Answer :
Let's tackle each problem one by one with a detailed, step-by-step approach.
### 1. Whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex]
To determine whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex], we can use the factor theorem. According to the factor theorem, if [tex]\( x + \sqrt{2} \)[/tex] is a factor, then [tex]\( f(-\sqrt{2}) \)[/tex] should be equal to zero, where [tex]\( f(x) = 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex].
Evaluate [tex]\( f(-\sqrt{2}) \)[/tex]:
[tex]\[ f(x) = 2\sqrt{2} x^2 + 5x + \sqrt{2} \][/tex]
Substitute [tex]\( x = -\sqrt{2} \)[/tex]:
[tex]\[ f(-\sqrt{2}) = 2\sqrt{2}(-\sqrt{2})^2 + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 2\sqrt{2}(2) + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\( f(-\sqrt{2}) = 0 \)[/tex], [tex]\( x + \sqrt{2} \)[/tex] is indeed a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex].
### 2. Prove that [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is exactly divisible by [tex]\( x^2 - 3x + 2 \)[/tex] without actual division
The roots of [tex]\( x^2 - 3x + 2 \)[/tex] are found by solving:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
So, the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]. If [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is divisible by [tex]\( x^2 - 3x + 2 \)[/tex], then both [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex] should be the roots of [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex].
Evaluate the polynomial at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2(1)^4 - 6(1)^3 + 3(1)^2 + 3(1) - 2 \][/tex]
[tex]\[ = 2 - 6 + 3 + 3 - 2 \][/tex]
[tex]\[ = 0 \][/tex]
Evaluate the polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2(2)^4 - 6(2)^3 + 3(2)^2 + 3(2) - 2 \][/tex]
[tex]\[ = 2(16) - 6(8) + 3(4) + 3(2) - 2 \][/tex]
[tex]\[ = 32 - 48 + 12 + 6 - 2 \][/tex]
[tex]\[ = 0 \][/tex]
Both evaluations yield 0, thus proving that [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is exactly divisible by [tex]\( x^2 - 3x + 2 \)[/tex].
### 3. Find the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] if [tex]\( x-2 \)[/tex] and [tex]\( x+1 \)[/tex] are the factors of [tex]\( x^3 + mx^2 - nx + 4 \)[/tex]
Using the factor theorem:
- For [tex]\( x-2 \)[/tex] to be a factor, [tex]\( f(2) = 0 \)[/tex]
- For [tex]\( x+1 \)[/tex] to be a factor, [tex]\( f(-1) = 0 \)[/tex]
Evaluate [tex]\( f(2) \)[/tex]:
[tex]\[ 2^3 + m(2)^2 - n(2) + 4 = 0 \][/tex]
[tex]\[ 8 + 4m - 2n + 4 = 0 \][/tex]
[tex]\[ 12 + 4m - 2n = 0 \quad \text{(1)} \][/tex]
Evaluate [tex]\( f(-1) \)[/tex]:
[tex]\[ (-1)^3 + m(-1)^2 - n(-1) + 4 = 0 \][/tex]
[tex]\[ -1 + m + n + 4 = 0 \][/tex]
[tex]\[ m + n + 3 = 0 \quad \text{(2)} \][/tex]
Solving equations (1) and (2):
From (2):
[tex]\[ m + n = -3 \quad (2) \][/tex]
Substitute [tex]\( n = -3 - m \)[/tex] into (1):
[tex]\[ 12 + 4m - 2(-3 - m) = 0 \][/tex]
[tex]\[ 12 + 4m + 6 + 2m = 0 \][/tex]
[tex]\[ 18 + 6m = 0 \][/tex]
[tex]\[ 6m = -18 \][/tex]
[tex]\[ m = -3 \][/tex]
Substitute [tex]\( m = -3 \)[/tex] into [tex]\( n = -3 - m \)[/tex]:
[tex]\[ n = -3 - (-3) \][/tex]
[tex]\[ n = 0 \][/tex]
Thus, [tex]\( m = -3 \)[/tex] and [tex]\( n = 0 \)[/tex].
### 4. Find the remainder when [tex]\( p(x) = ax^3 + 4x^2 + 3x - 4 \)[/tex] and [tex]\( q(x) = x^3 - 4x + a \)[/tex] are divided by [tex]\( x-3 \)[/tex]
We need to find the remainders of [tex]\( p(x) \)[/tex] and [tex]\( q(x) \)[/tex] when divided by [tex]\( x-3 \)[/tex] and set them equal.
Evaluate [tex]\( p(3) \)[/tex]:
[tex]\[ p(3) = a(3)^3 + 4(3)^2 + 3(3) - 4 \][/tex]
[tex]\[ = 27a + 36 + 9 - 4 \][/tex]
[tex]\[ = 27a + 41 \][/tex]
Evaluate [tex]\( q(3) \)[/tex]:
[tex]\[ q(3) = (3)^3 - 4(3) + a \][/tex]
[tex]\[ = 27 - 12 + a \][/tex]
[tex]\[ = 15 + a \][/tex]
Set [tex]\( p(3) \)[/tex] equal to [tex]\( q(3) \)[/tex]:
[tex]\[ 27a + 41 = 15 + a \][/tex]
[tex]\[ 27a - a = 15 - 41 \][/tex]
[tex]\[ 26a = -26 \][/tex]
[tex]\[ a = -1 \][/tex]
Therefore, substituting [tex]\( a = -1 \)[/tex]:
The remainder when [tex]\( p(x) \)[/tex] is divided by [tex]\( x-3 \)[/tex]:
[tex]\[ 27(-1) + 41 = -27 + 41 = 14 \][/tex]
[tex]\( \boxed{14} \)[/tex] is the remainder.
