Mathematics

Key unit competence

By the end of this unit, I will be able to perform operations, factorise polynomials, and solve related problems

Unit outline

• Definition and classifications of polynomials including homogenous polynomials.

• Operations on polynomials.

• Numerical values of polynomials.

• Algebraic identities.

• Factorisation of polynomials

2.1 Introduction to polynomials

With reference to algebra, a polynomial is an expression that consist of variables and coefficients combined by the operations of addition, subtraction, multiplication, and non-negative integer exponents. Polynomials describe an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). It is the second unit of the book and the learner will be able to learn about the definition, classification, and operations of polynomials. Similarly, the learner will be able to look into numerical values of polynomials, algebraic identity, and factorisation of quadratic expressions relating them to the real life situation.

Consider the following algebraic expression

(a) 2x 

Answer the following questions based on the expressions above.

1. How many terms does each of the expression have?

2. State the highest power of x in each expression.

3. Using a dictionary or internet, find out the classification of  polynomials based on the number of terms.

(a) A monomial: is an algebraic expression which consists of only one term.

(b) A binomial: is an algebraic expression which contain (or is made up of) two terms only.

Examples: 2a – 3b and 2x2 + 5 are binomials.

(c) A trinomial: is an algebraic expression which is made up of three terms. Examples: 2a + 3b + c, 2x² – 3x + 5

(d) A polynomial: is any algebraic expression containing more than two terms of different positive powers of the same variable or variables.

The highest power of the variables in a polynomial defines the degree or order of the polynomial.

For example x⁴ + 2x³ + 3x² – x + 5 and 3x⁵ – 4x⁴ + 2x³ – x² + x + 3  are examples of polynomials of order 4 and 5 respectively. A zero degree polynomial is known as a consonant e.g 2x° = 2  is a constant.

The general form of a polynomial order n

The power of the variable in any term of a polynomial must be positive.

For example, an expression of the type

is not a polynomial Because some of the powers of are not positive.

Generally, a quadratic polynomial will take the form ax²+bx+c, a cubic polynomial takes the form ax³+bx²+cx+d and so on.

Homogenous Polynomial

A polynomial containing two or more variables are said to be homogenous if every term is of the same degree. 

For example, xy² + x2y + 3x³ and 3x + 2y – 4z are homogenous polynomials of degree 3 and 1 respectively.

xy² + x²y + 3x³, its equivalent form isx1y² + x2y¹ + 3x³

 In every term, the sum of the powers of the variables is equal to 3. Therefore, this is called a homogenous polynomial of order 3.

Similarly, the expression 3x + 2y – 4z, is a homogenous polynomial of first degree

Exercise 2.1

1. For each of the following write the (i) numerical coefficient (ii)variable part

2. Identify each of the following expression as a monomial, binomial, trinomial or polynomial

(a) 5x³ + x² – 3x – 4

(b) 3ab²c – 6b

(c) 4x²y²

(d) 4 – 6ab²

(e)  x² + y³

(f) a²b³ – 25

(g)  –yx²z³

2a + 3a²c – b

(i)  –5x² + 6x + 3

3. Which of the following are homogenous? State the degree of those that are homogenous.

(a) zx + xy

(b) a²b² + 2a + 2b

(c) x³ + y³ – z³

(d) x² – 3xy – 40y²

(e) 6x – 5y + 6z

(f) ab + ac + bc

(g) x³ + y³ + z³ + 3a²c + 3ac²

2a2 – 7ab – 30y2

(i) 5x³ + 6x²y – 7xy² + 6y³

4.  Which of the following are homogenous polynomials? State their degree where possible.

(a) x² + y² + z²

(b) 3xy + zx – 2yz

 (c) x³ + y³ – 2x – 2y

(d) x³ + y²x + z³x

(e) x³ + y²z – x²yz

(f) a²b + ab² – x²y + xy²

2.2 Operations on polynomials

The rule that governs the basic operations on numbers also apply in polynomials. In this unit, we are going to concentrate on addition and subtraction, multiplication and division of polynomials.

