Question

Factorise (a^2-4^a)^2-(a^2-4a)-20

Answer 1

Answer:

Step-by-step explanation:

$x = a^{2}-4a\\then, x^{2} -x-20\\= x^{2} -5x+4x-20\\=x(x-5)+4(x-5)\\=(x+4)(x-5)\\hence on putting value of x = a^{2}-4a,\\we get,\\(a^{2} -4a+4)(a^{2}-4a-5)$

this will be your solution after factorization

i hope it helps you.

Answer 2

Answer:

step 1: given

(a^2-4a)^2-(a^2-4a)-20 =0

Let    'a^2-4a'   be x

step 2:  Solve x

x^2-x-20=0

x^2-5x+4x-20=0

x(x-5)+4(x-5)=0

(x-5)(x+4)=0

x=5 or -4

step3:  solve a

Case 1: if x=5

a^2-4a=5

a^2-4a-5=0

(a-5)(a+1)=0

a=5 or -1

case 2:  if x= -4

a^2-4a+4=0

(a-2)^2=0

a-2=0

a=2

∴a= 5,2,-1