Factorisation Of Algebraic Expressions

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have x² + 5x + 6.

First: Multiply the coefficient of x² and the constant.

Coefficient of x² = 1

Constant = 6

⇒ 1 × 6 = 6 …(a)

Now, Observe the coefficient of x = 5

We need to split the value obtained at (a), such that the sum\difference of the split numbers comes out to be 5.

To split: We need to find the factors of 6.

Factors of 6 = 2, 3

Add 2 and 3 = 2 + 3 = 5

Multiply 2 and 3 = 2 × 3 = 6

Since, the sum of 2 and 3 is 5 and multiplication is 6. So, we can write

x² + 5x + 6 = x² + (3x + 2x) + 6

⇒ x² + 5x + 6 = x² + 3x + 2x + 6

⇒ x² + 5x + 6 = (x² + 3x) + (2x + 6)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ x² + 5x + 6 = x(x + 3) + 2(x + 3)

Now, take out the common number or variable again from the two pairs.

⇒ x² + 5x + 6 = (x + 3)(x + 2)

Thus, the factorization of x² + 5x + 6 = (x + 3)(x + 2).

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have x² + x – 6.

First: Multiply the coefficient of x² and the constant.

Coefficient of x² = 1

Constant = -6

⇒ 1 × -6 = -6 …(a)

Now, Observe the coefficient of x = 1

We need to split the value obtained at (a), such that the sum/difference of the split numbers comes out to be 1.

To split: We need to find the factors of -6.

Factors of -6 = 2, 3, -1

Add 3 and (2 × -1 =) -2 = 3 + (-2) = 1

Multiply 3 and -2 = 3 × -2 = -6

Since, the sum of -2 and 3 is 1 and multiplication is -6. So, we can write

x² + x – 6 = x² + (3x – 2x) – 6

⇒ x² + x – 6 = x² + 3x – 2x – 6

⇒ x² + x – 6 = (x² + 3x) + (-2x – 6)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ x² + x – 6 = x(x + 3) – 2(x + 3)

Now, take out the common number or variable again from the two pairs.

⇒ x² + x – 6 = (x + 3)(x – 2)

Thus, the factorization of x² + x – 6 = (x + 3)(x – 2).

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have x² – x – 6.

First: Multiply the coefficient of x² and the constant.

Coefficient of x² = 1

Constant = -6

⇒ 1 × -6 = -6 …(a)

Now, Observe the coefficient of x = -1

We need to split the value obtained at (a), such that the sum/difference of the split numbers comes out to be -1.

To split: We need to find the factors of -6.

Factors of -6 = 2, 3, -1

Add (3 × -1 =) -3 and 2 = -3 + 2 = -1

Multiply -3 and 2 = -3 × 2 = -6

Since, the sum of 2 and -3 is 1 and multiplication is -6. So, we can write

x² – x – 6 = x² – (2x – 3x) – 6

⇒ x² – x – 6 = x² – 2x + 3x – 6

⇒ x² – x – 6 = (x² – 2x) + (3x – 6)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ x² – x – 6 = x(x – 2) + 3(x – 2)

Now, take out the common number or variable again from the two pairs.

⇒ x² – x – 6 = (x – 2)(x + 3)

Thus, the factorization of x² – x – 6 = (x – 2)(x + 3).

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have y² + 23y + 102.

First: Multiply the coefficient of y² and the constant.

Coefficient of y² = 1

Constant = 102

⇒ 1 × 102 = 102 …(a)

Now, Observe the coefficient of y = 23

We need to split the value obtained at (a), such that the sum/difference of the split numbers comes out to be 23.

To split: We need to find the factors of 102.

Factors of 102 = 2 × 3 × 17

Add 17 and (2 × 3 =) 6 = 17 + 6 = 23

Multiply 17 and 6 = 17 × 6 = 102

Since, the sum of 17 and 6 is 23 and multiplication is 102. So, we can write

y² + 23y + 102 = y² + (17y + 6y) + 102

⇒ y² + 23y + 102 = y² + 17y + 6y + 102

⇒ y² + 23y + 102 = (y² + 17y) + (6y + 102)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ y² + 23y + 102 = y(y + 17) + 6(y + 17)

Now, take out the common number or variable again from the two pairs.