2.2.1 Addition and subtraction

Activity 2.2

Consider the expressions;

(i) 3x³ – 13x² + 4x – 2

(ii)  x² + 3x + 4

(iii) 5x²– 3x + 3

(a) Identify the terms that are alike in (i),(ii) and (iii) and group them together.

(b) Combine the sets of terms in (a) above.

(c) Simplifying each group to a single term.

(d) Now add all the expressions together  (3x³ – 13x² + 4x – 2)  + (x2 + 3x + 4) + (5x² – 3x + 3)

To be able to add or to subtract polynomials you need to identify:

(i) Like terms: These are terms which have the same variables to the same power.

                      

 4x and 6x are like terms:  and 7a and 4a are like terms.

(ii) Unlike terms - these are terms which either have different variables or same variable to different powers.

2x and – 3y, 5a and 3b cannot combine because the variable parts are different i.e are unlike terms.

3x² and 4x are unlike terms because they have different powers of x.

The following are more example of like and unlike terms

2a and 3a are like terms,

2ab and 4ab are like terms,

2a and 5b are unlike terms,

2a and 2ab are unlike terms, a

and a2 are unlike terms.

When simplifying algebraic expressions, first collect the like terms together. Simplification is usually easier if the positive like terms are separated from the negative ones.

Example 2.1

Simplify the expression

2x – 4y + 5x – 3y

Solution

To simplify the expression, you first collect like terms together for example,

2x – 4y + 5x – 3y  

= 2x + 5x – 4y – 3y (like term together) 

= 7x – 7y (combine like terms)       

= 7x – 7y (this expression cannot be simplified further       because 7x and –7y are unlike terms)

2.2.2 Substitution and evaluation

Activity 2.3

1. Consider the polynomial expressions 

(a) x² + y + 1

(b) 3x² + 2y - 3

2. Given x = 2 and y = 3, solve the expressions.

3. Compare your findings with other classmates.

Polynomials can be evaluated numerically if some numerical values are attached to the variable or variables. We use the method of substitution in the given expression and then simplify.Substitution involves replacing variables, in an algebraic expression, with specific values. The expression may then be evaluated.

Exercise 2.2

1. In each of the following, pick out the term which is unlike the others.

(a)  2x,  4x,  6x,  3x2,  8x

(b)  m3,  5m²,   6m³,  3m³,  10m³

(c)   x2y,  xy,  yx2,  3x2y,  4yx2

(d)  3mn,  nm,  –mn,  m2n, 2 /3 mn

2. Simplify the following expressions.

(a)  y + y + y

(b)  n + n – n + n + n – n – n

(c)   f – f + f – f + f 

(d)  d + d + d – d – d + d

3. Simplify:

(a)  3a + 3a  

(b)  4b – b

(c)   6z – 2z  

(d) k – k

(e) q + 2q  

(f)   5p + 7p

(g) 9r – 8r  

  w – 5w

4. Simplify:

(a)  5a + a + 3a 

(b)   c + 2c + 4c

(c)   3b + 3b + 3b 

(d)   5y – 4y + y

(e)   12w – 6w + 6w 

(f)   9n – 3n + 2n

(g)  2m + 8m – 4m + m – 2m

  8t – 2t – 3t + 4t – 7t

5. Simplify the following by first collecting like terms together and grouping those with  the same signs.

(a)   x + y + y + x    

(b)  3w + 8w + 9z – 4z

(c)   11n + 11 + n – 10

(d)   4s – 2t + 5t – 3s  

(e)   2p – 7 – 4 + 5p  

(f)   14b – 9c – 6b 

(g)   4m – m + 5n – 4n

   10 – 5d + 2d – 15 + 4d 

6. Simplify

(a)  x² – 3x – 2 + 4x² – 2x + 5

(b)  3y² – 4y – 6 – 3 – 2y – 3y²

(c)  (2x² – 3x)+ (5x – 8)

(d)  (3y² – 2y) – (3y + 4)

