Find the third degree polynomial which vanishes when x =−1 and x = 2, which has a value 8 when x =0 and leaves a remainder ((16)/3) when divided by 3x + 2.


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Question Number 67298 by Rio Michael last updated on 25/Aug/19

$F i n d t h e t h i r d d e g r e e p o l y n o m i a l w h i c h v a n i s h e s w h e n$ $x = - 1 a n d x = 2, w h i c h h a s a v a l u e 8 w h e n x = 0 a n d l e a v e s a r e m a i n d e r \frac{16}{3} w h e n$ $d i v i d e d b y 3 x + 2 .$
Commented by Prithwish sen last updated on 25/Aug/19

$let the polynomial be (x + 1) (x - 2) (ax - b) = 0$ $putting x = 0 we get b = 4$ $putting x = - \frac{2}{3} we get a = 3$ $∴ the polynomial is (x + 1) (x - 2) (3 x - 4) = 0$ $i . e 3 x^{3} - 7 x^{2} - 2 x + 8 = 0 please check .$
Commented by Rio Michael last updated on 25/Aug/19

$t h a n k s, p e r f e c t$ $h e l p m e s i r w i t h t h e o n e a b o v e$
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