Find all real zeros of the polynomial f(x)= x^4 -130x^2+3969 and determine the mutiplicity of each.
let z = x^2
z^2 -130 z + 3969 = 0
(z- )(z- ) = 0
start factoring 3969
3969 = 3*3*3**3 *49 = 81*49
81+49 = 130 !!!!
(z-81)(z-49) = 0
so
x^2-49 = 0 ---> x = +7 or x = -7
x^2-81 = 0 ---> x = +9 or x = -9
Using the polynomial function f(x)=3x^4-5x^3+8x-6. According to the rational zero theorem, which of the following could not be a zero of the f(x)?
To find the real zeros of a polynomial, such as f(x) = x^4 - 130x^2 + 3969, we need to factor it or use other techniques. For this particular polynomial, we can use a substitution to simplify the equation.
Let's make a substitution: let u = x^2. Then our equation becomes u^2 - 130u + 3969 = 0.
Now we have a quadratic equation in terms of u. We can solve it using factoring, completing the square, or the quadratic formula. In this case, let's use factoring.
The equation u^2 - 130u + 3969 = 0 can be factored as (u - 63)(u - 63) = 0.
Setting each factor equal to zero, we have two solutions: u - 63 = 0, which gives us u = 63.
Now, recall that u = x^2. So, substituting back, we have x^2 = 63. Taking the square root of both sides, we get x = ±√63.
Thus, the real zeros of the polynomial f(x) = x^4 - 130x^2 + 3969 are x = ±√63.
To determine the multiplicity of each zero, we can look at the factors in the factored form of the polynomial.
In this case, we have one factor (u - 63) repeated twice, which corresponds to x = ±√63 being a double zero.
Therefore, the multiplicity of each zero is 2.