Question

Factor: 3x4y3 – 48y3.


Solution:

3x4y3– 48y3.

= 3y3(x4 – 16).

= 3y3[(x2)2 - 42].

= 3y3(x2 + 4)(x2 - 4).

= 3y3(x2 + 4)(x2 - 22).

= 3y3(x2 + 4)(x + 2)(x -2).
whats the answer?

Answer 1

Answer:

The answer is $\mathbf{3y^3(x^2 + 4)(x + 2)(x -2)}$

Step-by-step explanation:

We need to factor $3x^4y^3 - 48y^3$

Solution:

$3x^4y^3- 48y^3= 3y^3(x^4 – 16)= 3y^3[(x^2)^2 - 4^2]= 3y^3(x^2 + 4)(x^2 - 4)= 3y^3(x^2 + 4)(x^2 - 2^2)= 3y^3(x^2 + 4)(x + 2)(x -2)$

So, the factorisation of $3x^4y^3 - 48y^3$ is $\mathbf{3y^3(x^2 + 4)(x + 2)(x -2)}$

The answer is $\mathbf{3y^3(x^2 + 4)(x + 2)(x -2)}$