Factor the high-degree polynomial 50x5−32x to solve the polynomial equation 50x5−32x=0.(1 point)
To factor the polynomial 50x^5 - 32x, we can start by factoring out the greatest common factor of 2x:
2x(25x^4 - 16)
Now, let's factor the expression inside the parentheses. Notice that it is a difference of squares:
2x(5x^2)^2 - 4^2
Using the formula for factoring difference of squares (a^2 - b^2 = (a + b)(a - b)), we have:
2x(5x^2 + 4)(5x^2 - 4)
Now, we can factor the expression further using a difference of squares again:
2x(5x^2 + 4)(√5x)^2 - 2^2
This gives us the final factored form:
2x(5x^2 + 4)(√5x + 2)(√5x - 2)
To solve the polynomial equation 50x^5 - 32x = 0, we set each factor equal to zero:
2x = 0, 5x^2 + 4 = 0, √5x + 2 = 0, √5x - 2 = 0
Solving these equations, we find the possible solutions:
x = 0, x = ±√(-4/5), x = ±2/√5
Note that x = ±√(-4/5) has no real solutions since we cannot take the square root of a negative number. So, the real solutions to the equation 50x^5 - 32x = 0 are:
x = 0, x = ±2/√5.
To factor the high-degree polynomial 50x^5 - 32x, we can first factor out the common factor, which is 2x:
2x(25x^4 - 16)
Now let's focus on the expression inside the parentheses: 25x^4 - 16. This is a difference of squares, since 25x^4 can be written as (5x^2)^2 and 16 can be written as (4)^2.
Using the difference of squares formula, we have:
2x(5x^2 + 4)(5x^2 - 4)
Now we can factor each of the quadratic expressions:
2x(5x + 2)(5x - 2)(5x^2 - 4)
Therefore, the factored form of 50x^5 - 32x is 2x(5x + 2)(5x - 2)(5x^2 - 4).