Answer :
Certainly! Let's solve each polynomial function step by step to find all other zeros, given one zero for each polynomial.
### 33. [tex]\( f(x) = x^3 - x^2 - 4x - 6 \)[/tex]
Given zero: [tex]\( 3 \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - 3) \cdot Q(x) \)[/tex] where [tex]\( Q(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - 3) \)[/tex], we can find [tex]\( Q(x) \)[/tex].
3. Solving the quadratic equation [tex]\( Q(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros, found from the calculations, are:
[tex]\[ -1 - i, \quad -1 + i \][/tex]
### 34. [tex]\( f(x) = x^3 + 4x^2 - 5 \)[/tex]
Given zero: [tex]\( 1 \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - 1) \cdot R(x) \)[/tex] where [tex]\( R(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex], we can find [tex]\( R(x) \)[/tex].
3. Solving the quadratic equation [tex]\( R(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -\frac{5}{2} - \frac{\sqrt{5}}{2}, \quad -\frac{5}{2} + \frac{\sqrt{5}}{2} \][/tex]
### 35. [tex]\( f(x) = x^3 - 7x^2 + 17x - 15 \)[/tex]
Given zero: [tex]\( 2 - i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( 2 + i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - (2 - i))(x - (2 + i)) \cdot P(x) \)[/tex] where [tex]\( P(x) \)[/tex] is a linear polynomial.
3. Solving [tex]\( P(x) = 0 \)[/tex] will give the remaining zero.
The calculations lead to these other zeros:
[tex]\[ 3, \quad 2 + i \][/tex]
### 36. [tex]\( f(x) = 4x^3 + 6x^2 - 2x - 1 \)[/tex]
Given zero: [tex]\( \frac{1}{2} \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - \frac{1}{2}) \cdot S(x) \)[/tex] where [tex]\( S(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - \frac{1}{2}) \)[/tex], we can find [tex]\( S(x) \)[/tex].
3. Solving the quadratic equation [tex]\( S(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -1.70710678118655, \quad -0.292893218813452 \][/tex]
### 37. [tex]\( f(x) = x^4 + 5x^2 + 4 \)[/tex]
Given zero: [tex]\( -i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x + i)(x - i) \cdot T(x) \)[/tex] where [tex]\( T(x) \)[/tex] is a quadratic polynomial.
3. Solving [tex]\( T(x) = 0 \)[/tex] will give the remaining zeros.
The calculations result in these other zeros:
[tex]\[ -2i, \quad i, \quad 2i \][/tex]
### 38. [tex]\( f(x) = x^4 + 26x^2 + 25 \)[/tex]
Given zero: [tex]\( i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( -i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - i)(x + i) \cdot U(x) \)[/tex] where [tex]\( U(x) \)[/tex] is a quadratic polynomial.
3. Solving [tex]\( U(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -5i, \quad -i, \quad 5i \][/tex]
And there you have it! All other zeros for each given polynomial function.
### 33. [tex]\( f(x) = x^3 - x^2 - 4x - 6 \)[/tex]
Given zero: [tex]\( 3 \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - 3) \cdot Q(x) \)[/tex] where [tex]\( Q(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - 3) \)[/tex], we can find [tex]\( Q(x) \)[/tex].
3. Solving the quadratic equation [tex]\( Q(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros, found from the calculations, are:
[tex]\[ -1 - i, \quad -1 + i \][/tex]
### 34. [tex]\( f(x) = x^3 + 4x^2 - 5 \)[/tex]
Given zero: [tex]\( 1 \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - 1) \cdot R(x) \)[/tex] where [tex]\( R(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - 1) \)[/tex], we can find [tex]\( R(x) \)[/tex].
3. Solving the quadratic equation [tex]\( R(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -\frac{5}{2} - \frac{\sqrt{5}}{2}, \quad -\frac{5}{2} + \frac{\sqrt{5}}{2} \][/tex]
### 35. [tex]\( f(x) = x^3 - 7x^2 + 17x - 15 \)[/tex]
Given zero: [tex]\( 2 - i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( 2 + i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - (2 - i))(x - (2 + i)) \cdot P(x) \)[/tex] where [tex]\( P(x) \)[/tex] is a linear polynomial.
3. Solving [tex]\( P(x) = 0 \)[/tex] will give the remaining zero.
The calculations lead to these other zeros:
[tex]\[ 3, \quad 2 + i \][/tex]
### 36. [tex]\( f(x) = 4x^3 + 6x^2 - 2x - 1 \)[/tex]
Given zero: [tex]\( \frac{1}{2} \)[/tex]
1. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - \frac{1}{2}) \cdot S(x) \)[/tex] where [tex]\( S(x) \)[/tex] is a quadratic polynomial.
2. Dividing [tex]\( f(x) \)[/tex] by [tex]\( (x - \frac{1}{2}) \)[/tex], we can find [tex]\( S(x) \)[/tex].
3. Solving the quadratic equation [tex]\( S(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -1.70710678118655, \quad -0.292893218813452 \][/tex]
### 37. [tex]\( f(x) = x^4 + 5x^2 + 4 \)[/tex]
Given zero: [tex]\( -i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x + i)(x - i) \cdot T(x) \)[/tex] where [tex]\( T(x) \)[/tex] is a quadratic polynomial.
3. Solving [tex]\( T(x) = 0 \)[/tex] will give the remaining zeros.
The calculations result in these other zeros:
[tex]\[ -2i, \quad i, \quad 2i \][/tex]
### 38. [tex]\( f(x) = x^4 + 26x^2 + 25 \)[/tex]
Given zero: [tex]\( i \)[/tex]
1. Complex roots come in conjugate pairs, so [tex]\( -i \)[/tex] is also a root.
2. We can factor [tex]\( f(x) \)[/tex] as [tex]\( (x - i)(x + i) \cdot U(x) \)[/tex] where [tex]\( U(x) \)[/tex] is a quadratic polynomial.
3. Solving [tex]\( U(x) = 0 \)[/tex] will give the remaining zeros.
The other zeros are:
[tex]\[ -5i, \quad -i, \quad 5i \][/tex]
And there you have it! All other zeros for each given polynomial function.