⇒ y² + 23y + 102 = (y + 17)(y + 6)

Thus, the factorization of y² + 23y + 102 = (y + 17)(y + 6).

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have a² + a – 132.

First: Multiply the coefficient of a² and the constant.

Coefficient of a² = 1

Constant = -132

⇒ 1 × -132 = -132 …(a)

Now, Observe the coefficient of a = 1

We need to split the value obtained at (a), such that the sum/difference of the split numbers comes out to be 1.

To split: We need to find the factors of -132.

Factors of -132 = 2, 2, 3, 11, -1

Add (2 × 2 × 3 =) 12 and (11 × -1 =) -11 = 12 + (-11) = 1

Multiply 12 and -11 = 12 × -11 = -132

Since, the sum of 12 and -11 is 1 and multiplication is -132. So, we can write

a² + a – 132 = a² + (12a – 11a) – 132

⇒ a² + a – 132 = a² + 12a – 11a – 132

⇒ a² + a – 132 = (a² + 12a) + (-11a – 132)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ a² + a – 132 = a(a + 12) – 11(a + 12)

Now, take out the common number or variable again from the two pairs.

⇒ a² + a – 132 = (a + 12)(a – 11)

Thus, the factorization of a² + a – 132 = (a + 12)(a – 11).

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations.

And, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

Let’s factorize the given algebraic expression.

We have p² + 3p – 18.

First: Multiply the coefficient of p² and the constant.

Coefficient of p² = 1

Constant = -18

⇒ 1 × -18 = -18 …(a)

Now, Observe the coefficient of p = 3

We need to split the value obtained at (a), such that the sum/difference of the split numbers comes out to be 3.

To split: We need to find the factors of -18.

Factors of -18 = 2, 3, 3, -1

Add (2 × 3 =) 6 and (3 × -1 =) -3 = 6 + (-3) = 3

Multiply 6 and -3 = 6 × -3 = -18

Since, the sum of 6 and -3 is 3 and multiplication is -18. So, we can write

p² + 3p – 18 = p² + (6p – 3p) – 18

⇒ p² + 3p – 18 = p² + 6p – 3p – 18

⇒ p² + 3p – 18 = (p² + 6p) + (-3p – 18)

Now, take out common number or variable in first two pairs and the last two pairs subsequently.

⇒ p² + 3p – 18 = p(p + 6) -3(p + 6)

Now, take out the common number or variable again from the two pairs.

⇒ p² + 3p – 18 = (p + 6)(p – 3)

Thus, the factorization of p² + 3p – 18 = (p + 6)(p – 3).

(i) To resolve it into factors, we first factorize 129

129= 3×43

Given, x² + (p + q) x + pq

= (x + p)(x + q) … (2)

Comparing 1 and 2

p + q = - 40

= -43 + 3

pq = - 129

= -43 × 3

∴ (1) can be written as

x²-43x + 3x-129

x(x-43) + 3(x-43)

= (x-43) (x + 3)

∴ p=-43, q=3 and the resolved factors are (x-43) (x + 3).

(ii) To resolve it into factors, we first factorize 60

60=2×2×3×5

Given, x² + (p + q) x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 19

= 15 + 4

pq = 60

= 15× 4

∴ (1) can be written as

m² + 15m + 4m + 60

= (m + 15) (m + 4)

∴ p=15, q=4 and the resolved factors are (m + 15) (m + 4).

(iii) To resolve it into factors, we first factorize 6

6= 3×2

Given, x² + (p + q) x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = -1

= -3 + 2

pq =-6

= -3 × 2

∴ (1) can be written as

x²-3x + 2x-6

=(x-3)(x + 2)

∴ p=-3, q=2 and the resolved factors are (x-3)(x + 2).

