + 6x + 9
: squaring a binomial means multiplying by itself (remember
the square exponent does not mean square each separate
term inside the bracket!)
(b) (x - 5)2
= (x – 5)(x – 5)
= x2 – 5x – 5x + 25
Unit 7 – Polynomials
(c) (2x + 1)2
= (2x + 1)(2x + 1)
= 4x2 + 2x + 2x + 1
- 18 -
= x2 – 10x + 25
= 4x2 + 4x = 1
Factoring Binomials (Difference of Squares)
Once again factoring is the reverse of foiling.
EX.1 Factor
(a) x2 – 25
= (x + 5) (x – 5)
: note the difference of two perfect squares
(b) x2 – 9
= (x – 3)(x + 3)
: It does not matter what order the factors are written.
(c) 49 – x2
= (7 – x)(7 + x)
(d) 9x2 – 16
= (3x – 4)(3x + 4)
(e) y2 + 49
: Can NOT be done! There is no such thing as factoring sum of squares.
Seatwork/ Homework: Assign worksheet
Unit 7 – Polynomials
- 19 -
Perimeter and Area
EX.1 (a) Find an expression for the perimeter of the rectangle.
P = 2l + 2w
= 2(5x – 1) + 2(3x)
= 10x – 2 + 6x
= 16x – 2
3x
5x - 1
(b) What is the perimeter if x = 2m?
P = 16x – 2
= 16(2) – 2
= 32 – 2
= 30
 the perimeter is 30 m.
(c) Find an expression for the area of the above rectangle
A=l×w
= (5x – 1)(3x)
= 15x2 – 3x
(d) What is the area if x = 2m?
A = 15x2 – 3x
= 15(2)2 – 3(2)
= 15(4) – 3(2)
= 60 – 6
= 54 m2
: simply substitute the 2 in for x and simplify
: don’t forget the units!!!
EX.2 What is the total length of all edges in the figure shown?
3x
4(x) + 4(x + 2) + 4(3x)
= 4x + 4x + 8 + 12x
= 20x + 8
: use distributive property
: group like terms
x
x+2
Unit 7 – Polynomials
- 20 -
2x
EX.3 Consider the figure
2x - 2
4x
3x + 1
6x + 3
(A) Determine the perimeter
First it is necessary to determine the measures of the missing sides
2x
2x - 2
4x
4x
3x + 1
6x + 3
6x + 3
a
2x
Perimeter
b
c
3x + 1
2x - 2
= 2(2x) + 2(3x + 1) + 2(2x – 2) + 2(4x) + 2(6x + 3)
= 4x + 6x + 2 + 4x – 4 + 8x + 12x + 6
= 34x + 4
Found by
(6x + 3) – (4x)
(B) Find the total area of the figure.
It is clearly easier to find the area of the 3 rectangles (a, b, and c) then add these to get total.
A (rect a) = L × W
= 2x(6x + 3)
= 12x2 + 6x
A (rect b) = L × W
= (3x + 1)(2x + 3)
= 6x2 + 8x + 2x + 3
= 6x2 + 10x + 3
Total Area = Areaa + Areab + Areac
= 12x2 + 6x + 6x2 + 10x + 3 + 12x2 – 6x – 6
= 30x2 + 10x – 3
Unit 7 – Polynomials
- 21 -
A(rect c) = L × W
= (2x – 2)(6x + 3)
= 12x2 + 6x – 12x - 6
= 12x2 – 6x – 6
: group like terms
EX.4 TRY THIS ONE!
Consider this shape
4x + 6
3x
3x + 1
2x + 2
(A) Find the perimeter
(B) Find the total area
EX.5 AND THIS ONE!
Determine the total perimeter and area of the figure below.
x-4
3x
2x + 1
Unit 7 – Polynomials
- 22 -
EX.6 (a) Find an expression for the area of the shaded region. Both shapes are rectangles.
2x + 1
2x + 5
x
2x + 10
Area of shaded region = Area of large rectangle – Area of small rectangle
=L×W
L×W
= (2x + 10) (2x + 5)
(2x+ 1)(x)
2
= 4x + 10x + 20x + 50 (2x2 + x)
2
= 4x + 30x + 50
(2x2 + x)
= 4x2 + 30x + 50 – 2x2 – x
= 2x2 + 29x + 50
(b) If x = 1m, find the area of the shaded region
Areashaded = 2x2 + 29x + 50
= 2(1)2 + 29(1) + 50
= 2 + 29 + 50
= 81 m2
Seatwork/Homework:
Unit 7 – Polynomials
: don’t forget units!
p. 324 #2
- 23 -