### 1. Whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex]
To determine whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex], we can use the factor theorem. According to the factor theorem, if [tex]\( x + \sqrt{2} \)[/tex] is a factor, then [tex]\( f(-\sqrt{2}) \)[/tex] should be equal to zero, where [tex]\( f(x) = 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex].
Evaluate [tex]\( f(-\sqrt{2}) \)[/tex]:
[tex]\[ f(x) = 2\sqrt{2} x^2 + 5x + \sqrt{2} \][/tex]
Substitute [tex]\( x = -\sqrt{2} \)[/tex]:
[tex]\[ f(-\sqrt{2}) = 2\sqrt{2}(-\sqrt{2})^2 + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 2\sqrt{2}(2) + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\( f(-\sqrt{2}) = 0 \)[/tex], [tex]\( x + \sqrt{2} \)[/tex] is indeed a factor of [tex]\( 2\sqrt{2} x^2 + 5x + \sqrt{2} \)[/tex].
### 2. Prove that [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is exactly divisible by [tex]\( x^2 - 3x + 2 \)[/tex] without actual division
The roots of [tex]\( x^2 - 3x + 2 \)[/tex] are found by solving:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
So, the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]. If [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is divisible by [tex]\( x^2 - 3x + 2 \)[/tex], then both [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex] should be the roots of [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex].
Evaluate the polynomial at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2(1)^4 - 6(1)^3 + 3(1)^2 + 3(1) - 2 \][/tex]
[tex]\[ = 2 - 6 + 3 + 3 - 2 \][/tex]
[tex]\[ = 0 \][/tex]
Evaluate the polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2(2)^4 - 6(2)^3 + 3(2)^2 + 3(2) - 2 \][/tex]
[tex]\[ = 2(16) - 6(8) + 3(4) + 3(2) - 2 \][/tex]
[tex]\[ = 32 - 48 + 12 + 6 - 2 \][/tex]
[tex]\[ = 0 \][/tex]
Both evaluations yield 0, thus proving that [tex]\( 2x^4 - 6x^3 + 3x^2 + 3x - 2 \)[/tex] is exactly divisible by [tex]\( x^2 - 3x + 2 \)[/tex].
### 3. Find the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] if [tex]\( x-2 \)[/tex] and [tex]\( x+1 \)[/tex] are the factors of [tex]\( x^3 + mx^2 - nx + 4 \)[/tex]
Using the factor theorem:
- For [tex]\( x-2 \)[/tex] to be a factor, [tex]\( f(2) = 0 \)[/tex]
- For [tex]\( x+1 \)[/tex] to be a factor, [tex]\( f(-1) = 0 \)[/tex]
Evaluate [tex]\( f(2) \)[/tex]:
[tex]\[ 2^3 + m(2)^2 - n(2) + 4 = 0 \][/tex]
[tex]\[ 8 + 4m - 2n + 4 = 0 \][/tex]
[tex]\[ 12 + 4m - 2n = 0 \quad \text{(1)} \][/tex]
Evaluate [tex]\( f(-1) \)[/tex]:
[tex]\[ (-1)^3 + m(-1)^2 - n(-1) + 4 = 0 \][/tex]
[tex]\[ -1 + m + n + 4 = 0 \][/tex]
[tex]\[ m + n + 3 = 0 \quad \text{(2)} \][/tex]
Solving equations (1) and (2):
From (2):
[tex]\[ m + n = -3 \quad (2) \][/tex]
Substitute [tex]\( n = -3 - m \)[/tex] into (1):
[tex]\[ 12 + 4m - 2(-3 - m) = 0 \][/tex]
[tex]\[ 12 + 4m + 6 + 2m = 0 \][/tex]
[tex]\[ 18 + 6m = 0 \][/tex]
[tex]\[ 6m = -18 \][/tex]
[tex]\[ m = -3 \][/tex]
Substitute [tex]\( m = -3 \)[/tex] into [tex]\( n = -3 - m \)[/tex]:
[tex]\[ n = -3 - (-3) \][/tex]
[tex]\[ n = 0 \][/tex]
Thus, [tex]\( m = -3 \)[/tex] and [tex]\( n = 0 \)[/tex].
### 4. Find the remainder when [tex]\( p(x) = ax^3 + 4x^2 + 3x - 4 \)[/tex] and [tex]\( q(x) = x^3 - 4x + a \)[/tex] are divided by [tex]\( x-3 \)[/tex]
We need to find the remainders of [tex]\( p(x) \)[/tex] and [tex]\( q(x) \)[/tex] when divided by [tex]\( x-3 \)[/tex] and set them equal.
Evaluate [tex]\( p(3) \)[/tex]:
[tex]\[ p(3) = a(3)^3 + 4(3)^2 + 3(3) - 4 \][/tex]
[tex]\[ = 27a + 36 + 9 - 4 \][/tex]
[tex]\[ = 27a + 41 \][/tex]
Evaluate [tex]\( q(3) \)[/tex]:
[tex]\[ q(3) = (3)^3 - 4(3) + a \][/tex]
[tex]\[ = 27 - 12 + a \][/tex]
[tex]\[ = 15 + a \][/tex]
Set [tex]\( p(3) \)[/tex] equal to [tex]\( q(3) \)[/tex]:
[tex]\[ 27a + 41 = 15 + a \][/tex]
[tex]\[ 27a - a = 15 - 41 \][/tex]
[tex]\[ 26a = -26 \][/tex]
[tex]\[ a = -1 \][/tex]
Therefore, substituting [tex]\( a = -1 \)[/tex]:
The remainder when [tex]\( p(x) \)[/tex] is divided by [tex]\( x-3 \)[/tex]:
[tex]\[ 27(-1) + 41 = -27 + 41 = 14 \][/tex]
[tex]\( \boxed{14} \)[/tex] is the remainder.
