Physics, asked by Hitlerdidi02, 1 month ago

find the reminder when x²+3x²+3x+1 is divided by

________
1) x+1
2)x-¹/2
3)x
4(x+π
5)5+2x

_____RULES...
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Answers

Answered by singhsaarthak9
0

Answer:

(i) = 0

(ii) = 27/8

(iii) = 1

(iv) = -\pi ^3+3\pi ^2-3\pi +1−π

3

+3π

2

−3π+1

(v) = -27/8

Step-by-step explanation:

\begin{gathered}(i)x+1=0\\x=-1\\\end{gathered}

(i)x+1=0

x=−1

put x = -1 in equation (i)

\begin{gathered}(-1)^3+3\times(-1)^2+3\times(-1)+1\\-1+3-3+1\\=0\end{gathered}

(−1)

3

+3×(−1)

2

+3×(−1)+1

−1+3−3+1

=0

\begin{gathered}(ii)x-\frac{1}{2}=0\\x=\frac{1}{2}\\\end{gathered}

(ii)x−

2

1

=0

x=

2

1

put x=1/2 in equation (i)

\begin{gathered}(\frac{1}{2}) ^3+3\times(\frac{1}{2})^2+3\times(\frac{1}{2})+1\\\frac{1}{8}+3\times\frac{1}{4}+\frac{3}{2}+1\\\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+\frac{1}{1}\\\frac{1+6+12+8}{8}=\frac{27}{8}\end{gathered}

(

2

1

)

3

+3×(

2

1

)

2

+3×(

2

1

)+1

8

1

+3×

4

1

+

2

3

+1

8

1

+

4

3

+

2

3

+

1

1

8

1+6+12+8

=

8

27

\begin{gathered}(iii)x=0\\\end{gathered}

(iii)x=0

put x=0 in equation (i)

\begin{gathered}0^3+3\times0^2+3\times0+1\\0+0+0+1\\=1\end{gathered}

0

3

+3×0

2

+3×0+1

0+0+0+1

=1

\begin{gathered}(iv)x+\pi =0\\x=-\pi\end{gathered}

(iv)x+π=0

x=−π

put x=-\pix=−π in equation (i)

\begin{gathered}(-\pi)^3+3\times(-\pi)^2+3\times(-\pi)+1\\-\pi^3+3\pi^2-3\pi+1\\\end{gathered}

(−π)

3

+3×(−π)

2

+3×(−π)+1

−π

3

+3π

2

−3π+1

\begin{gathered}(v)5+2x=0\\2x=-5\\x=\frac{-5}{2}\\\end{gathered}

(v)5+2x=0

2x=−5

x=

2

−5

put x=\frac{-5}{2}x=

2

−5

in equation-(i)

\begin{gathered}(\frac{-5}{2})^33\times(\frac{-5}{2})^2+3(\frac{-5}{2})+1\\=\frac{-27}{8}\end{gathered}

(

2

−5

)

3

3×(

2

−5

)

2

+3(

2

−5

)+1

=

8

−27

Answered by itsrishab31
0

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Correct Answer - { 3 } x

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