Question

please solve this one...


x^3 + 3x^2 + 3x + 1 को निम्नलिखित से भाग देने पर शेषफल ज्ञात कीजिए:

i) x + 1

ii) x - 1/2

iii) x

iv) x + π

v) 5 + 2x

​

Answer 1

Answer:

Step-by-step explanation:

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

1) let ( x + 1) = 0 ⇒ x = -1

Now substituting " x = -1 " ,we get

p ( - 1 ) = ( -1 )^3 + 3 ( -1 )^2 + 3 ( -1) + 1

= - 1 + 3 ( 1 ) - 3 + 1 ( ∵ + × - = - )

= - 1 + 1 + 3 - 3

= 0 .is the answer.

2) If ( x - 1/2 ) = 0 ⇒ x = 1/2.

Now substituting " x = 1/2 " in p ( x ) ,we get

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

= 1 / 8 + 3 ( 1/4) + 3/2 + 1

= 1/8 + 3/4 + 3/2 + 1

= 1 × 1 + 2 × 3 + 4 × 3 + 1 × 8 / 8

= 1 + 6 + 12 + 8 / 8

= 27 / 8.is the answer.

3) same as given p ( x ) is the answer.

4) if ( x + π )=0 ⇒ x = - π

Now substituting " x = -π " in p ( x ),we get

p ( - π ) = ( - π )^3 + 3 ( - π )^2 + 3 ( - π ) + 1

= - π + 3 ( π ) - 3 π + 1

= - π + 3 π - 3 π + 1

= - π + 1 . Is the answer.

5) lf 5 + 2x = 0 ⇒ 2x = - 5 ⇒x = - 5 /2

Now substituting " x = - 5 /2" in p ( x ), we get

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

= - 125 / 8 + 3 ( 25 / 4 ) - 15 /2 + 1

= - 125 / 8 + 75 / 4 - 15 / 2 + 1

= 1 ( - 125 ) + 2 × 75 - 4 × 15 + 1 × 8 / 8

= - 125 + 150 - 60 + 8 / 8

= - 185 + 158 / 8

= - 27 / 8.is the answer.

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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