1. 1. Introduction to Polynomials
Topic 11 Polynomials
1. Introduction to Polynomials
Definition
A polynomial is an algebraic expression that only contains positive integer
powers of a variable. The variable is usually x but it may be other letters such
as y or z.
For example each of the following algebraic expressions is a polynomial.
(a)
(d)
5x3 – 4x2 + 7
y3 – y2 + 1
(b)
(e)
4x45 – 6x3 + x
2x4 – 4z5
(c)
(f)
49 – x2 + x3
8k – 5k2
It is usual to rearrange a polynomial into what is called its Standard Form , this is where you
write down the given polynomial in order of its powers with its highest power of x first and then
continuing downwards from there.
So for example it would be better to write 49 + x3 – x2 in its standard form x3 – x
We call the numbers next to each power its coefficient so in the polynomial 5x3 – 4x we
would say 5 is the coefficient of x3 and – 4 is the coefficient of x2 while the number + 7 is called
the constant term.
The degree of a polynomial is the highest power of x it contains so in the above examples.
The degree of the polynomial 5x3 – 4x2 + 7 is 3.
The degree of the polynomial 4x45 – 6x3 + x is 45.
The degree of the polynomial 49 – x2 + x3 is 3.
Certain polynomials are given special names depending on the degree of the polynomial.
If the degree is 1, for example 3x – 7 we say that the expression is Linear.
If the degree is 2, for example 5x2 + x – 4 we say that the expression is a Quadratic.
If the degree is 3, for example x3 +7x we say that the expression is a Cubic.
If the degree is greater than 3 we call it a polynomial of degree n.
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2. Which of the following algebraic expressions are polynomials.
Topic 11 Polynomials
Example 1: Express each polynomial in descending order. Give the degree of each polynomial:
(a) 6x – 5 – 2x3
(d) 10y + y5
(b) 8 – y3
(e) x2 – 3x6 + 2x
(c) 6x2 – 7x – x4 + 4
(f) 5t – 10t2 – 6
Solution (a): 6x – 5 – 2x3 =
– 2x3 + 6x – 5
Degree = 3
Solution (b):
8 – y3 =
– y
Degree = 3
Solution (c): 6x2 – 7x – x4 + 4 =
– x4 + 6x2 – 7x + 4
Degree = 4
Solution (d):
10y + y5 =
y5 + 10y
Degree = 5
Solution (e): x2 – 3x6 + 2x
Solution (f): 5t – 10t2 – 6 =
– 3x6 + x2 + 2x
– 10t2 + 5t – 6
Degree = 6
Degree = 2
Exercise 1: 1.
Which of the following algebraic expressions are polynomials.
(a)
(d)
2x5 – x2 + 3x
2y5 – 3y6 + 9
(b)
(e)
4x4.5 – 6x3 + x
(c)
(f)
49 –
7t– t2
2. Express each polynomial in descending order (its Standard Form).
Give the degree of each polynomial:
(a) 4x – 4x4
(d) b + b6
(b) 7y – y2
(e) 7c2 – 49c5 + 6
(c) 2x3 – x + 3x2 + 4
(f) 4d – d2– 1
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3. Topic 11 Polynomials
2. Addition and Subtraction of polynomials
The process o adding or subtracting polynomials involves collecting like terms together and then
adding and subtracting them. What we mean when we say “like terms” is that we have terms
which contain the same powers of same variables. They may have different coefficients, but that
is the only difference.
You probably know that if you have an expression like 4 + 3x you cannot add those terms to
simplify it in any way. That’s because one term in a constant (the 4) and the other term has a
variable (the x). We say that these are not like terms and cannot be combined. On the other
hand it is possible to simplify the expression 4x + 5x to get 9x as we do have like terms and can
therefore combine them in the usual way.
It is also not possible to add 4x3 + 3x2 as x3 and x2 are completely different powers of x and are
not considered like terms on the other hand you can add 5x4 + 3x4 – 2x4 = 6x3 as these are all
the same powers of x and are considered like terms.
In order to add or subtract a polynomial you typically group like terms together and add or
subtract each group of like terms together.
