topic 11 polynomials 1. introduction to polynomials definition a polynomial is an algebraic...

Topic 11 Polynomials


1. Introduction to Polynomials

Definition A polynomial is an algebraic expression that only contains positive integerpowers of a variable. The variable is usually x but it may be other letters suchas 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 youwrite down the given polynomial in order of its powers with its highest power of x first and thencontinuing downwards from there.

So for example it would be better to write 49 + x3 – x2 in its standard form x3 – x2 + 49.

We call the numbers next to each power its coefficient so in the polynomial 5x3 – 4x2 + 7 wewould say 5 is the coefficient of x3 and – 4 is the coefficient of x2 while the number + 7 is calledthe 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 +7x2 + 2 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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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 = – y3 + 8 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.

4x4.5 – 6x3 + x(a)

(d)

2x5 – x2 + 3x

2y5 – 3y6 + 9

(b)

(e)

(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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2. Addition and Subtraction of polynomials

The process o adding or subtracting polynomials involves collecting like terms together and thenadding and subtracting them. What we mean when we say “like terms” is that we have termswhich contain the same powers of same variables. They may have different coefficients, but thatis the only difference.

You probably know that if you have an expression like 4 + 3x you cannot add those terms tosimplify it in any way. That’s because one term in a constant (the 4) and the other term has avariable (the x). We say that these are not like terms and cannot be combined. On the otherhand it is possible to simplify the expression 4x + 5x to get 9x as we do have like terms and cantherefore 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 arenot considered like terms on the other hand you can add 5x4 + 3x4 – 2x4 = 6x3 as these are allthe 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 orsubtract 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 andcombining 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


Example 3: Simplify the polynomial 5x3 – 5x4 + 2x3 – x4

Solution: 5x3 – 5x4 + 2x3 – x4 =

=

– 5x4 – x4 + 5x3 + 2x3

– 6x4 + 7x3



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



When you subtract polynomials we need to be careful to subtract all the terms and not just thefirst 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 forwardprocess.

Example 4: Simplify ( 2y3 + 8y ) – ( y3 – 9y)

Solution: ( 2y3 + 8y ) – ( y3 – 9y) =

=

=

2y3 + 8y – y3 + 9y

2y3 – y3 + 8y + 9y

y3 + 17y



Example 5: Subtract ( 2y3 + 8y ) from ( y3 – 9y)

Solution: ( y3 – 9y) – ( 2y3 + 8y ) =

=

=

y3 – 9y – 2y3 – 8y

y3 – 2y3 – 9y – 8y

– y3 – 17y



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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x – 5)

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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) – (3x2 + 3.5) (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)

(r) (2x3 – 4x2 + 5x – 7) – (3x + 3 25

(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)


3. Using the Distributive Law to simplify polynomials

The distributive law states that for all variables a,b and c

a(b + c) = ab + acThe 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: Use the distributive Law5(x + 3) =

=

5(x) + 5(3)

15x + 15

Example 2: Use the distributive law to remove the parenthesis form the expression – 5(4x2 – 3).

Solution: Use the distributive Law– 5(4x2 – 3) =

=

(– 5)(4x2) + (– 5)(– 3)

– 20x2 + 15

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: Use the distributive Law5x(2x2 + 4) =

=

5(2x2) + 5(4)

30x2 + 20

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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When you simplify a polynomial expression you may also have to use the distributive law torearrange the expression first and then you can simplify the expression. In some cases, you mayhave 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 expressionmay be simplified.

Use the distributive Law3b3 – 2(4b3 – 6b + 2) + 7 =

=

3b3 – 8b3 + 12b2 – 4 + 7

– 5b3 + 12b2 + 3

Example 8: Simplify 2x3(2x – 5) – 3x(x2 + 5)

Solution: Use the distributive Law


2x3(2x – 5) – 3x(x2 +5x3) =

=

=

2x4 – 10x3 –3x3 – 15x4

2x4 – 15x4 – 10x3 – 3x3

– 13x4 – 13x3

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


= – 3x3 – 35x2

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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)

(p) 4x5(2x2 – x + 5) (q) 0.3x(2xy + 5x – 6y) (r)1

33x)6 yy( y 2

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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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 termin the second parenthesis.

This is the product of the second term in the first parenthesis and the first termin the second parenthesis.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 + 5x + 6x2 + 3x + 2x + 6 =F

O I L

Solution (b): (a – 2)2 = (a – 2)(a – 2) = a2 – 2a – 2a + 4 = a2 – 4a + 4F O I L

Solution (c): (2r + 3)(r – 7) = = 2r2 – 11r – 212r2 – 14r + 3r – 21F

O I L

Solution (d): (2x + 2)(3x – 3) = 6x2 – 6x + 6x – 6 = 6x2 – 6F O I L


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 youperform the multiplication. The general multiplication is shown below.

Outer

First

(a + b)(c + d) = ac + ad + bc + dbInner

Last


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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 – 26d + 815d2 – 20d – 6d + 8F

O I L

Solution (g): (0.5x + 1.4)(3.5x – 1) = 1.75x2 – 0.5x + 4.9x – 1.4 = 1.75x2 + 4.4x – 1.4F O I L

Solution (h): ( y – 1)2 = ( y – 1) ( y – 1)

= = y2 – y + 1y2 – y – y + 1

F

O I L

Solution (i): ( d – 3)( d – 5) = = d2 – d + 15d2 – d – d + 15

F

O I L

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 + 36x2

Solution (d): (4x – y)2

=

=

25y2 – 30xy – 30xy + 36x2

F

O I L

(4x – y)(4x – y)= = 16x2 – 8xy + y216x2 – 4xy – 4xy + y2

F

O I L

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


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)

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5. Division of polynomials

In order to divide a polynomial by a constant we separate it into its component parts and divideeach 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.

