Factor the high-degree polynomial 50x5−32x to solve the polynomial equation 50x5−32x=0 .(1 point)
Responses
x=2/5√, x=−2/5√
x equals Start Fraction 2 over Start Root 5 End Root End Fraction , x equals negative Start Fraction 2 over Start Root 5 End Root End Fraction
x=4/5, x=−4/5
x equals Start Fraction 4 over 5 End Fraction , x equals negative Start Fraction 4 over 5 End Fraction
x=2/5√
x equals Start Fraction 2 over Start Root 5 End Root End Fraction
x=0, x=2/5√, x=−2/5√
To factor the polynomial 50x^5 - 32x, we can first factor out x:
x(50x^4 - 32)
Next, we can factor out 2 from the expression inside the parentheses:
x(2(25x^4 - 16))
Now, we can factor a difference of squares:
x(2(5x^2 - 4)(5x^2 + 4))
This gives us the factored form of the polynomial as:
x(2)(5x^2 - 4)(5x^2 + 4)
To solve the polynomial equation 50x^5 - 32x = 0, we can set each factor equal to zero:
x = 0
5x^2 - 4 = 0
5x^2 + 4 = 0
Solving for x in each equation, we get:
x = 0
For 5x^2 - 4 = 0:
5x^2 = 4
x^2 = 4/5
x = ±√(4/5)
x = ±(2/√5)
x = ±(2/5√5)
For 5x^2 + 4 = 0:
5x^2 = -4
x^2 = -4/5
There are no real solutions for this equation since the square of a real number cannot be negative.
Therefore, the solutions to the polynomial equation 50x^5 - 32x = 0 are:
x = 0, x = 2/5√5, x = -2/5√5
To factor the high-degree polynomial 50x^5 - 32x, we first need to find the common factor. In this case, it is 2x.
Step 1: Factor out 2x:
2x(25x^4 - 16)
Now we have a quadratic expression (25x^4 - 16) inside the parentheses. To solve the polynomial equation 50x^5 - 32x = 0, we set each factor to zero and solve for x.
Step 2: Set 2x = 0 and solve for x:
2x = 0
x = 0
Step 3: Set 25x^4 - 16 = 0 and solve for x:
To simplify the equation (25x^4 - 16), we can rewrite it as a difference of squares, since 16 can be written as 4^2 and 25x^4 can be written as (5x^2)^2:
(5x^2)^2 - 4^2 = 0
Now, we have a difference of squares, which can be factored as:
(5x^2 - 4)(5x^2 + 4) = 0
Step 4: Set each factor to zero and solve for x:
Setting (5x^2 - 4) = 0:
5x^2 - 4 = 0
5x^2 = 4
x^2 = 4/5
x = ±√(4/5)
x = ±(2/√5)
x = ±(2/5√5)
Setting (5x^2 + 4) = 0:
5x^2 + 4 = 0
5x^2 = -4
x^2 = -4/5 (No real solution, since the square of a real number cannot be negative)
Therefore, the solutions to the polynomial equation 50x^5 - 32x = 0 are:
x = 0 and x = ±(2/5√5)
To factor the high-degree polynomial 50x^5 - 32x, we first look for any common factors among the terms. In this case, we can factor out an x, which gives us:
x(50x^4 - 32)
Next, we can look for any further common factors within the parentheses. In this case, we can factor out a common factor of 2:
x(2)(25x^4 - 16)
Now, we have factored the polynomial as much as possible. To solve the polynomial equation 50x^5 - 32x = 0, we set the factored polynomial equal to zero and solve for x:
x(2)(25x^4 - 16) = 0
Setting each factor equal to zero, we get:
x = 0
2 = 0 (This is not possible, so we ignore it)
25x^4 - 16 = 0
To solve the equation 25x^4 - 16 = 0, we can use the difference of squares formula, which states that a^2 - b^2 can be factored as (a + b)(a - b). In this case, we have:
(5x^2)^2 - 4^2 = 0
Using the difference of squares formula, we can factor this equation as:
(5x^2 + 4)(5x^2 - 4) = 0
Now, we can set each factor equal to zero and solve for x:
5x^2 + 4 = 0
5x^2 = -4
x^2 = -4/5
x = ±√(-4/5)
5x^2 - 4 = 0
5x^2 = 4
x^2 = 4/5
x = ±√(4/5)
Therefore, the solutions to the polynomial equation 50x^5 - 32x = 0 are:
x = 0
x = ±√(-4/5)
x = ±√(4/5)