Example 1: Simplify the algebraic expression 2x4 + 3x2 – 2x + 3x4 + 6x2 + 8x + 8
Solution:
This expression can be simplified by identifying like terms and then grouping and
combining like terms, so we have three groups of like terms
+ 2x4 and + 3x4 are like terms, and can be combined to give + 5x4
+ 3x2 and + 6x2 are like terms, and can be combined to give + 9x2
– 2x and + 8x are like terms, and can be combined to give + 6x
the constant + 8 is on its own so it will stay as + 8
So after simplifying, this expression becomes:
2x4 + 3x2 – 2x + 3x4 + 6x2 + 8x + 8 =
2x4 + 3x4 + 3x2 + 6x2 – 2x + 8x + 8
5x4 + 9x2 + 6x + 8
Example 2: Simplify the polynomial (6x3 – 5x) + (3x3 – 4x2)
Solution:
(6x3 – 5x) + (3x3 – 4x2) =
6x3 + 3x3– 4x2– 5x
9x3– 4x2– 5x
Rearrange like terms in order
Add and subtract like terms
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4. y3 – 9y – 2y3 – 8y Topic 11 Polynomials
Example 3: Simplify the polynomial 5x3 – 5x4 + 2x3 – x4
Solution:
5x3 – 5x4 + 2x3 – x4 =
– 5x4 – x4 + 5x3 + 2x3
– 6x4 + 7x3
Rearrange like terms in order
Add and subtract like terms
Example 4: Add the polynomials ( 2y3 + 8y ) and ( y3 – 9y)
Solution:
( 2y3 + 8y ) + ( y3 – 9y) =
2y3 + 8y + y3 – 9y
2y3 + y3 + 8y – 9y
3y3 – y
Rearrange like terms in order
Add and subtract like terms
When you subtract polynomials we need to be careful to subtract all the terms and not just the
first term. For example (6x3 – 3x + 1) – (x3 + 5x – 2) =
6x3 – 3x + 1 – x3 – 5x + 2
This is because we are subtracting all the terms in this bracket so – (x3 + 5x – 2) is the same as
– (x3 + 5x – 2) =
– 1(x3 + 5x – 2) =
– x3 – 5x + 2.
As long as we pay attention to this fact subtraction of polynomials is a relatively straight forward
process.
Example 4: Simplify ( 2y3 + 8y ) – ( y3 – 9y)
Solution:
( 2y3 + 8y ) – ( y3 – 9y) =
2y3 + 8y – y3 + 9y
2y3 – y3 + 8y + 9y
y y
Rearrange like terms in order
Add and subtract like terms
Example 5: Subtract ( 2y3 + 8y ) from ( y3 – 9y)
Solution:
( y3 – 9y) – ( 2y3 + 8y ) =
y3 – 9y – 2y3 – 8y
y3 – 2y3 – 9y – 8y
– y3 – 17y
Rearrange like terms in order
Add and subtract like terms
Notice the order that we subtracted the two polynomials subtract ( 2y3 + 8y ) from ( y3 – 9y)
We have the polynomial ( 2y3 + 8y ) first and subtract the polynomial ( y3 – 9y) from it.
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5. (x3 – 4x) – (–2x3 + 3x) Topic 11 Polynomials Exercise 2: 1. 2.
Add the polynomials (5x3 – 4x) + (6x3 + 3x2)
Simplify the following.
(a)
(6x2 – 5x + 3) – (x2 + 8x – 1)
(b) 8x3 – 2x + x3 – 2x
(c) (5x4 + 2x2 ) + (–3x2 + x3)
(e) (5x2 – 3x – 1) – (x2 – 4x + 1)
(d)
(f)
(x3 – 4x) – (–2x3 + 3x)
(6x3 – 5x) + (3x3 – 4x2)
(g)
(6x2 – 5x + 3) – (x2 + 8x – 1)
(h) (4w2 – 4w – 8) – (2w2 + 3w – 6)
(i) (9x2 + 7x – 5) – (3x )
(j)
8x3 – 2x + x3 – 2x
(k) (5x4 + 2x2 ) + (–3x2 + x3)
(m) (5x2 – 3x – 1) – (x2 – 4x + 1)
(l) (x3 – 4x) – (–2x3 + 3x)
(m) (6x3 – 5x) + (3x3 – 4x2)
(n)
(6x2 – 5x + 3) – (x2 + 8x – 1)
(o) (5x2 – x – 1) – (–3x2 – 2x – 5)
(p) (– 5x2 – 2x ) – (2x2 – 7x + 9)
(q) (8x3 – 2x2 – 4x + 5) – (5x2 + 8)
3 2 5
(r) (2x3 – 4x2 + 5x – 7) – (3x +
x – 5)
(s) (5x3 + 2x2 – x + 1) + (2x3 – 6x2 – x )
(t) (d4 – 3d + 8) – (2d4 – 3d + 4)
(u) 12p2 – 5p + p2 +3p – 7p2 – 2 3 4. 5. 6. 7. 8.
