Question 1129941
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The statement of the problem is not complete.  Without the stated requirement that the polynomial have rational coefficients, the fourth root could be any complex number.<br>
Assuming rational coefficients, the four roots are -4, -4, 5+5i, and 5-5i.<br>
The quadratic polynomial with roots -4 and -4 is<br>
{{{(x+4)(x+4) = x^2+8x+16}}}<br>
The quadratic polynomial with roots 5+5i and 5-5i is<br>
{{{(x-(5+5i))(x-(5-5i)) = ((x-5)-5i)((x-5)+5i) = (x^2-10x+25)+25 = x^2-10x+50}}}<br>
An alternative way to find the quadratic polynomial with roots 5+5i and 5-5i is to use the fact that the quadratic polynomial ax^2+bx+c has roots whose sum is -b/a and whose product is c/a.<br>
The sum of the roots 5+5i and 5-5i is 10; so the linear coefficient in the quadratic polynomial is -10.<br>
The product of the roots 5+5i and 5-5i is 25-25i^2 = 25+25 = 50; so the constant in the quadratic polynomial is 50.<br>
Then the quadratic polynomial with roots 5+5i and 5-5i is x^2-10x+50.<br>
Finally the polynomial with roots -4, -4, 5+5i and 5-5i is<br>
{{{(x^2+8x+16)(x^2-10x+50) = x^4-2x^3-14x^2+240x+800}}}