Discover a wealth of knowledge and get your questions answered on IDNLearn.com. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