Subtract (5x + 4) from (8x + 2)
Add: (5x3 – 4x) + (6x3 + 3x2)
Subtract: (5x2 – 2x + 3) from (x2 + 4x – 1)
Subtract (5x – 6) from (2x2 – 4x + 8)
Subtract: (x2 + 4x – 1) from (5x2 – 2x + 3)
Subtract (2x2 – 4x + 8) from (5x – 6)
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6. Topic 11 Polynomials
3. Using the Distributive Law to simplify polynomials
The distributive law states that for all variables a,b and c
a(b + c) = ab + ac
The distributive law allows us a means of multiplying out parenthesis and to simplify the results.
Example 1: Use the distributive law to remove the parenthesis form the expression 5(x + 3).
Solution:
5(x + 3) = =
5(x) + 5(3)
15x + 15
Use the distributive Law
Example 2: Use the distributive law to remove the parenthesis form the expression – 5(4x2 – 3).
Solution:
– 5(4x2 – 3) = =
(– 5)(4x2) + (– 5)(– 3)
– 20x2 + 15
Use the distributive Law
Example 3: Use the distributive law to remove the parenthesis form the expression
– 2(4x2 + 3x – 7).
Solution:
– 2(4x2 + 3x – 7) = =
(– 2)(4x2) + (– 2)(+ 3x) + (– 2)( –7) Use the distributive Law
– 8x2– 6x + 14
Example 4: Use the distributive law to remove the parenthesis form the expression 5x(2x2 + 4).
Solution:
5x(2x2 + 4) = =
5(2x2) + 5(4)
30x2 + 20
Use the distributive Law
Example 5: Use the distributive law to remove the parenthesis form the expression
– x2(2x3 + x – 3).
Solution:
– x2(2x3 + x – 3) = =
(– x2)(2x3) + (– x2)(x) + (– x2)(– 3) Use the distributive Law
– 2x5 – x3 + 3x2
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7. 3b3 – 8b3 + 12b2 – 4 + 7 Topic 11 Polynomials
When you simplify a polynomial expression you may also have to use the distributive law to
rearrange the expression first and then you can simplify the expression. In some cases, you may
have to perform other simplification before you can combine like terms.
Example 6: Simplify 4(2x – 3) – 4(x + 5)
Solution:
4(2x – 3) – 4(x + 5) =
(4)(2x) +(4)(– 3) – 4(x) + (– 4)(5) Use the distributive Law
8x – 12 – 4x – 20
8x – 4x – 12 – 20
Rearrange like terms
4x – 32
Example 7: Simplify the algebraic expression 3b3 – 2(4b3 – 6b + 2) + 7
Solution:
The brackets in this expression can be removed first, and then the expression
may be simplified.
3b3 – 2(4b3 – 6b + 2) = =
3b3 – 8b3 + 12b2 – 4 + 7
– 5b3 + 12b2 + 3
Use the distributive Law
Example 8: Simplify 2x3(2x – 5) – 3x(x2 + 5)
Solution:
2x3(2x – 5) – 3x(x2 +5x3) = =
2x4 – 10x3 –3x3 – 15x4
2x4 – 15x4 – 10x3 – 3x3
– 13x4 – 13x3
Use the distributive Law
Rearrange like terms
Example 9: Simplify 7x(2 – 5x) – 3x3 + 5x
Solution:
7x(2 – 5x) – 3x3 – 14x =
14x – 35x2 –3x3 – 14x
– 3x3 – 35x2 + 14x – 14x
Use the distributive Law
Rearrange like terms =
– 3x3 – 35x2
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8. Express each of the following as polynomials in their standard form.
Topic 11 Polynomials
Exercise 3: 1.
Express each of the following as polynomials in their standard form.
