how do i solve the following problems?
a^5+6a^4+5a^3=0
t^5-256t=0
y^3-512=0
z=8(root)z+240
You have to factor to solve.
a^5 + 6a^4 + 5a^3 = 0
a^3(a + 1)(a + 5) = 0
a^3 = 0
a = 0
a + 1 = 0
a = -1
a + 5 = 0
a = -5
Do the other problems the same to solve.
i don't understand how to do them. ive tried getting help but i just don't understand them
To solve these problems, we need to apply some basic algebraic techniques. Let's go through each problem step by step:
1. To solve the equation: a^5 + 6a^4 + 5a^3 = 0, we can factor out an "a^3" from each term:
a^3(a^2 + 6a + 5) = 0
Now, we have two parts to solve individually:
a^3 = 0 or a^2 + 6a + 5 = 0
For the first part, a^3 = 0, the solution is simply a = 0.
For the second part, a^2 + 6a + 5 = 0, we can use factoring or the quadratic formula to find the solutions. Factoring gives:
(a + 5)(a + 1) = 0
Setting each factor equal to zero gives:
a + 5 = 0 or a + 1 = 0
So, a = -5 or a = -1 are the other two solutions.
Therefore, the solutions to the equation a^5 + 6a^4 + 5a^3 = 0 are: a = 0, -5, -1.
2. For the equation: t^5 - 256t = 0, we can again factor out a common term:
t(t^4 - 256) = 0
Now, we have two parts to solve individually:
t = 0 or t^4 - 256 = 0
For the first part, t = 0 is an obvious solution.
For the second part, we can recognize that t^4 - 256 represents a difference of squares:
(t^2)^2 - 16^2 = 0
This can be factored as:
(t^2 - 16)(t^2 + 16) = 0
Setting each factor equal to zero gives:
t^2 - 16 = 0 or t^2 + 16 = 0
Solving each equation gives:
t^2 - 16 = 0: t = ±4
t^2 + 16 = 0: No real solutions, as "t^2 + 16" is always positive.
Therefore, the solutions to the equation t^5 - 256t = 0 are: t = 0, ±4.
3. For the equation: y^3 - 512 = 0, we can rearrange it as:
y^3 = 512
Taking the cube root of both sides, we find:
y = ∛(512)
The cube root of 512 can be calculated as:
∛(512) = 8
Therefore, the solution to the equation y^3 - 512 = 0 is: y = 8.
4. For the equation: z = 8√z + 240, we can start by isolating the square root term:
z = 8√z + 240
Subtracting 8√z from both sides gives:
z - 8√z = 240
Now, let's introduce a substitution:
Let x = √z
The equation becomes:
x^2 - 8x = 240
Rearranging and factoring:
x^2 - 8x - 240 = 0
(x - 20)(x + 12) = 0
Setting each factor equal to zero gives:
x - 20 = 0 or x + 12 = 0
Solving for x gives:
x = 20 or x = -12
Now, substitute back √z for x:
√z = 20 or √z = -12
Square both sides to find the solutions for z:
z = 400 or z = 144
Therefore, the solutions to the equation z = 8√z + 240 are: z = 400, 144.