The number of cubic polynomials P(x) satisfyi | vedclass

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Question

(a)

Let the equation of a cubic polynomial

$P(x)=a x^3+b x^2+c x+d$

Now,

$P(1)=a+b+c+d=2 \ldots (i)$

$P(2)=8 a+4 b+2 c+d=4 \ldots (ii)$

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Vedclass · Answer

(a)

Let the equation of a cubic polynomial

$P(x)=a x^3+b x^2+c x+d$

Now,

$P(1)=a+b+c+d=2 \ldots (i)$

$P(2)=8 a+4 b+2 c+d=4 \ldots (ii)$

$P(3)=27 a+9 b+3 c+d=6 \ldots (iii)$

$P(4)=64 a+16 b+4 c+d=8 \ldots$ (iv)

From Eqs. $(i)$ and $(ii),$ we get

$7 a+3 b+c=2$

From Eqs.$(ii)$ and $(iii)$, we get

$19 a+5 b+c=2$

From Eqs. $(iii)$ and $(iv)$, we get

$37 a+7 b+c=2$

Now, from Eqs.$(v)$ and $(vi)$, we get

$12 a+2 b=0$

and from Eqs.$(vi)$ and $(vii)$, we get

$18 a+2 b=0$

From Eqs.$(viii)$ and $(ix)$, we get

$a=0 	ext { and } b=0 	ext {, }$

$c=2 	ext { and } d=0 	ext {. }$

So, $P(x)=2 x$

$	herefore$ no cubic polynomial is possible.

The number of cubic polynomials $P(x)$ satisfying $P(1)=2, P(2)=4, P(3)=6, P(4)=8$ is

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