Comparing the following algebraic expression with the identity x² + (p + q) x + pq= (x + p)(x + q), Let’s write the values of p and q and factorize them

(i) x²–40x–129 … (1)

(ii) m² + 19m + 60 … (1)

(iii) x²-x-6 … (1)

(iv) (a + b)² – 4(a + b) – 12 … (1)

(v) (x-y)² – x + y – 2

(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}].