Answer :
Alright, let's work through the problems step by step:
### Problem 40: [tex]\( f(t) = t^2 (3t^2 - 10t + 7) \)[/tex]
#### a. Finding all real zeros of the polynomial function
1. Factor the polynomial completely:
[tex]\[
f(t) = t^2 (3t^2 - 10t + 7)
\][/tex]
We can set each factor to zero to find the zeros:
[tex]\[
t^2 = 0 \quad \Rightarrow \quad t = 0
\][/tex]
[tex]\[
3t^2 - 10t + 7 = 0 \quad \Rightarrow \quad t = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm \sqrt{16}}{6} = \frac{10 \pm 4}{6}
\][/tex]
This gives us:
[tex]\[
t = \frac{14}{6} = \frac{7}{3} \quad \text{and} \quad t = \frac{6}{6} = 1
\][/tex]
So the zeros are [tex]\( t = 0 \)[/tex], [tex]\( t = \frac{7}{3} \)[/tex], and [tex]\( t = 1 \)[/tex].
#### b. Determine the multiplicity of each zero
- Multiplicity of [tex]\( t = 0 \)[/tex]:
- The factor [tex]\( t \)[/tex] appears squared, so the multiplicity is 2 (even).
- Multiplicity of [tex]\( t = \frac{7}{3} \)[/tex]:
- The factor [tex]\( (3t - 7) \)[/tex] appears once in [tex]\( 3t^2 - 10t + 7 \)[/tex], so the multiplicity is 1 (odd).
- Multiplicity of [tex]\( t = 1 \)[/tex]:
- The factor [tex]\( (t - 1) \)[/tex] appears once in [tex]\( 3t^2 - 10t + 7 \)[/tex], so the multiplicity is 1 (odd).
#### c. Determine the maximum possible number of turning points
- The degree of the polynomial [tex]\( f(t) = t^2 (3t^2 - 10t + 7) \)[/tex] is 4.
- The maximum number of turning points for a polynomial of degree [tex]\( n \)[/tex] is [tex]\( n - 1 \)[/tex].
[tex]\[
\text{Maximum turning points} = 4 - 1 = 3
\][/tex]
---
### Problem 46: [tex]\( f(t) = 2t^4 - 2t^2 - 40 \)[/tex]
#### a. Finding all real zeros of the polynomial function
1. Set the polynomial equal to zero:
[tex]\[
2t^4 - 2t^2 - 40 = 0 \quad \Rightarrow \quad t^4 - t^2 - 20 = 0
\][/tex]
Let [tex]\( u = t^2 \)[/tex], then the equation becomes:
[tex]\[
u^2 - u - 20 = 0
\][/tex]
Solving the quadratic equation:
[tex]\[
u = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2}
\][/tex]
This gives us:
[tex]\[
u = 5 \quad \text{and} \quad u = -4
\][/tex]
Since [tex]\( u = t^2 \)[/tex]:
[tex]\[
t^2 = 5 \quad \Rightarrow \quad t = \pm \sqrt{5}
\][/tex]
[tex]\[
t^2 = -4 \quad \Rightarrow \quad \text{no real solution}
\][/tex]
So the real zeros are [tex]\( t = \sqrt{5} \)[/tex] and [tex]\( t = -\sqrt{5} \)[/tex].
#### b. Determine the multiplicity of each zero
- Multiplicity of [tex]\( t = \sqrt{5} \)[/tex]:
- The factor [tex]\( (t - \sqrt{5}) \)[/tex] appears once, so the multiplicity is 1 (odd).
- Multiplicity of [tex]\( t = -\sqrt{5} \)[/tex]:
- The factor [tex]\( (t + \sqrt{5}) \)[/tex] appears once, so the multiplicity is 1 (odd).
#### c. Determine the maximum possible number of turning points
- The degree of the polynomial [tex]\( f(t) = 2t^4 - 2t^2 - 40 \)[/tex] is 4.
- The maximum number of turning points for a polynomial of degree [tex]\( n \)[/tex] is [tex]\( n - 1 \)[/tex].
