Math, asked by anandita1212, 1 year ago

how to factorise:

5-(3a²-2a)(6-3a²+2a)

Answers

Answered by Anonymous
38

5-(3a2-2a)(6-3a2+2a) 

= 5 – (18a^2 – 9a^4 + 6a^3 – 12a + 6a^3 – 4a^2)

 

= 5 – (– 9a^4 + 6a^3 + 6a^3+ 18a^2 – 4a^2– 12a)

 

= 5 – (– 9a^4 + 12a^3 + 14a^2 – 12a)

 

= 9a^4 - 12a^3 - 14a^2 + 12a + 5

 

= (a+1)(3a+1)(a−1)(3a−5)

 

 \huge \boxed{hope \: it \: helps \: u}

Answered by baljitdhillon1234443
0

Answer:

your ans was wrong

Similar questions