7. Given that x = 4, y = 3 and z = 2,

evaluate; (a)  2x – y + 7 

(b)  4x – 2y + 2z

(c)  5x – y – z   

(d)  3x – 3y + 4z

8. If m = 4 and n = 3,

evaluate: (a)  3m + 3     

(b) 4m – 5n 

(c)  1/2 m + n  

(d)  1/4 m – 1/3n  

(e)  5m – 5  

(f)  6m +2n

(g)  3m – 4n     

2/3 m – n

(i)  3n²   

(j) 2mn²    

(k)  mn – n   

(l) m(n– m)

(m) 2m³      

  1/2 mn

(o) m² – n²

10. If E = 1/2 mv²,

find E when m = 27 and  v = 1 3.

11. If xy = 5 and y = 2,

find: (a)  x       (b)  2(x + y)

2.2.3 Multiplication of polynomials

2.2.3.1 Multiplication of monomials

Activity 2.4

Each car has four wheels. Let w stand for one wheel.

1.  Write a mathematical expression for the number of wheels for one car.

2.  Write a mathematical expression with w and involving multiplication for finding the total number of wheels for 5 cars.

3.  Simplify the expression in step 2 above.

4. Discuss with your partner how to simplify the following expressions:

5. Compare your results in step 4 above with those of other classmates.

Study the following facts regarding multiplication. Consider:

Exercise 2.3

Simplify the following:

1. (a)  3 × 4a                 (b)  4m × 5

    (c)  6x × 9                  (d)  11 × 3q

2. (a)  3x × 2y                    (b)  2a × 7b

   (c)  8q × 5p                      (d)  y × 8x

3. (a)  a × 3ab                      (b)  2a × 7ab

   (c)  7y × 4yx                      (d)  13st × 11t

4. (a)  5 × 2m²                      (b)  15pq × p²

(c)   (3q)² × pq                       (d)  3p2 × 2q²

2.2.3.2 Multiplication of a  polynomials by a monomial

Removing  brackets in multiplication of polynomials

Usually, multiplication of polynomials involves removing brackets. The following activity prepares us for multiplicatio

Activity 2.5

1. Open the brackets correctly and simplify the following expressions.

(a) 4a + (5 + 3a)                             (b)  4a – (5 - 3a)

(c)   4a – 2(5 + 3a)                         (d)  4a – 2(- 3a – 5)  

2. Using your observation remove the brackets in

(a) a – (b + c)                                 (b) a – (–b – c).

3. Comment on your answers and compare your results to those of other members of the class.

The terms inside brackets are intended to be taken as one term.

For example, 9 + (7 + 3) means that 7 and 3 are to be added together, and their sum added to 9, so that,

9 + (7 + 3)  =  9 + 10  =  19

Also,  9 + 7 + 3  =  19

Hence, the bracket may be removed without changing the result,

i.e.  9 + (7 + 3)  =  9 + 7 + 3

In general,

x + (y + z)  =  x + y + z

Consider the expression 9 + (7 – 3)

9 + (7 – 3)  = 9 + 4 (First subtracting 3        from 7)  9 + 4      =  13

Also, 9 + 7 – 3  =  16 – 3  =  13

Hence, the bracket may again be removed without changing the result.

In general,

x + (y – z)  =  x + y – z

Consider the expression 9 – (7 + 3) 9 – (7 + 3) = 9 – 10   (First adding 3 to        = –1     7) =  10

Also,  9 – 7 – 3  =  2 – 3  =  –1

Hence, in this case, when the bracket is removed, the sign of each term inside the bracket is changed.

i.e.  9 – (7 + 3) = 9 – 7 – 3

In general,

x – (y + z)  =  x – y – z

Caution  !

9 – 7 + 3  =  2 + 3  =  5

Note that this is not the same as 9 – (7 + 3). This is a common mistake, which must be avoided.

In general,

x – (y + z)  ≠  x – y + z

Consider the expression 9 – (7 – 3) 9 – (7 –  3)  =  9 – 4

(First subtracting 3  from 7)=  5

Also,  9 – 7 + 3  =  2 + 3  = 5.

