Question 1126946
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In order for the problem to make sense, I will assume that the last root you show is the square ROOT of 3, not the square of 3.<br>
Then, since the polynomial is to have integer coefficients, the 5th root must be negative square root of 3.<br>
The polynomial with those 5 roots is<br>
{{{a(x+3)(x+1)(x-3)(x-sqrt(3))(x+sqrt(3))}}}<br>
The constant term of the polynomial is<br>
{{{a(3)(1)(-3)(-3) = 27a}}}<br>
Since the constant term is 54, a must be 2.  Then the polynomial (still in factored form) is<br>
{{{2(x+3)(x+1)(x-3)(x-sqrt(3))(x+sqrt(3))}}}<br>
In expanded form the polynomial is<br>
{{{2(x^2-9)(x^2-3)(x+1) = 2(x^4-12x^2+27)(x+1) = 2(x^5+x^4-12x^3-12x^2+27x+27) = 2x^5+2x^4-24x^3-24x^2+54x+54}}}