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2. Polynomials
hard
$x^{3}+3 x^{2}+3 x+1$ को निम्नलिखित से भाग देने पर शेषफल ज्ञात कीजिए
$x+\pi$
A
$-\pi^{3}+3 \pi^{2}-3 \pi+1$
B
$\pi^{3}-3 \pi^{2}-3 \pi-1$
C
$-\pi^{3}+3 \pi^{2}+3 \pi-1$
D
$\pi^{3}-3 \pi^{2}+3 \pi-1$
Solution
We have $p ( x )= x ^{3}+3 x ^{2}+3 x +1$ and zero of $x +\pi=(-\pi)$
$[\because x+\pi=0 \Rightarrow x=-\pi]$
$\therefore$ $p (-\pi)=(-5)^{3}+3(-\pi)^{2}+3(-\pi)+1$
$=-\pi^{3}+3\left(\pi^{2}\right)+(-3 \pi)+1=-\pi^{3}+3 \pi^{2}-3 \pi+1$
Thus, the required remainder is $-\pi^{3}+3 \pi^{2}-3 \pi+1$.
Standard 9
Mathematics
