find the real zeros of the polynomial function and state their multiplicities. explain how you arrived at your answer.

`f(x)=-1.4x^4-2.8x^3`

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You need to find roots of `-1.4x^4-2.8x^3=0`

Let's rewrite this as

`-x^3(1.4x+2.8)=0`

Now we have product and product will be equal to 0 if at least one of the terms is equal to 0 and thus we get two equations

`x^3=0=>x=0`

This root has multiplicity of 3 because that is equation of third degree which has 3 roots and each of them is 0.

`1.4x+2.8=0`

`1.4x=-2.8`

`x=-2`

And this root has multiplicity of 1 because we got it by solving linear equation which has only one root.

Since we had equation of fourth degree and we got one root with multiplicity 3 and one with multiplicity 1 which means we have all of the solutions to the equation.

`-14x^4-28x^3=0` (multiplied by 10)

`14x^3(x+2)=0` (Changed sign)

`14x=0 ` `x=0` (its a three value root)

`x+2=0` `x=-2`

So equation has roots:

`x_1=0` `x_2=0` `x_3=0` `x_4=-2`

It's looks like function hasonly two root, for doesn not report the multiciply of root `x=0`

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