Question

Please solve this from question no 10 to 11​

Answer 1

Answer:

Let p(u) =3u³-4u²-12u+16

Factors of 16 are 1,2,4,8,16

On checking we get

P(2)=24-16-24+10

=0

So u-2 is the one factor

Now

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)