Even though there are three roots given, the polynomial has four roots because the conjugate of (3+rt2)/4 is also a root. Therefore, the smallest (minimal degree) is 4, and the leading coefficient will be on the x^4 term.
Since we know the roots, and we know they are solutions to a polynomial equation, we can write the polynomial equation as:
(x - 4) * (x + 7)) * (x - [(3+rt2)/4] * (x - [(3-rt2)/4] = 0
The first two binomials multiply out to (x^2 + 3x - 28).
The next two are a little tougher; however, they multiply out to (x^2 - (3/2)x + (7/16)),
Since each of the two polynomials above can be set to zero, we can set the second to zero:
x^2 - (3/2)x + (7/16) = 0
Multiply both sides by 16 to get
16x^2 - 24x + 7 = 0
This means that the two problem can be written like this:
(x^2 + 3x - 28) * (16x^2 - 24x + 7) = 0
If we multiply the two polynomials out (we don't have to), the leading term would be 16x^4.
Therefore, the coefficient of the leading term is 16.
The hardest part of this problem is multiplying out (x - [(3+rt2)/4] * (x - [(3-rt2)/4] to get x^2 - (3/2)x + (7/16). This step involves knowing how to multiply conjugates, and is a little too much to put into a typewritten explanation.
Hope this helps!
Kaela W.
Yes thank you !10/12/21