Question

Factorise : 3u cube - 4u square - 12u + 16

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

p(u) = 3u³ - 4u² - 12u + 16

Here,

(u - 1) or maybe (u + 1) will not factor of p(u)

Then,

Constant (K) = 16

K = |c|

K = 16

Prime factor of 16

= 1 × 2 × 2 × 2 × 2

If :-

u = 2

p(2) = 3(2)³ - 4(2)² - 12(2) + 16

= 24 - 16 - 24 + 16

= 0

(u - 2) is factor of p(u)

Now,

= 3u³ - 4u² - 12u + 16

= 3u³ - 6u² + 2u² - 4u - 8u + 16

= 3u²(u - 2) + 2u(u - 2) - 8(u - 2)

= (u - 2)(3u² + 2u - 8)

= (u - 2)(3u² + 6u - 4u - 8)

= (u - 2){3u(u + 2) - 4(u + 2)}

= (u - 2)(u + 2)(3u - 4)