(a) 3(x + 8)
(d) 4x3(x – 6)
(g) –2d3(d – 7d2)
(j) 2x2(4x – 3)
(m) –2x2(3x2 – 4x + 2)
(b) 2(4y2 + 9y)
(e) a( 4 + 5a)
(h) 2(x2 + 7)
(k) –6x(2 – x3)
(n) –x2(4x2 – 2x + 1)
(c) 2(3a3 + 4a2 + 9)
(f) –5c( 1 – 4c)
(i) x( 5x + 2x6)
(l) 4x5(2x2 – x + 5)
(o) 5x3(3x2 – x + 2) 1 3
(p) 4x5(2x2 – x + 5)
(q) 0.3x(2xy + 5x – 6y)
(r)
y( y 2
6 y
3x) 2.
Express each of the following as polynomials in their standard form.
(a) 4(x3 – 6x) + 34x(x2 – 8)
(b) 5(3y2 + 4y) – 2(10y + 9)
(c) 2t(3t + 6) – 12(t2 + 3t + 8)
(d) 5t3(t – 3t2) + 5t2(t2 – t3)
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9. Inner Topic 11 Polynomials 4. Multiplication using FOIL.
In order to expand the multiplication of two parentheses we use the algebraic technique of FOIL.
This Mnemonic stands for First Outer Inner Last and gives us the systematic order in which you
perform the multiplication. The general multiplication is shown below.
Outer
First
(a + b)(c + d) = ac + ad + bc + db
Inner
Last F O I L
First:
Outer:
Inner:
Last:
This is the product of the first term in each parenthesis.
This is the product of the first term in the first parenthesis and the second term
in the second parenthesis.
This is the product of the second term in the first parenthesis and the first term
This is the product of the second term in the each parenthesis.
Example 1: Expand each of the following
(a) (x + 2)(x + 3)
(d) (2x + 2)(3x – 3)
(g) (0.5x + 1.4)(3.5x – 1)
(b) (a – 2)2
(e) (4b – 3)2
(h) ( y – 1)2
(c) (2r + 3)(r – 7)
(f) (5d – 2)(3d – 4)
(i) ( d – 3)( d – 5)
Solution (a):
(x + 2)(x + 3) =
x2 + 3x + 2x + 6 = F L O I
x2 + 5x + 6
Solution (b):
(a – 2)2 =
(a – 2)(a – 2) =
a2 – 2a – 2a + 4 =
a2 – 4a + 4 F O I L
Solution (c):
(2r + 3)(r – 7) =
2r2 – 14r + 3r – 21 F L O I =
2r2 – 11r – 21
Solution (d):
(2x + 2)(3x – 3) =
6x2 – 6x + 6x – 6 =
6x2 – 6 F O I L
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10. y2 – y – y + 1 d2 – d – d + 15 Topic 11 Polynomials Solution (e):
(4b – 3)2 =
(4b – 3)(4b – 3) =
16b2 – 12b – 12b + 9 =
16b2 – 24b + 9 F O I L
Solution (f):
(5d – 2)(3d – 4) =
15d2 – 20d – 6d + 8 F L O I =
15d2 – 26d + 8
Solution (g):
(0.5x + 1.4)(3.5x – 1) =
1.75x2 – 0.5x + 4.9x – 1.4 =
1.75x x – 1.4 F O I L
Solution (h):
( y – 1)2 =
( y – 1) ( y – 1) =
y2 – y – y + 1 F
I L O =
y2 – y + 1
Solution (i):
( d – 3)( d – 5) =
d2 – d – d + 15 F L O I =
d2 – d + 15
Example 2: Expand each of the following
(a) (x + 3y)(x + 4y)
(d) (4x – y)2
(b) (3x – b)(x + 4b)
(e) (x2 – 3)(x – 3x2)
(c) (5y – 6x)2
(f) (5xy + 4x)(5yx – 4x)
Solution (a): (x + 3y)(x + 4y) =
x2 + 4xy + 3xy + 12y2 =
x2 + 7xy + 12y2 F O I L
Solution (b): (3x – b)(x + 4b) =
3x2 + 12bx – bx – 4b2 =
3x2 + 1bx – 4b2 F O I L
Solution (c): (5y – 6x)2 =
(5y – 6x) (5y – 6x) =
25y2 – 30xy – 30xy + 36x2 F
I L O
(4x – y)(4x – y) =
25y2 – 30xy + 36x2
Solution (d): (4x – y)2 =
16x2 – 4xy – 4xy + y2 F
I L O =
16x2 – 8xy + y2
Solution (e): (x2 – 3)(x – 3x2) =
x3 – 3x4 – 3x + 9x2 F O I L
Solution (f): (5xy + 4x)(5yx – 4x) =
25x2y2 – 20x2y + 20x2y – 16x2 =
25x2y2 – 16x2 F O I L
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11. Multiply out and simplify the following.