(a)6 x 3

2

12 x 2(b)

2

42 x(c) (–3x – 8) 4

(d)46 x

2(e)

5

45x(f)

6x 12 x 2

3

(g)8x 12

4(h)

6 x 4 10x 3 8x 2

2 x 2(j)

3x 1

3x 2

6 x 2

(k)8x 12

4 x 2(l) 5x 2x 3

2 x 3

4 x 4 (m)3x 1

x 4

6 x 6

(f) 9x – 48x + 64

2.(m) 4x + x – 2 (n) 9x – 4x – 5x 5x – 13x + 2

3.(e) x – 12x + 36

(h) 25y – 60xy + 36x

(f) –5c + 20c

2.(a) – 4x + 4x Degree = 4 (b) – y + 7y Degree = 2

(v) 10x – 21x – 10 (w) 3x – 18x + 151.(u) 10a – 13a – 3

(o) – 2y – y + 6

x + 5x – 2

1.(m) 25x – 16

(f) 6x – 7x – 24

1.(m) – 6x + 8x – 4x (n) – 4x + 2x – x (o) 15x – 5x + 10x3

1.(e) 12x – 30x – 8

4.(g) 16x – 8xy + 64y

2.(e) 4x + x – 2 9x – 4x – 5x 5x – 13x + 4

1 3

4.(m) 25x y – 20xy + 20x y – 4xy (n) 12x – 11x – 15

(n) 12x – 11x – 15

(b) 3x + 6xb – 4b4.(a) x + 7xy + 12y

(g) 16x – 40x + 25

(g) – 4x + 32x – 28

1.(e) 4a + 5a (g) – 2d + 14d

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SolutionsExercise 1:

1.(a) Yes (b) No (c) No (d) Yes (e) No (f) Yes4 2 (c) 2x3 + 3x2 – x + 4 Degree = 3

(e) – 49c5 + 7c2 + 6 Degree = 5 (f) – d2 + 4d– 1 Degree = 22.(d) b6 + 6 Degree = 6Exercise 2:1. 11x3 + 3x2– 4x2.(a) 11x2 – 13x + 4 (b) 9x3 – 4x (c) 5x4 + x3– x2 (d) 3x3 – 7x

2 (f) 3 2 (g) 2 (h) 2w2 – 7w – 22.(i) 6x2 + 7x – 8.5 (j) 9x3 – 4x (k) 5x4 + x3– x2 (l) 5x3 – 7x

2 3 2 (o) 2 (p) 8x2 + x + 4

2.(q) – 7x2 + 5x – 9 (r) 8x3 – 7x2 – 4x – 3 (s) 2x3 –17 25

2.(t) 7x3 – 4x2 – 2x + 2 (u)– d4 – 3d + 4 (v) 6p2 – 2p – 24. 11x3 + 3x2 – 4x 5. –4x2 + 6x – 4 6. 2x2 – 9x + 14 7. 4x2 – 6x + 4 8. –2x2 +3. 3x – 2

9x – 14Exercise 3:1.(a) 3x + 24 (b) 8y2 + 18y (c) 6a3 + 8a2 + 18 (d) 4x4 – 24 x3

2 2 4 2 (h) 2x2 + 141.(i) 5x2 + 2x7 (j) 8x3 – x2 (k) – 12x + 6x4 (l) 4x5(2x2 – x + 5)

4 3 2 4 3 2 5 4

1.(p) 8x7– 6x6 + 20x5 (q) 0.6x2y + 1.5x2 – 1.8xy (r) xyy 6 y 23

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.(i) 6x2 – 5x – 6 (j) 2x2 + 4.7x – 0.24 (k) x2 – 25 (l) x2 – 6x + 92

1.(q) 2z2 + 7z + 6

2

(r) 25r2 – 49

2

(s) 6x2 – 7x + 2(p) 3p2 + 3p – 6(t) 12x2 – 11x – 15

2 2 2 (x) 2a2 + 6a + 42. A3.(a) x2 + 6x + 9

2(b) x2 – 2x + 1

2(c) 9x2 – 30x + 25

2(d) 9x2 + 24x + 16(h) 36x2 – 60x + 25

3.(j) 4x2 – 0.4x + 0.01 (k) x2 + 10x + 25 (l) x2 – 6x + 9 (m) 4x2 – 12x + 92 2

4.(d) 9x2 + 6xy – 8y2

2 2

(e) 12x2 – 30xb + 12b2

(c) 30c2 – 8xc – 6x2

(f) 6x2 – 7xy – 24y222 2

4.(j) 4x2 – 0.4xy + 0.01y2

2

(k) x2 + 10xy + 25y2

(i) 4x2 – 12xy + 9y2

(l) x3 + 18x2 – 3x(o) 8y4z2 + 16y2z + 6

2 2 2 2 2

Exercise 5:

(a) 6x2 – 3x + (b) x + 2 (c) (d) – 3x + 2 (e) – x + (f) – 2x + 4x2 (g) 2x – 3

(h) 3x2 – 5x + 4 (j) (k) (l) (m)

topic 11 polynomials 1. introduction to polynomials definition a polynomial is an algebraic...

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