Discover a world of knowledge and get your questions answered at IDNLearn.com. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.
Sagot :
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]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
