factorise:3u3-4u2-12u+16
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Answered by
92
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)
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)
Answered by
4
Step-by-step explanation:
use hidden trail method instead of alogrithim only 2steps full marks u will get
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