(iv) to resolve it into factors we first resolve 12

12=2×2×3

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = -4

= -6 + 2

pq =-12

= - 6 × 2

∴ (1) can be written as

(a + b)² – 6(a + b) + 2(a + b)-12

=[{(a + b)-6}{(a + b) + 2}]

∴ p=-6, q=2 and the resolved factors are [{(a + b)-6} {(a + b) + 2}]

(v) (x-y)² – x + y – 2

= (x-y)² – (x-y) – 2 … (1)

To resolve it into factors, we first factorize 2

2= 1×2

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = -1

= -2 + 1

pq = -2

= -2 × 1

∴ (1) can be written as

(x-y)² – 2(x-y) + (x-y) – 2

= [{(x-y)-2}{(x-y) + 1}]

∴ p=-2, q=1 and the resolved factors are [{(x-y)-2}{(x-y) + 1}].

(a + b)²-5a-5b + 6

= (a + b)²-5(a + b) + 6 … (1)

To resolve it into factors we first resolve 6

6=2×3

We know, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = -5

= - 3 -2

pq = 6

= -3 × -2

∴ (1) can be written as

(a + b)² – 3(a + b) - 2(a + b) + 6

(a + b) { a + b – 3} – 2 { a + b -3}

=[{a + b- 2}{a + b - 3}]

∴ the resolved factors are [{a + b- 2}{a + b - 3}]

(x²-2x)² + 5(x²-2x)-36

To resolve it into factors we first resolve 36

36=9×4

We know, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2 we get,

p + q = 5

= 9 -4

pq = - 36

= 9 × (-4)

∴ 1 can be written as:

(x²-2x)² + 9(x²-2x) - 4(x²-2x) -36

= (x²-2x)[ (x²-2x) + 9] – 4 [(x²-2x) + 9]

= (x²-2x +9) (x²-2x - 4)

To resolve it into factors we first resolve 63

63=7×3×3

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n = -16

mn =63

∴ (1) can be written as

(p²-3q²)² - 9(p²-3q²) - 7(p²-3q²) + 63

=[{(p²-3q²) - 9}{(p²-3q²) - 7}]

∴ the resolved factors are [{(p²-3q²) - 9}{(p²-3q²) - 7}]

(a²)² + 4a²-5 … (1)

To resolve it into factors we first resolve 5

5=1×5

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =4

mn =-5

∴ (1) can be written as

(a²)² + 5a²- a²-5

= (a² + 5)( a²-1)

∴ the resolved factors are (a² + 5)( a²-1)

To resolve it into factors, we first factorize 420

420= 2×2×3×5×7

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 23

pq =-420

∴ (1) can be written as

x²y² + 35xy-12xy-420

= (xy + 35)(xy-12)

∴ The resolved factors are (xy + 35)(xy-12).

(x²)²-7x² + 12 … (1)

To resolve it into factors we first resolve 12

12=1×2×2×3

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =-7

mn =12

∴ (1) can be written as

(x²)²-3x²-4x² + 12

= x²(x² – 3) – 4(x² – 3)

= (x²-4)( x²-3)
We know a² – b² = (a – b)(a+b)

= (x-2)(x+2)( x²-3)

∴ The resolved factors are (x-2)(x+2)( x²-3)

To resolve it into factors we first resolve 12

12=1×2×2×3

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =1

mn =-12

∴ (1) can be written as

a² + 4ab-3ab + 12b²

= (a + 4b)( a-3b)

∴ The resolved factors are (a + 4b)( a-3b).

To resolve it into factors we first resolve 108

108=2×2×3×3×3

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =31

mn =108

∴ (1) can be written as

p² + 27pq + 4pq + 108q²

= (p + 27q)( p + 4q)

∴ The resolved factors are (p + 27q)( p + 4q)

(a³)² + 3a³b³-40(b³)²

To resolve it into factors we first resolve 40

40=2×2×2×5

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =3

mn =-40

∴ (1) can be written as

(a³)² + 8a³b³-5a³b³-40(b³)²

= (a³ + 8b³)( a³-5b³)

∴ The resolved factors are (a³ + 8b³)( a³-5b³)

Apply the formula a³ + b³ = (a+b)(a² + b² – ab) in (a³ + 8b³)

Now (a³ + 8b³) = (a³ + (2b)³)

= (a² + 4b² – 2ab)( a³-5b³)

∴ The resolved factors are (a² + 4b² – 2ab)( a³-5b³).