Similarly, when the bracket is removed, the sign of each term inside the bracket is changed to the opposite sign,

i.e.  9 – (7 – 3)  =  9 – 7 + 3.

In general,

x – (y – z)  =  x – y + z

Caution  !

9 – (7 – 3)  ≠  9 – 7 – 3 !

∴  in general,

x – (y – z)  ≠  x – y – z

The rules, therefore, are:

1. If there is a positive (plus) sign just before a bracket, the sign of each term inside the bracket is unchanged when the bracket is removed  (i.e. when the expression is expanded).

2.  If there is a negative (minus) sign just before a bracket, the sign of each term inside the bracket must be changed to the opposite sign when the bracket is removed. Removing the bracket is like multiplying each term by –1

Example 2.6

Remove the brackets and simplify:

(a) 7g + (3g – 4h) – (2g – 9h)

(b) (6x – y + 3z) – (2x + 5y  –  4z)

 Solution

(a) 7g + (3g – 4h) – (2g – 9h) 

=  7g + 3g – 4h – 2g + 9h  

=  7g + 3g – 2g + 9h – 4h  

=  8g + 5h

(b) (6x – y + 3z) – (2x + 5y – 4z) 

=  6x – y + 3z – 2x – 5y + 4z

=  6x – 2x – y – 5y + 3z + 4z 

=  4x – 6y + 7z 

Note:  In (b) above, there is no sign before the first bracket so a positive sign is assumed.

In the expression 9 × (7 + 3), the bracket means that 7 and 3 are to be added together, and the result multiplied by 9. Thus,  9 × (7 + 3)  =  9 × 10  =  90.

But

9 × 7 + 9 × 3  =  63 + 27  = 90. ∴ 9 × (7 + 3)  = 9 × 7 + 9 × 3, which means that 7 and 3 may each be multiplied by 9 and the products added together. The multiplication sign is usually omitted, so that 9(7 + 3) means exactly the same as  9 × (7 + 3), just as a × b  = ab.

In general, when expression in a bracket is multiplied by a number in order to remove the brackets, every term inside the bracket must be multiplied by that number.

  Thus, 

a(x + y) = a × x +  a × y = ax + ay   and a(x – y) = a × x – a × y = ax – ay

Example 2.7

Remove the brackets and simplify:
 2(3x – y) + 4(x + 2y) – 3(2x – 3y)
Solution

2(3x – y) + 4(x + 2y) – 3(2x – 3y)

= 2 × 3x – 2 ×  y + 4 × x + 4 × 2y – 3 × 2x + 3 × 3y

=  6x – 2y + 4x + 8y – 6x + 9y

=  6x + 4x – 6x + 8y + 9y – 2y

=  4x + 15

Notice how the signs obey the rules obtained earlier

When simplifying expressions containing brackets enclosed in other brackets, remove the innermost bracket first and collect the like terms (if any) before removing the next outer bracket.

Example 2.8

Simplify   {3y – (x – 2y)} – {5x – (y + 3x)}

Solution

{3y –(x – 2y)} – {5x – (y + 3x)}

=  {3y – x + 2y} – {5x – y – 3x}  

=  {5y – x} – {2x – y} =  5y – x – 2x + y

=  6y – 3x

Note: Different shapes of brackets are usually used to make the meaning of the hyphen expression more easily understood.

Exercise 2.4 1.

Remove the brackets and simplify the following:

(a)  5(2x + 3)                                  (b)  4(3m – 2n)

(c)  7(2b – 3c + 1)                           (d)  3w(4x – 1)

(e)  6(3x – 5y – 1)                               (f)  2(4r – 3) + 3(s – 1)

(g)  3a(2b + c) – 2a(2x + y)                    5(c + 4) – 2(3c – 8)

(i)  –(x + y) + x                                       (j)  (7a + 5b) – (3a – 10b)

(k)  (x – 3y)9y + 2(y² – 3xy)                    (l)  xy(x – xy) – x(xy – x2)

2. Write algebraic expressions for the following. Do not remove brackets.

(a) Add a to 2y and multiply the result by 4.

