Question

find the remainder when x^3+3x^2+3x+1 is divided by (i) x+1 (ii) x-1/2 (iii) x (iv) x+π (v) 5+2x​

Answer 1

Answer:

Let `p(x)=x^(3)+3x^(2)+3x+1`

(i) The zero of x+1 is -1. `" " (because x+1=0 implies x=-1)`

So, `p(-1)=(-1)^(3)+3(-1)^(2)+3(-1)+1`

`implies p(-1)=-01+3-3+1`

`implies p(-1)=0`

`therefore` Required remainder =0

(ii) The zero of `x-(1)/(2) " is" (1)/(2)`

So, `p((1)/(2))=((1)/(2))^(3)+3((1)/(2))^(2)+3((1)/(2))+1`

`=(1)/(8)+(3)/(4)+(3)/(2)+1=(1+6+12+88)/(8)`

`implies p((1)/(2))=((27)/(8))`

`therefore` Required remainder `=(27)/(8)`

(iii) The zero of x is 0.

So, `p(0)=(0)^(3)+3(0)+1=0+0+0+1`

`implies p(-pi)=-pi^(3)+3pi^(2)-3pi+1`

`therefore` Required remainder `=-pi^(3)+3pi^(2)-3pi+1`

(v) The zero of 5+2x is `x=-(5)/(2)`.

So, `p(-(5)/(2))=(-(5)/(2))^(3)+3(-(5)/(2))^(2)+3(-(5)/(2))+1`

`=-(125)/(8)+(75)/(4)-(15)/(2)+1=(-125+150-60+8)/(8)=-(27)/(8)`

Answer 2

Step-by-step explanation:

हल :

माना p(x) = x³ + 3x² + 3x +1

(i) (x + 1) का शून्यक (-1) है।

[∵ x + 1 = 0 , x = - 1]

∴ p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1

⇒ p(-1) = -1 + 3 - 3 + 1

⇒ p(-1) = 2 - 2 = 0

⇒ p(-1) = 0

∴ अभीष्ट शेषफल = 0 (शेषफल प्रमेय से)

(ii) (x - 1/2) का शून्यक (1/2) है।

[∵ x - 1/2 = 0 , x = 1/2]

∴ p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1

⇒ p(1/2) = 1/8 + ¾ + 3/2 + 1

⇒ p(1/2) = (1 + 6 + 12 + 8)/8

⇒ p(1/2) = 27/8

∴ अभीष्ट शेषफल = 27/8 (शेषफल प्रमेय से)

(iii) x का शून्यक (0) है।

∴ p(0) = (0)³ + 3(0)² + 3(0) + 1

⇒ p(0) = 0 + 0 + 0 + 1

⇒ p(0) = 1

∴ अभीष्ट शेषफल = 1 (शेषफल प्रमेय से)

(iv) (x + π) का शून्यक (-π) है।

[∵ x + π = 0 , x = - π]

∴ p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1

⇒ p(-π) = -π³ + 3π² - 3π + 1

∴ अभीष्ट शेषफल = -π³ + 3π² - 3π + 1 (शेषफल प्रमेय से)

(v) (5 + 2x) का शून्यक (-5/2) है।

[∵ 2x + 5 = 0 , x = - 5/2]

∴ p(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1

⇒ p(-5/2) = -125/8 + 75/4 - 15/2 + 1

⇒ p(-5/2) = (-125 + 150 - 60 + 8)/8

⇒ p(-5/2) = -27/8

∴ अभीष्ट शेषफल = -27/8 (शेषफल प्रमेय से)

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