[tex]\[
\text{Maximum turning points} = 4 - 1 = 3
\][/tex]
### Problem 40: [tex]\( f(t) = t^2 (3t^2 - 10t + 7) \)[/tex]
#### a. Finding all real zeros of the polynomial function
1. Factor the polynomial completely:
[tex]\[
f(t) = t^2 (3t^2 - 10t + 7)
\][/tex]
We can set each factor to zero to find the zeros:
[tex]\[
t^2 = 0 \quad \Rightarrow \quad t = 0
\][/tex]
[tex]\[
3t^2 - 10t + 7 = 0 \quad \Rightarrow \quad t = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm \sqrt{16}}{6} = \frac{10 \pm 4}{6}
\][/tex]
This gives us:
[tex]\[
t = \frac{14}{6} = \frac{7}{3} \quad \text{and} \quad t = \frac{6}{6} = 1
\][/tex]
So the zeros are [tex]\( t = 0 \)[/tex], [tex]\( t = \frac{7}{3} \)[/tex], and [tex]\( t = 1 \)[/tex].
#### b. Determine the multiplicity of each zero
- Multiplicity of [tex]\( t = 0 \)[/tex]:
- The factor [tex]\( t \)[/tex] appears squared, so the multiplicity is 2 (even).
- Multiplicity of [tex]\( t = \frac{7}{3} \)[/tex]:
- The factor [tex]\( (3t - 7) \)[/tex] appears once in [tex]\( 3t^2 - 10t + 7 \)[/tex], so the multiplicity is 1 (odd).
- Multiplicity of [tex]\( t = 1 \)[/tex]:
- The factor [tex]\( (t - 1) \)[/tex] appears once in [tex]\( 3t^2 - 10t + 7 \)[/tex], so the multiplicity is 1 (odd).
#### c. Determine the maximum possible number of turning points
- The degree of the polynomial [tex]\( f(t) = t^2 (3t^2 - 10t + 7) \)[/tex] is 4.
- The maximum number of turning points for a polynomial of degree [tex]\( n \)[/tex] is [tex]\( n - 1 \)[/tex].
[tex]\[
\text{Maximum turning points} = 4 - 1 = 3
\][/tex]
---
### Problem 46: [tex]\( f(t) = 2t^4 - 2t^2 - 40 \)[/tex]
#### a. Finding all real zeros of the polynomial function
1. Set the polynomial equal to zero:
[tex]\[
2t^4 - 2t^2 - 40 = 0 \quad \Rightarrow \quad t^4 - t^2 - 20 = 0
\][/tex]
Let [tex]\( u = t^2 \)[/tex], then the equation becomes:
[tex]\[
u^2 - u - 20 = 0
\][/tex]
Solving the quadratic equation:
[tex]\[
u = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2}
\][/tex]
This gives us:
[tex]\[
u = 5 \quad \text{and} \quad u = -4
\][/tex]
Since [tex]\( u = t^2 \)[/tex]:
[tex]\[
t^2 = 5 \quad \Rightarrow \quad t = \pm \sqrt{5}
\][/tex]
[tex]\[
t^2 = -4 \quad \Rightarrow \quad \text{no real solution}
\][/tex]
So the real zeros are [tex]\( t = \sqrt{5} \)[/tex] and [tex]\( t = -\sqrt{5} \)[/tex].
#### b. Determine the multiplicity of each zero
- Multiplicity of [tex]\( t = \sqrt{5} \)[/tex]:
- The factor [tex]\( (t - \sqrt{5}) \)[/tex] appears once, so the multiplicity is 1 (odd).
- Multiplicity of [tex]\( t = -\sqrt{5} \)[/tex]:
- The factor [tex]\( (t + \sqrt{5}) \)[/tex] appears once, so the multiplicity is 1 (odd).
#### c. Determine the maximum possible number of turning points
- The degree of the polynomial [tex]\( f(t) = 2t^4 - 2t^2 - 40 \)[/tex] is 4.
- The maximum number of turning points for a polynomial of degree [tex]\( n \)[/tex] is [tex]\( n - 1 \)[/tex].
[tex]\[
\text{Maximum turning points} = 4 - 1 = 3
\][/tex]