Question

11. Divide (x+ + x3) by (x + 1).
(x4 + x3) को (x + 1) से भाग दीजिए ।​

Answer 1

Given :

• $\sf {p(x) =x^4+x^3 }$
• $\sf {g(x)=x+1}$

To Find :

• Remainder

Solution :-

By Remainder theorem :

Let g (x) = 0.

$:\implies{\sf{x+1=0}}\\ \\ :\implies{\sf{x=0-1}}\\ \\ :\implies {\tt{x=-1}}$

$\rule {71}{1}$

$: \implies {\sf {p (x)=x^4+x^3}}\\ \\ : \implies{\sf {p (-1)=(-1)^4+(-1)^3}}\\ \\ : \implies {\sf {p (-1)=1+(-1) }}\\ \\ : \implies {\sf {p (-1)=1-1}}\\ \\ :\implies{\boxed{\tt{p (-1)=0}}}$

$\underline{\bf {\therefore{Remainder =0.}}}$

$\rule {181}{2}$

दिया गया है:

• $\sf {p(x) =x^4+x^3 }$
• $\sf {g(x)=x+1}$

निकालना है :

• शेषफल ।

हल :-

शेषफल प्रमेय द्वारा :

मान लीजिए कि g (x) = 0.

$:\implies{\sf{x+1=0}}\\ \\ :\implies{\sf{x=0-1}}\\ \\ :\implies {\tt{x=-1}}$

$\rule {71}{1}$

$: \implies {\sf {p (x)=x^4+x^3}}\\ \\ : \implies{\sf {p (-1)= (-1)^4+(-1)^3}}\\ \\ :\implies {\sf {p (-1)=1+(-1) }}\\ \\ : \implies {\sf {p (-1)=1-1}}\\ \\ :\implies{\boxed{\tt{p(-1)=0}}}$

∴ शेषफल = 0.

$\rule {181}{2}$

Answer 2

Answer:

$\huge \bf \red{GIVEN :- }$

f(x) = x⁴ - x³ - 2x² + x + 1.

$\huge \sf \gray{TO \: FIND :- </p><p>}$

The Remainder.

$\huge \tt \green{SOLUTION :- }$

◉ Let x - 1 = 0

➣ x = 1

BY REMAINDER THEOREM ,

➣ f(x) = x⁴ - x³ - 2x² + x + 1.

➣ f(1) = (1)⁴ - (1)³ - 2 × (1)² + 1 + 1

➣ 1 - 1 - 2 × 1 + 1 + 1

➣ 1 - 1 - 2 + 2

➣ 1 + 2 - 2 -1

➣ 3 - 3

➣ 0

Hence the Remainder is 0

$\sf \pink{ADDITIONAL \: INFORMATION :-}$

◉ Remember theorem :- If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial x - a, Then the Remainder is p(a)

◉ Factor theorem :- x - a is a factor of polynomial p(x) , If p(a) = 0. Also if x ' a is a factor of p(x), Then p(a) = 0.