Question

When (x-1) divides x²-x+3 remainder obtained is?​

Answer 1

Answer:

3

Step-by-step explanation:

As per the provided information in the given question, we have been asked to calculate the reminder when (x ― 1) divides (x² ― x + 3).

Let,

⠀⠀⠀⠀⠀★ g(x) = (x ― 1)

⠀⠀⠀⠀⠀★ p(x) = (x² ― x + 3)

Here we'll use remainder theorem to calculate the remainder. Firstly we'll have to calculate the value of x and in order to find that we'll find the zero of g(x).

$\underline{\Large \sf { Zero \; of \; g(x) \: :}}$

$\implies\sf{g(x) = 0}$

$\implies\sf{x - 1= 0}$

$\implies\sf{x= 1}$

Therefore, the value of x is 1. Now, on substituting the value of x in p(x) the result we get will be the remainder.

$\implies\sf{p(x)=x^2 - x +3 }$

$\implies\sf{p(1)=(1)^2 - 1 +3 }$

$\implies\sf{p(1)=1 - 1 +3 }$

$\implies\sf{p(1)= 0+3 }$

$\implies\underline{\boxed{\textbf{\textsf{p(1)= 3 }}}}$

Therefore, the reminder when (x ― 1) divides (x² ― x + 3) is 3.

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Answer 2

$\bold{ANSWER≈}$

Let p(x) be any polynomial of degree greater than or equal to one and let 'a' be any real number.

If a polynomial p(x) is divided by x - a then the remainder is p(a).

Let p(x) = x3 + 3x2 + 3x + 1

(i) The root of x + 1 = 0 is -1

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

= -1 + 3 - 3 + 1

= 0

Hence by the remainder theorem, 0 is the remainder when x3 + 3x2 + 3x + 1 is divided by x + 1. We can also say that x + 1 is a factor of x3 + 3x2 + 3x + 1.

(ii) The root of x - (1/2) = 0 is 1/2.

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

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

= (1 + 6 + 12 + 8)/8 = 27/8

Hence by the remainder theorem, 27 / 8 is the remainder when x3 + 3x2 + 3x + 1 is divided by x.

(iii) The root of x = 0 is 0

p(0) = (0)3 + 3(0)2 + 3(0) +1

= 0 + 0 + 0 + 1

= 1

Hence by the remainder theorem, 1 is the remainder when x3 + 3x2 + 3x + 1 is divided by x.

(iv) The root of x + π = 0 is - π

p(-π) = (-π)3 + 3(-π)2 + 3(-π) + 1

= -π3 + 3π2 - 3π + 1

Hence by the remainder theorem, -π3 + 3π2 - 3π + 1 is the remainder when x3 + 3x2 + 3x +1 is divided by x + π.

(v) Now, the root of 5 + 2x = 0 is -5/2

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

= (-185 + 158) / 8

= -27/8

Hence by remainder theorem, -27/8 is the remainder when x3 + 3x2 + 3x + 1 is divided by 5 + 2x.