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F(x)=x^4+2x^3-9x^2-2x+6 and f(-4)=0 algebraically find all the zeros.

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If we know that f(-4) = 0, we can say that:

x + 4 is a factor of the function f(x).

This is because when we substitute x = -4 into the function, we get:

f(-4) = (-4)^4 + 2(-4)^3 - 9(-4)^2 - 2(-4) + 6 = 0

This means that (x + 4) is one of the factors of the function.

Now, we can use polynomial division or synthetic division to find the other factors. If we divide the function f(x) by (x + 4), we obtain:

x^3 - 2x^2 - 17x + 3

We can use the Factor theorem on this cubic function f(x) = x^3 - 2x^2 - 17x + 3.

If we try to substitute a = 1 we find that f(1) = -15 which is not zero. If we try to substitute a = -1, we find that f(-1) = 23 which is not zero.

Now we can solve as follows:

Let x = t - 2/3. After substitution we get a cubic equation.

t^3 -25/3 t - 85/27 = 0

We can find one root of this cubic by rational root theorem which is t = 5 or t = -17/3.

Now, we can use synthetic division again to divide f(x) by the binomial (x - 5). This gives us:

(x - 5)(x^2 + 3x - 1)

Therefore, the complete factorization of f(x) is:

f(x) = (x + 4)(x - 5)(x^2 + 3x - 1)

Therefore, the zeros of the function are:

x = -4, 5, (-3 ± √13)/2

Thus, these are the zeros of the polynomial.

User Matthew Gertner
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