= (x + 1)(x-4)(x + 3)(x-6) + 24

= (x²-4x + x-4)(x²–6x + 3x-18) + 24

= (x²-3x-4)(x²–3x-18) + 24

Let x² – 3x = a

= (a-4)(a-18) + 24

= a²-18a-4a + 72 + 24

= a²-22a + 96

= a²-16a-6a + 96

= a(a-16)-6(a-16)

= (a-6)(a-16)

Putting back the values of a we get,

= (x² – 3x -6)( x² – 3x -16)

= (x² + 9x + x + 9) (x + 5)² + 63

Apply the formula (a + b)² = a² + b² + 2ab in (x + 5)²

= (x² + 10x + 9) (x² + 10x + 25) + 63

Let x² + 10x = a

= (a + 9)(a + 25) + 63

= a² + 25a + 9a + 225 + 63

= a² + 34a + 288

= a² + 18a + 16a + 288

= a(a + 18) + 16(a + 18)

= (a + 16)(a + 18)

Putting back the values of a we get,

= (x² + 10x + 16)( x² + 10x + 18)

= (x² + 8x + 2x + 16)(x² + 10x + 18)

= [x(x + 8) + 2(x + 8)] (x² + 10x + 18)

= (x + 8)(x + 2) (x² + 10x + 18)

x(x + 9)(x + 3)(x + 6) + 56

= (x² + 9x) (x² + 6x + 3x + 18) + 56

= (x² + 9x) (x² + 9x + 18) + 56

Let x² + 9x = a

= a (a + 18) + 56

= a² + 18a + 56

= a² + 14a + 4a + 56

=a(a + 14) + 4(a + 14)

= (a + 4)(a + 14)

Put back the values to get,

= (x² + 9x + 4)( x² + 9x + 14)

= (x² + 9x + 4)( x² + 7x + 2x + 14)

= (x² + 9x + 4)[( x(x + 7) + 2(x + 7)]

= (x² + 9x + 4)(x + 2)(x + 7)

x²-2ax + (a + b)(a-b) … (1)

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q =-2a … (3)

pq =(a + b)(a-b) … (4)

to make the factors of 3 from 4

-2a=-(a-b)-(a + b)

∴ (1) can be written as

x²-(a-b)x-(a + b)x + (a + b)(a-b)

=[{x-(a-b)}{x-(a + b)}]

∴ the resolved factors are [{x-(a-b)}{x-(a + b)}]

x²-bx-(a + 3b)(a + 2b) … (1)

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q =-b … (3)

pq =-(a + 3b)(a + 2b) … (4)

to make the factors of 3 from 4

-b= (a + 2b) – (a + 3b)

∴ (1) can be written as

x² + (a + 2b)x – (a + 3b)x-(a + 3b)(a + 2b)

=[{x + (a + 2b)}{x-(a + 3b)}]

∴ the resolved factors are [{x + a + 2b)}{x- a - 3b)}]

(a + b)²-5(a + b) + 6 … (1)

To resolve it into factors we first resolve 6

6=2×3

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = -5

pq =6

∴ (1) can be written as

(a + b)² – 3(a + b) - 2(a + b) + 6

= (a+b)(a+b-3) -2(a+b-3)

=[{(a + b)-3}{(a + b) -2}]

∴ the resolved factors are [{(a + b)-6}{(a + b) + 1}]

x² + 4abx-(a²-b²)² … (1)

(a²-b²)²=[(a + b)²(a-b)²]²

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 4ab

pq =-(a²-b²)²

we know that (a + b)²=a² + 2ab + b² … (3)

similarly, (a-b)²=a²-2ab + b² … (4)

subtracting 4 from 3

(a + b)²-(a-b)² =a² + 2ab + b²-( a²-2ab + b²)

=4ab

∴ (1) can be written as

x² + (a + b)²x-(a-b)²x-[(a + b)²(a-b)²]²

=[{x + (a + b)²}{x-(a-b)²}]

Apply the formula (a+b)² = a² + b² + 2ab

(a-b)² = a² + b² - 2ab

=[{x + a² + b² + 2ab }{x-( a² + b² - 2ab )}]

=[{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

∴ the resolved factors are [{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

∴ the factors of are

(x³y³)²-9x³y³ + 8 … (1)

