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Answer :
We start by solving each equation step by step.
----------------------------------
1. For the equation
[tex]$$
4h^5 - 8h^4 = 0,
$$[/tex]
we first notice that both terms have the common factor [tex]$4h^4$[/tex]. Factoring it out gives
[tex]$$
4h^4 (h - 2) = 0.
$$[/tex]
By the zero-product property, we set each factor equal to zero:
- From [tex]$4h^4 = 0$[/tex], we obtain
[tex]$$
h^4 = 0 \quad \Longrightarrow \quad h = 0.
$$[/tex]
(Note that [tex]$h=0$[/tex] is a solution of multiplicity 4.)
- From [tex]$h - 2 = 0$[/tex], we have
[tex]$$
h = 2.
$$[/tex]
Thus, the solutions are [tex]$h = 0$[/tex] (with multiplicity 4) and [tex]$h = 2$[/tex].
----------------------------------
2. For the equation
[tex]$$
m^2 - 2m - 3 = 0,
$$[/tex]
we look for two numbers that multiply to [tex]$-3$[/tex] and add to [tex]$-2$[/tex]. The factors are [tex]$m - 3$[/tex] and [tex]$m + 1$[/tex], so the equation factors as
[tex]$$
(m - 3)(m + 1) = 0.
$$[/tex]
Setting each factor equal to zero gives:
- From [tex]$m - 3 = 0$[/tex], we get
[tex]$$
m = 3.
$$[/tex]
- From [tex]$m + 1 = 0$[/tex], we get
[tex]$$
m = -1.
$$[/tex]
Thus, the solutions are [tex]$m = 3$[/tex] and [tex]$m = -1$[/tex].
----------------------------------
3. For the equation
[tex]$$
5c^2 + 17c + 14 = 0,
$$[/tex]
we seek factors of the quadratic. It factors as
[tex]$$
(c + 2)(5c + 7) = 0.
$$[/tex]
Setting each factor equal to zero:
- From [tex]$c + 2 = 0$[/tex], we obtain
[tex]$$
c = -2.
$$[/tex]
- From [tex]$5c + 7 = 0$[/tex], we obtain
[tex]$$
5c = -7 \quad \Longrightarrow \quad c = -\frac{7}{5}.
$$[/tex]
Thus, the solutions for [tex]$c$[/tex] are [tex]$c = -2$[/tex] and [tex]$c = -\frac{7}{5}$[/tex].
----------------------------------
In summary, the final results are:
- For [tex]$4h^5-8h^4=0$[/tex]:
- Factorization:
[tex]$$
4h^4 (h - 2)=0
$$[/tex]
- Solutions: [tex]$h = 0$[/tex] (with multiplicity 4) and [tex]$h = 2$[/tex].
- For [tex]$m^2-2m-3=0$[/tex]:
- Factorization:
[tex]$$
(m - 3)(m + 1)=0
$$[/tex]
- Solutions: [tex]$m = 3$[/tex] and [tex]$m = -1$[/tex].
- For [tex]$5c^2+17c+14=0$[/tex]:
- Factorization:
[tex]$$
(c + 2)(5c + 7)=0
$$[/tex]
- Solutions: [tex]$c = -2$[/tex] and [tex]$c = -\frac{7}{5}$[/tex].
This completes the detailed step-by-step solution.
----------------------------------
1. For the equation
[tex]$$
4h^5 - 8h^4 = 0,
$$[/tex]
we first notice that both terms have the common factor [tex]$4h^4$[/tex]. Factoring it out gives
[tex]$$
4h^4 (h - 2) = 0.
$$[/tex]
By the zero-product property, we set each factor equal to zero:
- From [tex]$4h^4 = 0$[/tex], we obtain
[tex]$$
h^4 = 0 \quad \Longrightarrow \quad h = 0.
$$[/tex]
(Note that [tex]$h=0$[/tex] is a solution of multiplicity 4.)
- From [tex]$h - 2 = 0$[/tex], we have
[tex]$$
h = 2.
$$[/tex]
Thus, the solutions are [tex]$h = 0$[/tex] (with multiplicity 4) and [tex]$h = 2$[/tex].
----------------------------------
2. For the equation
[tex]$$
m^2 - 2m - 3 = 0,
$$[/tex]
we look for two numbers that multiply to [tex]$-3$[/tex] and add to [tex]$-2$[/tex]. The factors are [tex]$m - 3$[/tex] and [tex]$m + 1$[/tex], so the equation factors as
[tex]$$
(m - 3)(m + 1) = 0.
$$[/tex]
Setting each factor equal to zero gives:
- From [tex]$m - 3 = 0$[/tex], we get
[tex]$$
m = 3.
$$[/tex]
- From [tex]$m + 1 = 0$[/tex], we get
[tex]$$
m = -1.
$$[/tex]
Thus, the solutions are [tex]$m = 3$[/tex] and [tex]$m = -1$[/tex].
----------------------------------
3. For the equation
[tex]$$
5c^2 + 17c + 14 = 0,
$$[/tex]
we seek factors of the quadratic. It factors as
[tex]$$
(c + 2)(5c + 7) = 0.
$$[/tex]
Setting each factor equal to zero:
- From [tex]$c + 2 = 0$[/tex], we obtain
[tex]$$
c = -2.
$$[/tex]
- From [tex]$5c + 7 = 0$[/tex], we obtain
[tex]$$
5c = -7 \quad \Longrightarrow \quad c = -\frac{7}{5}.
$$[/tex]
Thus, the solutions for [tex]$c$[/tex] are [tex]$c = -2$[/tex] and [tex]$c = -\frac{7}{5}$[/tex].
----------------------------------
In summary, the final results are:
- For [tex]$4h^5-8h^4=0$[/tex]:
- Factorization:
[tex]$$
4h^4 (h - 2)=0
$$[/tex]
- Solutions: [tex]$h = 0$[/tex] (with multiplicity 4) and [tex]$h = 2$[/tex].
- For [tex]$m^2-2m-3=0$[/tex]:
- Factorization:
[tex]$$
(m - 3)(m + 1)=0
$$[/tex]
- Solutions: [tex]$m = 3$[/tex] and [tex]$m = -1$[/tex].
- For [tex]$5c^2+17c+14=0$[/tex]:
- Factorization:
[tex]$$
(c + 2)(5c + 7)=0
$$[/tex]
- Solutions: [tex]$c = -2$[/tex] and [tex]$c = -\frac{7}{5}$[/tex].
This completes the detailed step-by-step solution.
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