Find the real zeroes of the polynomial:
f(x)=2x(x+5)^2-8x
I know you can pull the factors from the leading coefficient and the constant:
constant = p= 5 factors 1, 5
leading coefficient= q= 2 factors 1, 2
so would my rational zeros be: +/-{1, 1/2, 5, 5/2} and at that point whats my next order of operation?
for the zeros of 2x(x+5)^2-8x
2x(x+5)^2-8x = 0
2x( (x+5)^2 - 4) = 0
x = 0 or (x+5)^2 = 4
x = 0 or x+5 = ±2
x = 0, -3, -7
proof:
http://www.wolframalpha.com/input/?i=solve+2x(x%2B5)%5E2-8x+%3D0
I have no idea where you got your answers from, but they are not the zeroes of
your given function.
To find the real zeros of the polynomial, you can use the rational root theorem to check which of the possible rational zeros are actually roots of the polynomial.
The rational root theorem states that if a rational number p/q is a root of a polynomial with integer coefficients, then p must be a factor of the constant term, and q must be a factor of the leading coefficient.
In your case, the constant term is -8x and the leading coefficient is 2x. The factors of the constant term are 1 and 8, and the factors of the leading coefficient are 1 and 2.
So, the possible rational zeros are +/-{1, 2, 8}.
Next, you can use synthetic division or polynomial long division to check each possible zero. Start with 1 and perform the division. If the remainder is 0, then 1 is a zero of the polynomial. Repeat this process for each possible zero until you find the actual zeros.
Alternatively, you can solve the polynomial equation f(x) = 0 by factoring. Start by factoring out a common factor of 2x from the polynomial:
f(x) = 2x(x+5)^2 - 8x
= 2x[(x+5)^2 - 4]
Next, you can factor the quadratic term (x+5)^2 - 4 by recognizing it as a difference of squares:
f(x) = 2x[(x+5)^2 - 4]
= 2x[(x+5+2)(x+5-2)]
= 2x(x+7)(x+3)
Now set each factor equal to zero and solve for x:
2x = 0 --> x = 0
x + 7 = 0 --> x = -7
x + 3 = 0 --> x = -3
So, the real zeros of the polynomial f(x) are 0, -7, and -3.