Topic 11 Polynomials
Exercise 4: 1.
Multiply out and simplify the following.
(a) (x + 3)(x + 4)
(d) (3x + 4)(3x – 2)
(g) (4x – 4)(7 – x)
(j) (2x – 0.1)(x + 2.4)
(m) (5x + 4)(5x – 4)
(p) (3p + 6)(p – 1)
(s) (3x – 2)(2x – 1)
(b) (3x – 1)(x + 4)
(e) (3x – 6)(4x – 2)
(h) (5 – 6x)(2x – 7)
(k) (x + 5)(x – 5)
(n) (4x + 3)(3x – 5)
(q) (2z + 3)(z + 2)
(t) (4x + 3)(3x – 5)
(c) (5 – 3x)(6 + 2x)
(f) (3x – 8)(2x + 3)
(i) (2x – 3)(3x + 2)
(l) (x – 3)(x – 3)
(o) (3 – 2y)(2 + y)
(r) (5r – 7)(5r +7)
(u) (2a – 3)(5a + 1)
(v)
(5x + 2)(2x – 5)
(w) (3x
3)(x 5)
(x) (a + 1)(2a + 4) 2.
Which of the following is (2n + 2)(2n – 2) equivalent to
A. 4n2 – 4
B. 4n2 + 2n – 4
C. 4n2 – 4n – 4
D. 4n2 + 4n – 4 3.
Multiply out and simplify the following squares.
(a) (x + 3)2
(e) (x – 6)2
(j) (2x – 0.1)2
(b) (x – 1)2
(f) (3x – 8)2
(k) (x + 5)2
(c) (5 – 3x)2
(g) (4x – 5)2
(l) (3 – x)2
(d) (3x + 4)2
(h) (5 – 6x)2
(m) (3 – 2x)2 4.
Multiply out and simplify the following.
(a) (x + 3y)(x + 4y)
(d) (3x + 4y)(3x – 2y)
(g) (4x – y)2
(j) (2x – 0.1y)2
(m) (5xy + 4x)(5yx – 4y)
(b) (3x – b)(x + 4b)
(e) (3x – 6b)(4x – 2b)
(h) (5y – 6x)2
(k) (x + 5y)2
(n) (4x + 3)(3x – 5)
(c) (5c – 3x)(6c + 2x)
(f) (3x – 8y)(2x + 3y)
(i) (2x – 3y)2
(l) (x2 – 3)(x – 3x2)
(0) (2y2z + 3)(4y2z + 2)
Page | 11

12. 6 x 3 3x 1 x 3 3x 1 6 x 4 10x 3 8x 2 Topic 11 Polynomials
5. Division of polynomials
In order to divide a polynomial by a constant we separate it into its component parts and divide
each separately with the constant.
Example 1: Simplify the following
(a)
(b) (–32x – 8) 4
(c)
Solution(a): = =
4x3 – 14x + 8
Solution(b): (–32x – 8) 4 =
= – 8x – 2
Solution(c): = =
– 2x2 + 5x
Example 2: Simplify the following
(a)
(b) (–32x2 – 2) 4x
(c)
Solution(a): = =
6x2 – 4 +
Solution(b): (–32x2 – 2)
4x =
= – 8x
Solution(c): = =
– 8+
Exercise 5: Simplify the following.
