Answer :
Alright, let's go through each part of the problem step-by-step. We are given the polynomial [tex]\( P(x) = x^3 + 3x^2 - 3x \)[/tex] and we need to find the remainders when this polynomial is divided by the given divisors.
### (i) Division by [tex]\( x + 1 \)[/tex]
To find the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x + 1 \)[/tex], we can use the Remainder Theorem. According to the Remainder Theorem, the remainder of the division of a polynomial [tex]\( P(x) \)[/tex] by a linear divisor [tex]\( x - a \)[/tex] is [tex]\( P(a) \)[/tex].
Here, the divisor is [tex]\( x + 1 \)[/tex], which can be rewritten as [tex]\( x - (-1) \)[/tex]. So, we need to evaluate [tex]\( P(-1) \)[/tex]:
[tex]\[ P(-1) = (-1)^3 + 3(-1)^2 - 3(-1) = -1 + 3 + 3 = 5 \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x + 1 \)[/tex] is [tex]\( 5 \)[/tex].
### (ii) Division by [tex]\( x - \frac{1}{2} \)[/tex]
Next, we need the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - \frac{1}{2} \)[/tex]. Again, using the Remainder Theorem, we need to evaluate [tex]\( P\left(\frac{1}{2}\right) \)[/tex]:
[tex]\[ P\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) = \frac{1}{8} + 3\left(\frac{1}{4}\right) - \frac{3}{2} = \frac{1}{8} + \frac{3}{4} - \frac{3}{2} \][/tex]
First, let's convert all terms to have a common denominator of 8:
[tex]\[ P\left(\frac{1}{2}\right) = \frac{1}{8} + \frac{6}{8} - \frac{12}{8} = \frac{1 + 6 - 12}{8} = \frac{-5}{8} \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - \frac{1}{2} \)[/tex] is [tex]\( -\frac{5}{8} \)[/tex].
### (iii) Division by [tex]\( x \)[/tex]
For the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x \)[/tex], we can again use the Remainder Theorem. Here, the divisor is [tex]\( x \)[/tex], which can be written as [tex]\( x - 0 \)[/tex]. We need to evaluate [tex]\( P(0) \)[/tex]:
[tex]\[ P(0) = 0^3 + 3(0)^2 - 3(0) = 0 \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].
### (iv) Division by [tex]\( 5 + 2x \)[/tex]
Finally, to find the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( 5 + 2x \)[/tex], we need to rewrite the divisor in the form [tex]\( x - a \)[/tex]. Therefore, we consider [tex]\( x - \left(-\frac{5}{2}\right) \)[/tex]. We need to evaluate [tex]\( P\left(-\frac{5}{2}\right) \)[/tex]:
[tex]\[ P\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^3 + 3\left(-\frac{5}{2}\right)^2 - 3\left(-\frac{5}{2}\right) \][/tex]
Calculate each term step by step:
[tex]\[ \left(-\frac{5}{2}\right)^3 = -\frac{125}{8} \][/tex]
[tex]\[ 3\left(-\frac{5}{2}\right)^2 = 3 \cdot \frac{25}{4} = \frac{75}{4} = \frac{150}{8} \][/tex]
[tex]\[ -3\left(-\frac{5}{2}\right) = \frac{15}{2} = \frac{60}{8} \][/tex]
Adding these together:
[tex]\[ P\left(-\frac{5}{2}\right) = -\frac{125}{8} + \frac{150}{8} + \frac{60}{8} = \frac{-125 + 150 + 60}{8} = \frac{85}{8} \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( 5 + 2x \)[/tex] is [tex]\( \frac{85}{8} \)[/tex].
