Find all the zeros of f(x)=2x^4-5x^3-x^2-5x-3 urgent 20 points show work

The rational root theorem tells you possible rational roots will be of the form ...

... ±(divisor of 3)/(divisor of 2)

so will be from the set {-3, -3/2, -1, -1/2, 1/2, 1, 3/2, 3}.

By adding coefficients in various ways, we can find

... f(0) = -3

... f(1) = -12

... f(-1) = 8

so there is a root between -1 and 0. When we try -1/2, we find it is a root and we get a quotient of the cubic

... x^3 -3x^2 +x -3.

The pattern in the coefficients tells us we can factor by grouping to get

... x^2(x -3) + (x -3)

So the factorization of f(x) is

... f(x) = (2x +1)(x -3)(x^2 +1)

The last factor has zeros of ±√(-1) = ±i.

The zeros of f(x) are {-1/2, 3, -i, +i}.

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step-by-step explanation:

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(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: