Question

Factorise 3z^3 - 4z^2 - 12z + 16

Answer 1

Answer:

$\huge\underline\textsf{Question:- }$

$\boxed{\sf3z ^{2} - 4x {}^{2} - 12x + 16}$

$\huge\underline\textsf{Explantion:- }$

$\large\mathtt{z^2(3z-4)-4(3z-4)}$

$\large\mathtt{(3z-4)(z^2-4)}$

$\large\underline\textsf{ans.(3z-4)(z+2)(z-2) }$

Answer 2

Given,

• $3z^3 - 4z^2 - 12z + 16$ is given.

To find,

• factors of $3z^3 - 4z^2 - 12z + 16$

Solution,

The factors of $3z^3 - 4z^2 - 12z + 16$ are $(3z-4)(z+2)(z-2)$  .

We can simply find the factors of $3z^3 - 4z^2 - 12z + 16$ by taking out the common terms.

                       =  $3z^3 - 4z^2 - 12z + 16$

Taking z² common from first two terms and -4 common from last two terms, we get

                       = $z^2(3z-4)-4(3z-4)$

Now, taking (3z-4) common, we get

                      = $(3z-4)(z^2-4)$

Using (a²-b²) = (a+b)(a-b), we get

                      = $(3z-4)(z+2)(z-2)$

Hence, the factors of $3z^3 - 4z^2 - 12z + 16$ are $(3z-4)(z+2)(z-2)$.