1. Find the remainder when ` \x^3+3x^2+3x+1\ ` is divided by
(i) ` \x+1\ `
(ii) ` \x-\frac{1}{2}\ `
(iii) ` \x\ `
(iv) ` \x+π\ `
(v) ` \5+2x\ `
solution :
(i) ` \x+1\ `
Here, ` \p(x)=x^3+3x^2+3x+1\ `
The zero of ` \x+1\ ` is
` \x+1=0\ `
` \x=-1\ `
So, ` \P(−1)=(-1)^3+3(-1)^2+3(-1)+1\ `
` \=-1+3+(-3)+1\ `
` \=0\ `
So, by the Remainder Theorem, 0 is the remainder when ` \x^3+3x^2+3x+1\ ` is divided by ` \x+1\ `.
(ii) ` \x-\frac{1}{2}\ `
Here, ` \p(x)=x^3+3x^2+3x+1\ `
The zero of ` \x-\frac{1}{2}\ ` is
` \x-\frac{1}{2}=0\ `
` \x=\frac{1}{2}\ `
So, ` \P(\frac{1}{2})=(\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2})+1\ `
` \=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1\ `
` \=\frac{1+6+12+8}{8}\ `
` \=\frac{27}{8}\ `
So, by the Remainder Theorem, ` \frac{27}{8}\ ` is the remainder when ` \x^3+3x^2+3x+1\ ` is divided by ` \x-\frac{1}{2}\ `
(iii) ` \x\ `
Here, ` \p(x)=x^3+3x^2+3x+1\ `
The zero of ` \x\ ` is
` \x=0\ `
So, ` \P(0)=0^3+3×0^2+3×0+1\ `
` \=0+0+0+1\ `
` \= 1\ `
So, by the Remainder Theorem, 1 is the remainder when ` \x^3+3x^2+3x+1\ ` is divided by ` \x\ `.
(iv) ` \x+π\ `
Here, ` \p(x)=x^3+3x^2+3x+1\ `
The zero of ` \x+pi\ ` is
` \x+pi=0\ `
` \x=-pi\ `
So, ` \P(-pi)=-pi^3+3(-pi)^2+3(-pi)+1\ `
` \=\-pi^3+3pi^2-3pi+1\ `
So, by the Remainder Theorem, ` \=\-pi^3+3pi^2-3pi+1\ ` is the remainder when ` \x^3+3x^2+3x+1\ ` is divided by ` \x+pi\ `.
(v) ` \5+2x\ `
Here, ` \p(x)=x^3+3x^2+3x+1\ `
The zero of ` \5+2x\ ` is
` \5+2x=0\ `
` \2x=-5\ `
` \x=\frac{-5}{2}\ `
So, ` \P(\frac{-5}{2})=(\frac{-5}{2})^3+3(\frac{-5}{2})^2+3(\frac{-5}{2})+1\ `
` \=\(\frac{-125}{8})+(\frac{75}{8})+(\frac{-15}{8})+1\ `
` \=\frac{-27}{8}\ `
So, by the Remainder Theorem, ` \frac{-27}{8}\ ` is the remainder when ` \x^3+3x^2+3x+1\ ` is divided by ` \5 +2x\ `.
2. Find the remainder when ` \x^3-ax^2+6x-a\ ` is divided by ` \x-a\ `.
solution :
Let, ` \p(x)=x^3-ax^2+6x-a\ `
The zero of ` \x-a\ ` is
` \x-a=0\ `
` \x=a\ `
So, ` \P(a)=a^3-a×a^2+6a-a\ `
` \=a^3-a^3+5a\ `
` \=5a\ `
So, by the Remainder Theorem, ` \5a\ ` is the remainder when ` \x^3-ax^2+6x-a\ ` is divided by ` \x-a\ `.
3. Check whether ` \7+3x\ ` is a factor of ` \3x^3+7x\ `.
solution :
The zero of ` \7+3x\ ` is
` \7+3x=0\ `
` \3x=-7\ `
` \x=\frac{-7}{3}\ `
So,
` \p(\frac{-7}{3})=3(\frac{-7}{3})^3+7(\frac{-7}{3})\ `
` \=3(\frac{-343}{27})+(\frac{-49}{3})\ `
` \=(\frac{-343}{9})+(\frac{-49}{3})\ `
` \=(\frac{(-343)+(-147)}{9})\ `
` \=(\frac{-490}{9})\ne0\ `
` \=7+3x\ ` is not a factor of ` \3x^3+7x\ `.
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