12 x 2
6 x 3 2
2 x 4
(a)
(b)
(c) (–3x – 8) 4 2
6 x 2 4 5x 4
6x 12 x 2 3
(d)
(e)
(f) 5
6 x x x 2
2 x 2
6 x 2
8x 12 4
3x 1
3x 2
(g)
(h)
(j)
8x 12
4 x 2
4 x 4
x 3
2 x 3
5x 2
6 x 6
3x 1
x 4
(k)
(l)
(m)
Page | 12

13. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Topic 11 Polynomials Solutions
Exercise 1:
1.(a) Yes
(b) No
(c) No
(d) Yes
(e) No
(f) Yes
2.(a) – 4x + 4x Degree = 4 (b) – y + 7y Degree = 2
(c) 2x3 + 3x2 – x Degree = 3
2.(d) b Degree = 6
Exercise 2:
1. 11x3 + 3x2– 4x
(e) – 49c5 + 7c2 + 6 Degree = 5 (f) – d2 + 4d– 1
Degree = 2
2.(a)
11x2 – 13x + 4
(b) 9x3 – 4x
(c) 5x4 + x3– x2
(d)
3x3 – 7x
2.(e) 4x + x – 2 2
(f)
9x – 4x – 5x
(g)
5x – 13x + 4 2
(h) 2w2 – 7w – 2
2.(i) 6x2 + 7x – 8.5
(j)
9x3 – 4x
(k) 5x4 + x3– x2
(l) 5x3 – 7x
2.(m) 4x + x – 2 2
(n) 9x – 4x – 5x
(o)
5x – 13x + 2 2
(p) 8x2 + x + 4
2.(q) – 7x2 + 5x – 9
(r) 8x3 – 7x2 – 4x – 3
(s) 2x3 –
17 2 5
x + 5x – 2
2.(t) 7x3 – 4x2 – 2x + 2
(u)– d4 – 3d + 4
(v) 6p2 – 2p – 2
3. 3x – 2
9x – 14
Exercise 3:
4. 11x3 + 3x2 – 4x 5. –4x2 + 6x – x2 – 9x x2 – 6x + 4
8. –2x2 +
1.(a) 3x + 24
(b) 8y2 + 18y
(c) 6a3 + 8a2 + 18
(d) 4x4 – 24 x3
1.(e) 4a + 5a 2
(f) –5c + 20c 2
(g) – 2d + 14d 4 2
(h) 2x2 + 14
1.(i) 5x2 + 2x7
(j) 8x3 – x2
(k) – 12x + 6x4
(l) 4x5(2x2 – x + 5)
1.(m) – 6x + 8x – 4x 2
(n) – 4x + 2x – x 2
(o) 15x – 5x + 10x3
1 3
1.(p) 8x7– 6x6 + 20x5
(q) 0.6x2y + 1.5x2 – 1.8xy
(r) y
6 y 2 xy 3
2.(a) 38x3 – 296x
(b) 15y2 – 18
(c) – 30t3 – 12t2 + 12t – 96
(d) – 20t5 + 10t4
Exercise 4:
1.(a) x2 + 7x + 12 2
(b) 3x2 + 6x – 4 2
(c) – 6x2 – 8x + 30 2
(d) 9x2 + 6x – 8
(h) – 12x2 + 52x – 35
1.(e) 12x – 30x – 8
(f) 6x – 7x – 24
(g) – 4x + 32x – 28
1.(i) 6x2 – 5x – 6
(j) 2x x – (k) x2 – 25
(l) x2 – 6x + 9 2
1.(q) 2z2 + 7z + 6
1.(m) 25x – 16 2
(r) 25r2 – 49
(n) 12x – 11x – 15
(o) – 2y – y + 6 2
(s) 6x2 – 7x + 2
(p) 3p2 + 3p – 6
(t) 12x2 – 11x – 15
1.(u) 10a – 13a – 3 2
(v) 10x – 21x – 10 (w) 3x – 18x + 15
(x) 2a2 + 6a + 4
2. A
3.(a) x2 + 6x + 9 2
(b) x2 – 2x + 1 2
(c) 9x2 – 30x + 25 2
(d) 9x2 + 24x + 16
(h) 36x2 – 60x + 25
3.(e) x – 12x + 36
(f) 9x – 48x + 64
(g) 16x – 40x + 25
3.(j) 4x2 – 0.4x (k) x2 + 10x + 25
(l) x2 – 6x + 9
(m) 4x2 – 12x + 9
4.(d) 9x2 + 6xy – 8y2
4.(a) x + 7xy + 12y
(b) 3x + 6xb – 4b
(e) 12x2 – 30xb + 12b2
(c) 30c2 – 8xc – 6x2
(f) 6x2 – 7xy – 24y2
4.(j) 4x2 – 0.4xy y2
4.(g) 16x – 8xy + 64y 2
(k) x2 + 10xy + 25y2
(h) 25y – 60xy + 36x 2
(i) 4x2 – 12xy + 9y2
(l) x3 + 18x2 – 3x
Exercise 5:
4.(m) 25x y – 20xy + 20x y – 4xy (n) 12x – 11x – 15
(o) 8y4z2 + 16y2z + 6
(a) 6x2 – 3x +
(b) x + 2 (c)
(d) – 3x (e) – x +
(f) – 2x + 4x2
(g) 2x – 3
(h) 3x2 – 5x + 4
(j)
(k)
(l)
(m)
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