Question

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

Answer 1

Answer:

p(x) = x³+3x²+3x+1

(i) g(x) = x+1

x+1 = 0

x = -1

substitute in p(x) {according to remainder theorm}

(-1)³+3(-1)²+3(-1)+1

-1+3-3+1

= 0

(ii) g(x) = x-1/2

x-1/2 = 0

x = 1/2

substitute in p(x)

(-1/2)³+3(-1/2)²+3(-1/2)+1

-1/8+3/4-3/2+1

-1+6-12+8/8

-13+14/8

= 1/8

(iii) g(x) = x

no need to substitute because the question already has an X

==> x³+3x²+3x+1

(iv) g(x) = x+π

x = -π

substitute -π in the place of x

(-π)³+3(-π)²+3(-π)+1

-π³+3π²-3π+1

(v) g(x) = 5+2x

5+2x = 0

2x = -5

x = -5/2

substitute in place of x

(-5/2)³+3(-5/2)²+3(-5/2)+1

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

-125+150-60+8/8

-185+158/8

= 27/8

I'm sure they r correct answers because I completed that lesson in 9th class, lesson-polynomials

hope it helps.....

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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