(b) Divide 12e + 30d – 18 by 6.

(c) The product of 3 consecutive even numbers, the largest of which is p.

(d) The number by which a + b exceeds a – b. 3

3. Copy and complete the following:

(a) m + n – l  =  m + ( _____ )

(b) a – b – c  =  a – ( _____ )

(c) x – y + z  =  x – ( _____ )

(d) p – q – r + s  =  p – ( _____ )

(e) u – v + w + x = u – ( _____ )

(f) x – y + v – w  =   x – ( _____ )

(g) a + 2b – 2c  =  a + 2( _____ )

a – 3b – 6c  =  a – 3( _____ )

(i) 2a – 8c – 3x – 9z      

=  2( _____ ) – 3( _____ )

(j) k2  + 2kl – 3m2  + 4mn     

=  k( _____ ) –  m( _____ )

2.2.3.3  Multiplication of a  polynomial by a polynomia

We have just learned how to multiply expressions of the form a(x+y) using integers, and the result generalised. Remember:

• An expression such as a (2x + c) means  a × (2x) + a × c = 2ax + ac

• This skill will help us to multiply two or more polynomials, whatever the number of terms.

Now, consider the expression  p (x + y) we have just seen that: p (x + y) means p × x  +  p × y Multiplying (a + b)(x + y),suppose we let p represent (a + b)

So that (a + b)(x + y)

= p(x + y)……(1)           

= p × x + p × y            

= px + py            

= xp + yp..(2)

Substituting (a + b) for p in equation (2)   xp + yp = x(a + b) + y(a + b)    = ax + bx + ay + by (a + b)(x + y) = ax + bx + ay + by This is called a binomial expansion The result show that each term in one bracket, then the result added.

Example 2.9 Multiply and simplify

(a) (x + 2)(x + 3)

(b)  (3x – 2)(2x – 3)

The method used in examples 2.8 to 2.10 can be used to multiply polynomials of any degree. However when polynomials have more than two terms, it is easier to use an alternative arrangement; similar to the one used in Example 2.12 below

Exercise 2.5 1. 

Multiply

(a) (x – 3)(x – 2) 

(b) (a – 5)(a + 5

(c) (y + 4)(y – 4) 

(d) (x + 5)(x + 5)

2.   Simplify

(a)  3(x – 1)(x – 4) 

(b) (2y² – 1)(6y² + 7)

(c)  3(3x + 1)²     

(d) 2(3y – t)(2y – t)

3. (a) (x² – 8)(x² – 3)    

(b) (2y2 – 1)(6y2 + 7)

(c) (ab – 6)(ab + 6)

(d) (2x – 1/2 )²

4. (a) (a – 3)(2a – 2) + 2(2a + 3)(2a – 1)

(b) (3x – 5)² – 2(x – 5)(x + 5)

(c) 3(x – y)² – 3(x + y)(x – 2y)

(d) (3x – 2)(2x² – 2x + 1)

5. Expand and simplify the following

(a) (x + y)(x – y – 2)

(b) (3x – 2)(2x² – 2x + 1)

(c) (2x + 3 – y)²

(d) (a – 2b)(3a² – 2ab + b²)

(e)  –x(2x – 3x – 1)

(f) (x – y – 2)(2x + 3 – y)

2.2.4 Division of polynomials

2.2.4.1 Division of a monomial by a monomial

Activity 2.6

1.  With your partner, discuss the division rule of indices.

2.  Consider the polynomial expression

8x³y⁵ ÷ 4x²y³.

3.  State the denominator and the numerator of the expression in (2) above.

4. Simplify the expression by cancelling all the common factors.

5. Compare your results with those of other classmates

Consider a polynomial expression 36x³y⁵ ÷ 9xy².

To solve the expression, we first identify the numerator and denominator then cancelling the common terms. The division law of indices is applied i.e.

