Answer :
Let's break down and answer the questions step-by-step:
### Question 3: Determine for what [tex]\( d \)[/tex] the polynomial [tex]\( 1-2x \)[/tex] divides the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] where [tex]\( m=2 \)[/tex].
To determine if [tex]\( 1-2x \)[/tex] is a factor of the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] where [tex]\( m=2 \)[/tex], we will use the Factor Theorem. According to the Factor Theorem, [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( P(x) \)[/tex] if and only if [tex]\( P(a) = 0 \)[/tex].
First, rewrite [tex]\( 1-2x \)[/tex] as [tex]\( -2x + 1 \)[/tex]. To use the Factor Theorem, we need to set [tex]\( -2x + 1 = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ -2x + 1 = 0 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
Now substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex]:
[tex]\[ P\left(\frac{1}{2}\right)= 2\left(\frac{1}{2}\right)^4 - 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 \][/tex]
Given [tex]\( m = 2 \)[/tex], substitute [tex]\( m \)[/tex] and compute:
[tex]\[ P\left(\frac{1}{2}\right)= 2\left(\frac{1}{2}\right)^4 - 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ = 2\left(\frac{1}{16}\right) - 2\left(\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ = \frac{2}{16} - \frac{2}{8} + \frac{4}{4} + \frac{2}{2} + 1 \][/tex]
[tex]\[ = \frac{1}{8} - \frac{1}{4} + 1 + 1 + 1 \][/tex]
[tex]\[ = \frac{1}{8} - \frac{2}{8} + 3 \][/tex]
[tex]\[ = -\frac{1}{8} + 3 \][/tex]
[tex]\[ = 2.875 \][/tex]
For [tex]\( x = \frac{1}{2} \)[/tex] to be a root, [tex]\( P(\frac{1}{2}) \)[/tex] must be equal to 0. Since [tex]\( P(\frac{1}{2}) \neq 0 \)[/tex], thus [tex]\( 1-2x \)[/tex] is not a factor of the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] when [tex]\( m = 2 \)[/tex].
### Question 4: Use the factor theorem to determine whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex].
To determine if [tex]\( x + \sqrt{2} \)[/tex] is a factor of the polynomial [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex], we again use the Factor Theorem. According to the Factor Theorem, [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( P(x) \)[/tex] if and only if [tex]\( P(a) = 0 \)[/tex].
Rewrite [tex]\( x + \sqrt{2} \)[/tex] as [tex]\( x - (-\sqrt{2}) \)[/tex]. We need to substitute [tex]\( x = -\sqrt{2} \)[/tex] into the polynomial:
[tex]\[ P(x) = 2\sqrt{2}x^2 + 5x + \sqrt{2} \][/tex]
[tex]\[ P(-\sqrt{2}) = 2\sqrt{2}(-\sqrt{2})^2 + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 2\sqrt{2}(2) - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 4\sqrt{2} \][/tex]
[tex]\[ = 0 \][/tex]
Therefore, [tex]\( x + \sqrt{2} \)[/tex] is a factor of the polynomial [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex].
### Question 3: Determine for what [tex]\( d \)[/tex] the polynomial [tex]\( 1-2x \)[/tex] divides the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] where [tex]\( m=2 \)[/tex].
To determine if [tex]\( 1-2x \)[/tex] is a factor of the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] where [tex]\( m=2 \)[/tex], we will use the Factor Theorem. According to the Factor Theorem, [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( P(x) \)[/tex] if and only if [tex]\( P(a) = 0 \)[/tex].
First, rewrite [tex]\( 1-2x \)[/tex] as [tex]\( -2x + 1 \)[/tex]. To use the Factor Theorem, we need to set [tex]\( -2x + 1 = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ -2x + 1 = 0 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
Now substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex]:
[tex]\[ P\left(\frac{1}{2}\right)= 2\left(\frac{1}{2}\right)^4 - 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 \][/tex]
Given [tex]\( m = 2 \)[/tex], substitute [tex]\( m \)[/tex] and compute:
[tex]\[ P\left(\frac{1}{2}\right)= 2\left(\frac{1}{2}\right)^4 - 2\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ = 2\left(\frac{1}{16}\right) - 2\left(\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ = \frac{2}{16} - \frac{2}{8} + \frac{4}{4} + \frac{2}{2} + 1 \][/tex]
[tex]\[ = \frac{1}{8} - \frac{1}{4} + 1 + 1 + 1 \][/tex]
[tex]\[ = \frac{1}{8} - \frac{2}{8} + 3 \][/tex]
[tex]\[ = -\frac{1}{8} + 3 \][/tex]
[tex]\[ = 2.875 \][/tex]
For [tex]\( x = \frac{1}{2} \)[/tex] to be a root, [tex]\( P(\frac{1}{2}) \)[/tex] must be equal to 0. Since [tex]\( P(\frac{1}{2}) \neq 0 \)[/tex], thus [tex]\( 1-2x \)[/tex] is not a factor of the polynomial [tex]\( 2x^4 - mx^3 + 4x^2 + 2x + 1 \)[/tex] when [tex]\( m = 2 \)[/tex].
### Question 4: Use the factor theorem to determine whether [tex]\( x + \sqrt{2} \)[/tex] is a factor of [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex].
To determine if [tex]\( x + \sqrt{2} \)[/tex] is a factor of the polynomial [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex], we again use the Factor Theorem. According to the Factor Theorem, [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( P(x) \)[/tex] if and only if [tex]\( P(a) = 0 \)[/tex].
Rewrite [tex]\( x + \sqrt{2} \)[/tex] as [tex]\( x - (-\sqrt{2}) \)[/tex]. We need to substitute [tex]\( x = -\sqrt{2} \)[/tex] into the polynomial:
[tex]\[ P(x) = 2\sqrt{2}x^2 + 5x + \sqrt{2} \][/tex]
[tex]\[ P(-\sqrt{2}) = 2\sqrt{2}(-\sqrt{2})^2 + 5(-\sqrt{2}) + \sqrt{2} \][/tex]
[tex]\[ = 2\sqrt{2}(2) - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} \][/tex]
[tex]\[ = 4\sqrt{2} - 4\sqrt{2} \][/tex]
[tex]\[ = 0 \][/tex]
Therefore, [tex]\( x + \sqrt{2} \)[/tex] is a factor of the polynomial [tex]\( 2\sqrt{2}x^2 + 5x + \sqrt{2} \)[/tex].