To resolve it into factors we first resolve 8

8=2×2×2

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =-9

mn =8

∴ (1) can be written as

(x³y³)²-8x³y³-x³y³ + 8

= (x³y³-8)( x³y³-1)

Apply the formula a³ - b³ = (a - b)(a² + ab + b²) in (x³y³-8) and ( x³y³-1) as:

(x³y³-8) = ((xy)³ – 2³)

= (xy-2) (x²y² + 4 + 2xy)

(x³y³-1) = ((xy)³ – 1³)

= (xy-1) (x²y² + 1 + xy)

So,

(x³y³)²-8x³y³-x³y³ + 8 = (xy-2) (x²y² + 4 + 2xy) (xy-1) (x²y² + 1 + xy)

∴ the resolved factors are (xy-2) (x²y² + 4 + 2xy) (xy-1) (x²y² + 1 + xy)

(a² – a – 72)

= ( a – 9) (a + 8)

Now (2x² – x – 1)

This can be written as:

=2a² + 4a + a + 2

=2a(a + 2) + 1(a + 2)

=(2a + 1)(a + 2)

∴ the factors of 2a² + 5a + 2 are (2a + 1)(a + 2)

3x² + 14x + 8

=3x² + 12x + 2x + 8

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

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

∴ the factors of 3x² + 14x + 8 are (3x + 2)(x + 4)

2m² + 7m + 6

=2m² + 4m + 3m + 6

=2m(m + 2) + 3(m + 2)

=(2m + 3)(m + 2)

∴ the factors of 2m² + 7m + 6 are (2m + 3)(m + 2)

6x²-x-15

=6x²-10x + 9x-15

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

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

∴ the factors of 6x²-x-15 are (2x + 3)(3x-5)

9r² + r-8

=9r² + 9r-8r-8

=9r(r + 1)-8(r + 1)

=(9r-8)(r + 1)

∴ the factors of 9r² + r-8 are (9r-8)(r + 1)

6m²-11mn-10n²

=6m²-15mn + 4mn-10n²

=3m(2m-5n) + 2n(3m-5n)

=(3m + 2n)(2m-5n)

∴ the factors of 6m²-11mn-10n² are (3m + 2n)(2m-5n)

7x² + 48xy-7y²

=7x² + 49xy-xy-7y²

=7x(x + 7y)-y(x + 7y)

=(7x-y)(x + 7y)

∴ the factors of 7x² + 48xy-7y² are (7x-y)(x + 7y)

12 + x-6x²

=-(6x²-x-12)

=-(6x²-9x + 8x-12)

=-{3x(2x-3) + 4(2x-3)}

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

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

∴ the factors of 12 + x-6x² are (3x + 4)(3-2x)

=-(6a²-5a-6)
= - (6a²-9a + 4a-6)

= - [3a(2a-3) + 2(2a-3)]

= - (3a + 2)(2a-3)

= (3a + 2)(3-2a)

∴ The factors of 6 + 5a-6a² is (3a + 2)(3-2a).

6x²-13x + 6

=6x²-9x-4x + 6

=3x(2x-3)-2(2x-3)

=(3x-2)(2x-3)

∴ the factors of 6x²-13x + 6 are (3x-2)(2x-3)

99a²-202ab + 99b²

=99a²-121ab-81ab + 99b²

=11a(9a-11b)-9b(9a-11b)

=(11a-9b)(9a-11b)

∴the factors of 99a²-202ab + 99b² are (11a-9b)(9a-11b)

2a⁶-13a³-24

=2a⁶-16a³ + 3a³-24

=2a³(a³-8) + 3(a³-8)

= (2a³ + 3) (a³-8)

Apply the formula a³ - b³ = (a - b)(a² + ab + b²) in (a³-8),

(a³-8) = (a³-2³)

= (a-2)(a² + 4 + 2a)

So,

2a⁶-13a³-24= (2a³ + 3) (a-2)(a² + 4 + 2a)

∴ the factors of 2a⁶-13a³-24 are (2a³ + 3) (a-2)(a² + 4 + 2a).

