Q2 of 52 Page 127

Let’s resolve into factors

x2 + 4abx-(a2-b2)2


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 }]


More from this chapter

All 52 →
2

Let’s resolve into factors

x2-bx-(a + 3b)(a + 2b)


 2 O

Let’s resolve into factors

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


 2 Q

Let’s resolve into factors

x2-


 2 R

Let’s resolve into factors

x6y6-9x3y3 + 8