find all the zeros of the following polynomial. Write the polynomial in fraction form. Show all the work.
f(x)= x^4+3x^2-40
Let x^2 = y
y^2 +3y -40 = 0
(y+8)(y-5) = 0
y = -8 or 5
x = sqrty = +/- isqrt8 or +/- sqrt5
f(x) = (x - sqrt5)(x + sqrt5)(x^2 +8)
To find the zeros of the polynomial f(x) = x^4 + 3x^2 - 40, we can set the polynomial equal to zero and solve for x.
f(x) = x^4 + 3x^2 - 40 = 0
To make the equation easier to factor, let's rewrite it using a substitution. Let's say u = x^2, then the equation becomes:
u^2 + 3u - 40 = 0
Now we can factor this quadratic equation:
(u + 8)(u - 5) = 0
Let's set each factor equal to zero and solve for u:
u + 8 = 0 or u - 5 = 0
Solving for u in each equation:
u = -8 or u = 5
Now we need to substitute back the original variable x:
For u = -8:
x^2 = -8
Taking the square root of both sides:
x = ±√(-8)
Since the square root of a negative number is not a real number, there are no real solutions for this case.
For u = 5:
x^2 = 5
Taking the square root of both sides:
x = ±√5
So the zeros of the polynomial f(x) = x^4 + 3x^2 - 40 are x = ±√5.
To write the polynomial in fraction form, we can factor it using the zeros we found:
f(x) = (x - √5)(x + √5)(x^2 + 8)
Therefore, the polynomial f(x) in fraction form is:
f(x) = (x - √5)(x + √5)(x^2 + 8)