8a⁴ + 2a²-45

=8a⁴ + 20a²- 18a²-45

=4a²(2a² + 5)-9(2a² + 5)

=(4a²-9)(2a² + 5)

Apply the formula a² – b² = (a-b) (a+b) in(4a²-9)

(4a²-9)= ((2a)² – 3²)

= (2a-3)(2a+3)

So,

8a⁴ + 2a²-45 = (2a² + 5) (2a-3)(2a+3)

∴ the factors of 8a⁴ + 2a²-45 are (2a² + 5) (2a-3)(2a+3)

6(x-y)²-x + y-15

=6(x-y)²-(x-y)-15

=6(x-y)²-10(x-y) + 9(x-y)-15

=2(x-y){3(x-y)-5} + 3{3(x-y)-5}

=[{2(x-y) + 3}{3(x-y)-5}]

=[2x-2y+3][3x-3y-5]

∴ the factors of 6(x-y)²-x + y-15 are [2x-2y+3][3x-3y-5]

3(a + b)²-2a-2b-8

=3(a + b)²-2(a + b)-8

=3(a + b)²-6(a + b) + 4(a + b)-8

=3(a + b){a + b-2} + 4(a + b-2)

=(3a + 3b + 4)(a + b-2)

∴ the factors of 3(a + b)²-2a-2b-8 are (3a + 3b + 4)(a + b-2)

6(a + b)² + 5(a²-b²)-6(a-b)²

=6(a + b)² + 5(a-b)(a + b)-6(a-b)²

=6(a + b)² + 9(a-b)(a + b)-4(a-b)(a + b)-6(a-b)²

=3(a + b){2a + 2b + 3a-3b}-2(a-b){2a + 2b + 3a-3b}

=3(a + b)(5a-b)-2(a-b)(5a-b)

=(5a-b)(3a + 3b-2a + 3b)

=(5a-b)(a + 5b)

∴ the factors of 6(a + b)² + 5(a²-b²)-6(a-b)² are (5a-b)(a + 5b)

Add and subtract the half of square of coefficient of x.

=(x-1)² – 4

=(x-1)² – (2)²

Apply the formula a² – b² = (a + b)(a-b)

= (x-1-2)(x-1 + 2)

= (x-3)(x + 1)

Hence the factors of x²-2x-3 are (x-3)(x + 1).

Add and subtract the half of square of coefficient of x.

Apply the formula a² – b² = (a + b)(a-b)

= (x + 2)(x + 3)

Hence the factors of x² + 5x + 6 is (x + 2)(x + 3).

make the coefficient of x² as 1.

Add and subtract the half of square of coefficient of x.

Apply the formula a² – b² = (a + b)(a-b)

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

Hence the factors of 3x²-7x-6 are (x-3)(3x + 2).

make the coefficient of a² as 1.

Add and subtract the half of square of coefficient of “a”.

Apply the formula a² – b² = (a + b)(a-b)

= (3a-5)(a + 1)

Hence the factors of 3a²-2a-5 are (3a-5)(a + 1).

ax² + (a² + 1)x + a

=ax² + a²x + x + a

=ax(x + a) + 1(x + a)

=(ax + 1)(x + a)

∴ the factors of ax² + (a² + 1)x + a are (ax + 1)(x + a)

x² + 2ax + (a + b)(a-b)

=x² + (a + b)x + (a-b)x + (a + b)(a-b)

=x(x + a + b) + (a-b)(x + a + b)

=(x + a-b)(x + a + b)

∴ the factors of x² + 2ax + (a + b)(a-b) are (x + a-b)(x + a + b)

ax²-(a² + 1)x + a

=ax²-a²x-x + a

=ax(x-a)-1(x-a)

=(ax-1)(x-a)

∴ the factors of ax²-(a² + 1)x + a are (ax-1)(x-a)

ax² + (a²-1)x-a

=ax² + a²x-x-a

=ax(x + a)-1(x + a)

=(ax-1)(x + a)

∴ the factors of ax² + (a²-1)x-a are (ax-1)(x + a)

ax²-(a²-2)x-2a

=ax²-a²x + 2x-2a

=ax(x-a) + 2(x-a)

=(ax + 2)(x-a)

∴ the factors of ax²-(a²-2)x-2a are (ax + 2)(x-a)

∴ the factors of