Activity 2.7 Divide 16x²y³ + 8xy² – 2x³y³ by 2xy (a) Identify the number of terms in the numerator. (b) Divide each term by the divisor. (c) State your answer and compare with other classmate

Activity 2.7

Divide 16x²y³ + 8xy² – 2x³y³ by 2xy

(a) Identify the number of terms in the numerator.

(b) Divide each term by the divisor.

(c) State your answer and compare with other classmates

Consider the division: 12x³y² – 6xy² + 18x²y divided by 3xy

2.2.4.2  Division of a polynomial by a  polynomial

Activity 2.8

1. Through a discussion with your partner divide the polynomial 

x² + 9x + 18 by x + 3 Hint: Try the long division method.

2. Compare your result with those of other classmates.

For the division of one polynomial by another to work, the degree of the dividend must be higher than that of the divisor.

Remember:

Like in division of numbers, not all polynomials divide exactly, some will have remainders.

Suppose the polynomial to be divided is denoted by f(x), and the divisor by the polynomial g(x), we can denote the result of division as,

Where Q is the quotient and R is the remainder, the division process terminates as soon as the degree of R is less than the degree of division g(x)

In division, order of the terms is important;

(i) Both the dividend and the divisor must be written in descending powers of the variable.

(ii) If a term is missing, a zero term must be inserted in its place.

Example 2.15

Divide –20 + 6x³ – 4x² by 2x – 4 and state the quotient and the remainder.

Solution

(i) Rearrange the terms in the dividend and write them in descending order.

(ii) Insert 0x for the missing.

Example 2.16

Divide 2x⁴ + x³ – 3x² by x2 + 2 and state the quotient and the remainder.

Solution

In the dividend, the constant and the x term are missing, so we replace them with 0x, and 0 respectively. In the divisor, the term in x is missing, so we replace it with 0x.

Exercise 2.7

1. Arrange each of the polynomials in descending powers of the variable.

(a) –12x³ + 3x² – x⁴ + 7x + 10

(b) 4x³ – 3x⁵ + 6x²

(c) 4a + a² – 12 + 12a³

(d) 4a⁵ – 5a + 3a³ + a⁴ – 7a²

2. Arrange the following polynomials in descending order, inserting zero for any missing term.

(a) 8 + 4x³ – 2x

(b) –3x³ + 4x⁵ – 2x² + 7x

(c) 8x⁵ – 3x² + 4x³ – 7

(d) 7a + 3a⁴ – 8a²

3. Simplify the following:

(a) (m² + 5m) – (m² + m)

(b) (4x² + 6x) – (4x² – 10x)

(c) (–6x² – 3x) – (–6x – 8x)

(d) (11x² + 0x) – (11x² – 10x)

4. Verify whether the statements given below are true or false.

(a) a² – 7a – 18 = (a – 9) (a + 2)

(b) b³ – 8 = (b² + 2b + 4) (b – 2)

(c) 3x³ – 2x² + 16x – 8 = (3x² + 4x + 11) (x – 2) + 18

(d) 3p² – 3p + p³ – 8 = (p² + 5p + 7)  (p – 2) + 6

5. Divide the given polynomials and in each case state the quotient and the remainder.

(a) (13x + 14 + 3x²) ÷ (x + 2)

(b) (t + 20t² – 4) ÷ (4 + 5t)

(c) (17a + 14a² + 7) ÷ (2 – 7a)

(d) (9 – 16r²) ÷ (4r – 3)

(e) (x + 2 – 3x² – 2x³) ÷ (1 + 2x)

(f) (a + 2a⁴ – 14a² + 5) ÷ (a² + 5)

(g) (3y² + 4 – 10y) ÷ (3y – 2)

(3a³ – 1) ÷ (a – 1)

Exercise 2.11

1.Expand the following expression   using  the method used in Example   2.21.

