Q2 A of 52 Page 127

Let’s resolve into factors

(a + b)2-5a-5b + 6


(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}]


More from this chapter

All 52 →
1 F

I/we factorise the following algebraic expressions:

p2 + 3p – 18


 1

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


 2

Let’s resolve into factors

(x2-2x)2 + 5(x2-2x)-36


 2 C

Let’s resolve into factors

(p2-3q2)2 - 16(p2-3q2) + 63