So, the remainders for each divisor are:
(i) [tex]\( x + 1 \)[/tex]: [tex]\( 5 \)[/tex]
(ii) [tex]\( x - \frac{1}{2} \)[/tex] : [tex]\( -\frac{5}{8} \)[/tex]
(iii) [tex]\( x \)[/tex] : [tex]\( 0 \)[/tex]
(iv) [tex]\( 5 + 2x \)[/tex] : [tex]\( \frac{85}{8} \)[/tex]
### (i) Division by [tex]\( x + 1 \)[/tex]
To find the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x + 1 \)[/tex], we can use the Remainder Theorem. According to the Remainder Theorem, the remainder of the division of a polynomial [tex]\( P(x) \)[/tex] by a linear divisor [tex]\( x - a \)[/tex] is [tex]\( P(a) \)[/tex].
Here, the divisor is [tex]\( x + 1 \)[/tex], which can be rewritten as [tex]\( x - (-1) \)[/tex]. So, we need to evaluate [tex]\( P(-1) \)[/tex]:
[tex]\[ P(-1) = (-1)^3 + 3(-1)^2 - 3(-1) = -1 + 3 + 3 = 5 \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x + 1 \)[/tex] is [tex]\( 5 \)[/tex].
### (ii) Division by [tex]\( x - \frac{1}{2} \)[/tex]
Next, we need the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - \frac{1}{2} \)[/tex]. Again, using the Remainder Theorem, we need to evaluate [tex]\( P\left(\frac{1}{2}\right) \)[/tex]:
[tex]\[ P\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) = \frac{1}{8} + 3\left(\frac{1}{4}\right) - \frac{3}{2} = \frac{1}{8} + \frac{3}{4} - \frac{3}{2} \][/tex]
First, let's convert all terms to have a common denominator of 8:
[tex]\[ P\left(\frac{1}{2}\right) = \frac{1}{8} + \frac{6}{8} - \frac{12}{8} = \frac{1 + 6 - 12}{8} = \frac{-5}{8} \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x - \frac{1}{2} \)[/tex] is [tex]\( -\frac{5}{8} \)[/tex].
### (iii) Division by [tex]\( x \)[/tex]
For the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x \)[/tex], we can again use the Remainder Theorem. Here, the divisor is [tex]\( x \)[/tex], which can be written as [tex]\( x - 0 \)[/tex]. We need to evaluate [tex]\( P(0) \)[/tex]:
[tex]\[ P(0) = 0^3 + 3(0)^2 - 3(0) = 0 \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].
### (iv) Division by [tex]\( 5 + 2x \)[/tex]
Finally, to find the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( 5 + 2x \)[/tex], we need to rewrite the divisor in the form [tex]\( x - a \)[/tex]. Therefore, we consider [tex]\( x - \left(-\frac{5}{2}\right) \)[/tex]. We need to evaluate [tex]\( P\left(-\frac{5}{2}\right) \)[/tex]:
[tex]\[ P\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^3 + 3\left(-\frac{5}{2}\right)^2 - 3\left(-\frac{5}{2}\right) \][/tex]
Calculate each term step by step:
[tex]\[ \left(-\frac{5}{2}\right)^3 = -\frac{125}{8} \][/tex]
[tex]\[ 3\left(-\frac{5}{2}\right)^2 = 3 \cdot \frac{25}{4} = \frac{75}{4} = \frac{150}{8} \][/tex]
[tex]\[ -3\left(-\frac{5}{2}\right) = \frac{15}{2} = \frac{60}{8} \][/tex]
Adding these together:
[tex]\[ P\left(-\frac{5}{2}\right) = -\frac{125}{8} + \frac{150}{8} + \frac{60}{8} = \frac{-125 + 150 + 60}{8} = \frac{85}{8} \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( 5 + 2x \)[/tex] is [tex]\( \frac{85}{8} \)[/tex].
So, the remainders for each divisor are:
(i) [tex]\( x + 1 \)[/tex]: [tex]\( 5 \)[/tex]
(ii) [tex]\( x - \frac{1}{2} \)[/tex] : [tex]\( -\frac{5}{8} \)[/tex]
(iii) [tex]\( x \)[/tex] : [tex]\( 0 \)[/tex]
(iv) [tex]\( 5 + 2x \)[/tex] : [tex]\( \frac{85}{8} \)[/tex]
