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The polynomial of degree 3, p(x),has a root of multiplicity 2 at x=2 and root of multiplicity 1 at x=-4 the y intercept is -3.2

User Gskema
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We know that a polynomial of degree 3 has three roots. Since p(x) has a root of multiplicity 2 at x=2, we can write one factor of the polynomial as (x-2)^2. Similarly, since p(x) has a root of multiplicity 1 at x=-4, we can write another factor as (x+4). Thus, the polynomial can be written as:

p(x) = k(x-2)^2(x+4)

where k is some constant.

To find the value of k, we use the fact that the y-intercept is -3.2, which means that p(0) = -3.2. Substituting x=0 into the equation for p(x), we get:

p(0) = k(-2)^2(4) = 16k

Setting this equal to -3.2, we have:

16k = -3.2

k = -0.2

So the polynomial is:

p(x) = -0.2(x-2)^2(x+4)

Therefore, the polynomial of degree 3, p(x), with a root of multiplicity 2 at x=2 and root of multiplicity 1 at x=-4, and y-intercept of -3.2, is given by:

p(x) = -0.2(x-2)^2(x+4)
User Paralife
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