Trinomials

Expressions of the form ax² + bx + c are called trinomialsExpressions of the form ax² + bx + c are called trinomials

Factorising Trinomials

When the coefficient of the highest power is 1.

i.e., ax² ± bx ± c, when a = 1 and b and c are integers.

When two binomials are multiplied the product is a trinomial.

Thus (x + 4) (x + 5) = x² + 9x + 20 …(1)

(x - 4) (x - 5) = x² - 9x + 20 … (2)

In this chapter we try to express a trinomial into binomial factors. From the above results, we notice that

(i) Since the first term in both the factors is x, the first term in the trinomials is x².

(ii) The product of the second terms in the factors e.g., in (1), +4 and +5 (i.e., +20) is the third term in the trinomial.

(iii) The sum of the second terms of the factors e.g., in (2), the sum of -5 and -4 which is -9, is the coefficient of x in the trinomial.

We further illustrate these laws by the following examples.

Resolve into factors: x² + 8x + 15

x² + 8x + 15

= (x + 5) (x + 3)

The second terms of the factors must have a product of +15 and their sum equal to +8.

\ +5 and +3 are the second terms of the binomials.

Second Method:

x² + 8x + 15

= x² + 5x + 3x + 15 (after noticing that 5 + 3 = 8 and 5 3 = 15)

= x(x +5) + 3(x + 5)

= (x + 5) (x + 3)

Resolve into factors: x² - 15x + 56

x² - 15x + 56

Take factors of 56 having their sum = -15

They are -8, -7.

\ x² - 15x + 56 = x² - 7x - 8x + 56

\ x² - 15x + 56 = (x - 8) (x - 7)

Second Method:

x² - 7x - 8x + 56

Write -15x as -7x and -8x

= x (x - 7) - 8 (x - 7)

= (x - 7) (x - 8)

Now, we consider a case where the third term of the trinomial is negative.

Resolve into factors: x² + 3x - 28

Since the third term of the trinomial is -28, find two factors of 28 which differ by 3. The greater factor will be positive.

x² + 3x - 28                                28 = 7 4

= (x + 7) (x - 4)                             factors are +7 and -4.

Resolve into factors: x² + 5x - 24

x² + 5x - 24                                24 = 8 3

= (x + 8) (x - 3)                          factors are +8 and -3

Resolve into factors: x² - 2x - 48

x² - 2x - 48                                48 = 8 6

= (x - 8) (x + 6)                         factors are -8 and +6

Second Method:

x² - 2x - 48

= x² - 8x + 6x - 48

= x(x - 8) + 6(x - 8)

= (x - 8) (x + 6)

When the coefficient of the highest power is not unity.

i.e., type ax² ± bx ± c, when and b and c are integers.

Multiply (3x + 2) (x + 4) and consider the result so obtained.

= 3x (x + 4) +2 (x + 4)

= 3x² + 12x + 2x + 8

= 3x² + 14x + 8

When we want to factorise 3x² + 14x + 8, it becomes a converse problem i.e., we have to resolve the given expression into factors. We find the product of the coefficient of the 1^st term and the 3^rd term. The product is 3 x 8 = 24.

Since the third term is positive, the factors of 24 will have like signs and these signs will the same as the sign of the middle term, i.e., positive sign.

Find the factors of 24 with their sum = +14.

The factors are 12 and 2.

+14x will be written as +12x + 2x.

3x² + 14x + 8

= 3x² + 12x + 2x + 8

= 3x (x + 4) + 2 (x + 4)

= (3x + 2) (x + 4)

All the steps have occurred in the reverse order to the multiplication given above.

Resolve into factors: 6x² - 13x + 5

The product of 6 and 5 is 30. Since the third term is positive, both factors of 30 will have like signs.

Since the middle term is negative, both the factors will be negative.

6x² - 13x + 5

= 6x² - 10x - 3x + 5 [Find the factors of 30 having their sum = 13.

The factors are 10 and 3.

-13x will be written as -10x - 3x]

= 2x(3x - 5) - 1(3x - 5)

= (3x - 5)(2x - 1)

Resolve into factors: 7x² + 11x - 6

The product of 7 and -6 is -42. Since the third is negative, the factors of - 42 will have opposite signs. The bigger factor will have the sign of the middle term, and the smaller factor will have the opposite sign. The factors will differ by +11.

7x² + 14x - 3x - 6 [Find factors of 42 having their difference = 11.

The factors are 14 and 3.

\+11x will be written as +14 x - 3x]

= 7x (x + 2) -3 (x + 2)

= (7x - 3) (x + 2)

Resolve into factors: 5x² - 2xy - 16y²

The product of 5 and -16 is -80.

Since the third term is negative, the factors of 80 will have opposite signs, the bigger factor will have the same sign as the middle term and the smaller factor will have the opposite sign. The factors will differ by 2.

5x² - 2xy - 16y² [Find the factors of 80 having their difference = 2

The factors are 10 and 8.

-2xy will be written as -10xy + 8xy]

= 5x² - 10xy + 8xy - 16y²

= 5x (x - 2y) + 8y (x - 2y)

= 5x (x - 2y) + 8y (2 - 2y)

= (5x + 8y) (x - 2y)

It is not easy to select the proper factors at a glance. Practice alone will help the student to select the required pairs of factors with ease.

Factorising a³ b³

The product of a + b and a² - ab + b² is a³ + b³. Hence when a³ + b³ is factorised, we get:

a³ + b³= (a + b) (a² - ab + b²)

Similarly, a³ - b³= (a - b) (a² + ab + b²)

Factorise: x³ + 8

x³ + 8 = (x)³ + (2)³

= (x + 2) (x² - 2x + 4)

Factorise 64x³ - 125.

64x³ - 125 = (4x)³ - (5)³

= (4x - 5) (16x² + 20x + 25)

Factorise 64a⁶ - 343.

64a⁶ - 343 = (4a²)³ - (7)³

= (4a² - 7) (16a⁴ + 28a² + 49)