I need help with two Algebra problems!!! Please, if you have any idea on how to do this HELP!!!?

Question

1) Find a polynomial of degree 5 with -3 as a zero of multiplicity 3, 0 as a zero of multiplicity 1, and 4 as a zero of multiplicity 1.


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2) Find the rational zeros of x^5-3x^4-3x^3+9x^2-4x+12
note: ^ is raised to that power

Thank you so much! Anything can help!

Answer 1

1) degree of 5 so the maximum is x^5.
-3 as a zero with multiplicity of 3 means (x + 3)^3
0 as a multiplicity of 1 means x
4 as a zero with multiplicity of 1 means (x - 4)

so it is: x (x + 3)^3 (x - 4) if you expand this equation
the answer would be x^5 + 5x^4 - 9x^3 -81x^2 -108x

2) To find rational zero of x^5-3x^4-3x^3+9x^2-4x+12, you must first factor this expression
which is (x - 3)(x - 2)(x + 2)(x^2 + 1)
so for (x - 3) the first zero is 3
for (x - 2) the second zero is 2
for (x +2) the third zero is -2
since the zeros for (x^2 + 1) is imaginary +i and -i you can discard that

Answer: the rational zeros = 3, 2, -2

Answer 2

Heck, the answer to 1) is in its description! Write the polynomial in factored form, taking account of multiplicities

z1 is a zero of mult 3
z2 is a zero of mult 2

p(z) = (z-z1)^3*(z-z2)^2

They've given you almost everything! The only thing left is that you could multiply p(z) by a constant, that won't change the zeros, just the values everywhere else

Answer 3

1)it has factors
(x+3)^3
(x-0)^1
(x-4)^1
polynomial is=(x+3)^3*(x-0)^1*(x-4)^1