Answer :
Sure, let's solve these problems step by step using the remainder theorem.
Remainder Theorem: If a polynomial \( p(y) \) is divided by \( (y - c) \), the remainder is \( p(c) \).
### (i) \( p(y) = 4y^3 - 12y^2 + 5y - 4 \) and \( g(y) = 2y - 1 \)
To use the remainder theorem, we need to find the value of \( y \) that makes \( g(y) = 0 \). So we solve:
[tex]\[ 2y - 1 = 0 \][/tex]
[tex]\[ 2y = 1 \][/tex]
[tex]\[ y = \frac{1}{2} \][/tex]
Now, we substitute \( y = \frac{1}{2} \) into \( p(y) \):
[tex]\[ p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 4 \][/tex]
[tex]\[ = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 5\left(\frac{1}{2}\right) - 4 \][/tex]
[tex]\[ = \frac{4}{8} - \frac{12}{4} + \frac{5}{2} - 4 \][/tex]
[tex]\[ = \frac{1}{2} - 3 + \frac{5}{2} - 4 \][/tex]
[tex]\[ = \frac{1}{2} - 3 + 2.5 - 4 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = 0.5 + 2.5 - 7 \][/tex]
[tex]\[ = 3 - 7 \][/tex]
[tex]\[ = -4 \][/tex]
So, the remainder when \( p(y) \) is divided by \( g(y) \) is \(-4\).
### (ii) \( p(y) = y^3 - 4y^2 - 2y + 6 \) and \( g(y) = 1 - \frac{3}{4} \)
First, simplify \( g(y) \):
[tex]\[ g(y) = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4} \][/tex]
We need to find the root of \( 1 - \frac{3}{4} = 0 \):
[tex]\[ 1 - \frac{3}{4} = 0 \][/tex]
[tex]\[ y = \frac{1}{4} \][/tex]
Now, we substitute \( y = \frac{1}{4} \) into \( p(y) \):
[tex]\[ p\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^3 - 4\left(\frac{1}{4}\right)^2 - 2\left(\frac{1}{4}\right) + 6 \][/tex]
[tex]\[ = \frac{1}{64} - 4\left(\frac{1}{16}\right) - 2\left(\frac{1}{4}\right) + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{4}{16} - \frac{2}{4} + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{1}{4} - \frac{1}{2} + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{16}{64} - \frac{32}{64} + \frac{384}{64} \][/tex]
[tex]\[ = \frac{1 - 16 - 32 + 384}{64} \][/tex]
[tex]\[ = \frac{337}{64} \][/tex]
So, the remainder when \( p(y) \) is divided by \( g(y) \) is \( \frac{337}{64} \).
To summarize:
(i) The remainder when \( 4y^3 - 12y^2 + 5y - 4 \) is divided by \( 2y - 1 \) is \(-4\).
(ii) The remainder when [tex]\( y^3 - 4y^2 - 2y + 6 \)[/tex] is divided by [tex]\( 1 - \frac{3}{4} \)[/tex] is [tex]\( \frac{337}{64} \)[/tex].
Remainder Theorem: If a polynomial \( p(y) \) is divided by \( (y - c) \), the remainder is \( p(c) \).
### (i) \( p(y) = 4y^3 - 12y^2 + 5y - 4 \) and \( g(y) = 2y - 1 \)
To use the remainder theorem, we need to find the value of \( y \) that makes \( g(y) = 0 \). So we solve:
[tex]\[ 2y - 1 = 0 \][/tex]
[tex]\[ 2y = 1 \][/tex]
[tex]\[ y = \frac{1}{2} \][/tex]
Now, we substitute \( y = \frac{1}{2} \) into \( p(y) \):
[tex]\[ p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 4 \][/tex]
[tex]\[ = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 5\left(\frac{1}{2}\right) - 4 \][/tex]
[tex]\[ = \frac{4}{8} - \frac{12}{4} + \frac{5}{2} - 4 \][/tex]
[tex]\[ = \frac{1}{2} - 3 + \frac{5}{2} - 4 \][/tex]
[tex]\[ = \frac{1}{2} - 3 + 2.5 - 4 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = \frac{1}{2} + 2.5 - 7 \][/tex]
[tex]\[ = 0.5 + 2.5 - 7 \][/tex]
[tex]\[ = 3 - 7 \][/tex]
[tex]\[ = -4 \][/tex]
So, the remainder when \( p(y) \) is divided by \( g(y) \) is \(-4\).
### (ii) \( p(y) = y^3 - 4y^2 - 2y + 6 \) and \( g(y) = 1 - \frac{3}{4} \)
First, simplify \( g(y) \):
[tex]\[ g(y) = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4} \][/tex]
We need to find the root of \( 1 - \frac{3}{4} = 0 \):
[tex]\[ 1 - \frac{3}{4} = 0 \][/tex]
[tex]\[ y = \frac{1}{4} \][/tex]
Now, we substitute \( y = \frac{1}{4} \) into \( p(y) \):
[tex]\[ p\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^3 - 4\left(\frac{1}{4}\right)^2 - 2\left(\frac{1}{4}\right) + 6 \][/tex]
[tex]\[ = \frac{1}{64} - 4\left(\frac{1}{16}\right) - 2\left(\frac{1}{4}\right) + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{4}{16} - \frac{2}{4} + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{1}{4} - \frac{1}{2} + 6 \][/tex]
[tex]\[ = \frac{1}{64} - \frac{16}{64} - \frac{32}{64} + \frac{384}{64} \][/tex]
[tex]\[ = \frac{1 - 16 - 32 + 384}{64} \][/tex]
[tex]\[ = \frac{337}{64} \][/tex]
So, the remainder when \( p(y) \) is divided by \( g(y) \) is \( \frac{337}{64} \).
To summarize:
(i) The remainder when \( 4y^3 - 12y^2 + 5y - 4 \) is divided by \( 2y - 1 \) is \(-4\).
(ii) The remainder when [tex]\( y^3 - 4y^2 - 2y + 6 \)[/tex] is divided by [tex]\( 1 - \frac{3}{4} \)[/tex] is [tex]\( \frac{337}{64} \)[/tex].
