Q2 of 52 Page 127

Let’s resolve into factors

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


(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)


More from this chapter

All 52 →
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 A

Let’s resolve into factors

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


 2 C

Let’s resolve into factors

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


 2 D

Let’s resolve into factors

a4 + 4a2-5