(a)

(i) (a + 1) ²

(ii) (a + 6b) ²

(iii) (x + y) ²

(iv) (x + 9) ²

(v) (m + n) ²

(vi) (2a + 3b) ²

(vii) (3x + 4) ²

(viii) (3m + 2)²  

(ix) (4x + 3y) ²

 

(b)

(i) (b – 1) ²

(ii) (r – 3) ²

(iii) (x – y) ²

(iv) (4x – 3) ²

(v) (5x – 2) ²

(vi) (3x – 12) ²

  (vii) (5x – 3)²

(viii) (4z – 3b)²

(ix) (7x – 2y) ²

2. Expand the following using the method  used in Example 2.22.

(a) (a + 3)(a – 3)

(b) (a + 5)(a – 5) (

c) (x – 9)(x + 9) 

(d) (f + g)(f – g)

(e) (2p – 1)(2p + 1) 

(f) (4x – y)(4x + y)

(g) (7 + 2x)(7 – 2x) 

(2a + 3b)(2a – 3b)

(i) (5y + 3)(5y – 3) 

(j) (4x – 1)(4x + 1)

(k) (3x + 4)(3x – 4)

(l) (2x – 3y)(2x + 3y)

(m) (8 – 3x)(8 + 3x)

(3x + 7y)(3x – 7y)

 

2.4.3  Factorising quadratic expressions
2.4.3.1  General methodology  of factors in quadratic expressions

It is easy to see that 2x² – 16 = 2(x² – 8) and 3x2 – 5x = x(3x – 5).

However, it is not easy to see what the factors of x2 + 5x + 6 are.  Our experience in multiplying binomials is of great help here.

Now, consider the product (x + 3)(x + 2). (x + 3) and (x + 2) are prime binomial expressions, since the two terms in each bracket have no common factor.

 

Activity 2.17

1. Expand and simplify the expression (x + 4) (x – 2)

2.  Work back to factorise the expression you have obtained.

Given the expression; (x + 3)(x + 2) Expansion of the expression takes the steps below

(x + 3)(x + 2) =  x(x + 2) + 3(x + 2)    =  x² + 2x + 3x + 6    =  x² + 5x + 6       (since 2x and 3x      are like terms). This means that (x + 3) and (x + 2) are factors of x² + 5x  + 6. ⇒ x² + 5x + 6 = (x + 3)(x + 2)(in factor            form).

Note:  In x² + 5x + 6,

1. the coefficient of the quadratic term is 1,

2. the coefficient of the linear term is 5, the sum of the constant terms in the binomial factors, and

3. the constant term is 6, the product of the constant terms in the binomial factors.

Generally,

In a simple expression like ax² + bx + c, where a = 1, the factors are always of the form  (x + m)(x + n), where m and n are constants. Such an expression is factorisable only if there exists two integers m and n such that m × n = c and m + n = b.
To factorise a quadratic expression of the form ax² + bx + c, where a = 1, follow the steps below.

 

1. List all the possible pairs of integers whose product equals the constant term.

2. Identify the only pair whose sum equals the coefficient of the linear term.

3. Rewrite the given expression with the linear term split as per the factors in 2 above.

4. Factorise your new expression by grouping, i.e. taking two terms at a time.

5. Check that the factors are correct by expanding and simplifying.

 

Example 2.24

Factorise x² + 8x + 12.
Solution In this example, a = 1,  b = 8  and  c = 12.

1. List all the pairs of integers whose product is 12. These are:  1 × 12 3 × 4 2 × 6 1 × –12 –3 × –4 –2 × –6

2. Identify the pair of numbers whose sum is 8.  The numbers are 6 and 2.

3. Rewrite the expression with the middle term split. x² + 8x + 12  = x ² + 2x + 6x + 12

4. Factorise x² + 2x + 6x + 12 by grouping.  x² + 2x + 6x + 12 has 4 terms which we can group in twos so

that first and second terms make one group and third and fourth terms make another group

i.e  x²+ 2x + 6x + 12 In each group, factor out the common factor.

 Thus,

x²+ 2x + 6x + 12 =  x(x + 2) + 6(x + 2) We now have two terms, i.e. x(x + 2) and  6(x + 2), whose common factor is (x + 2) .

∴  x² + 8x + 12  =  (x + 2)(x + 6)  (Factor out the common factor (x + 2)) Check that (x + 2)(x + 6)  =  x² + 8x + 12