Factor the high-degree polynomial 50x5−32x to solve the polynomial equation 50x5−32x=0 .(1 point)
Responses
x=25√, x=−25√
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=45, x=−45
x equals Start Fraction 4 over 5 End Fraction , x equals negative Start Fraction 4 over 5 End Fraction
x=25√
x equals Start Fraction 2 over Start Root 5 End Root End Fraction
x=0, x=25√, x=−25√
To factor the polynomial 50x^5 - 32x, we can first factor out the greatest common factor, which is 2x:
2x(25x^4 - 16)
Next, we notice that we have a difference of squares within the parentheses. We can factor that as well:
2x(5x^2 + 4)(5x^2 - 4)
Now we have factored the polynomial as much as possible.
To solve the polynomial equation 50x^5 - 32x = 0, we set the factored expression equal to 0 and solve for x:
2x(5x^2 + 4)(5x^2 - 4) = 0
Setting each factor equal to 0, we get the following solutions:
2x = 0 --> x = 0
5x^2 + 4 = 0 --> x^2 = -4/5 --> x = ± √(-4/5) = ± √(4/5)i = ± 2/√5i = ± (2/√5)i = ± (2√5/5)i
5x^2 - 4 = 0 --> x^2 = 4/5 --> x = ± √(4/5) = ± 2/√5 = ± 2√5/5
Therefore, the solutions to the polynomial equation 50x^5 - 32x = 0 are:
x = 0, ± (2√5/5), ± (2√5/5)i
To factor the high-degree polynomial 50x^5 - 32x, we need to factor out the common factor first. In this case, the common factor is 2x, so we can rewrite the polynomial as:
2x(25x^4 - 16)
Next, we can factor the expression inside the parentheses. The expression 25x^4 - 16 is a difference of squares, which can be factored as:
(5x^2 - 4)(5x^2 + 4)
So, the fully factored form of the polynomial is:
2x(5x^2 - 4)(5x^2 + 4)
To solve the polynomial equation 50x^5 - 32x = 0, we can set each factor equal to zero and solve for x:
2x = 0 (implies x = 0)
5x^2 - 4 = 0 (implies x = ±2/√5 or x = ±2/√5)
5x^2 + 4 = 0 (implies no real solutions)
Therefore, the solutions to the polynomial equation 50x^5 - 32x = 0 are:
x = 0, x = ±2/√5, or in decimal form, approximately x = 0, x ≈ 0.894, x ≈ -0.894
To factor the high-degree polynomial 50x^5 - 32x, we can first look for common factors. In this case, we can factor out 2x:
50x^5 - 32x = 2x(25x^4 - 16)
Now, we can see that the polynomial inside the parentheses is a difference of squares, because 25x^4 = (5x^2)^2 and 16 = 4^2. Using the difference of squares formula, we can factor it further:
25x^4 - 16 = (5x^2)^2 - 4^2 = (5x^2 + 4)(5x^2 - 4)
Now, we have factored the high-degree polynomial as:
50x^5 - 32x = 2x(5x^2 + 4)(5x^2 - 4)
To solve the polynomial equation 50x^5 - 32x = 0, we can set each factor equal to zero and solve for x:
2x = 0 gives x = 0
5x^2 + 4 = 0 gives x = ±√(-4/5) = ±√(-4)/√5 = ±2/√5 = ±2√5/5
5x^2 - 4 = 0 gives x = ±√(4/5) = ±√4/√5 = ±2/√5 = ±2√5/5
So, the solutions to the polynomial equation are x = 0, x = ±2√5/5.