Answer:
x^4 - 7x^3 +7x^2 +21x-30
Explanation:
If the roots of the function are 2 sqrt(3) and 5
We know that there is one more root
Irrational roots come in pairs, if we have a +sqrt (b), we have a - sqrt(b)
Since we have 0+sqrt(3) we will also have 0-sqrt(3)
We can write the polynomial using the roots
2, sqrt(3) (-sqrt(3) 5 are our roots
0= (x-a) (x-b) (x-c) (x-d) where a,b,c,d are the 4 roots
0 = (x-2) (x-5) (x-sqrt(3)) (x--sqrt(3))
0 = (x-2) (x-5) (x-sqrt(3)) (x+sqrt(3))
FOIL the first two terms
(x-2) (x-5) = x^2 -5x-2x+10 = x^2 -7x+ 10
FOIl the last two terms
(x-sqrt(3)) (x+sqrt(3)) = x^2 - xsqrt(3) + xsqrt(3) - 3 = x^2 -3
0 = (x-2) (x-5) (x-sqrt(3)) (x+sqrt(3))
0 = (x^2 -7x+10) (x^2 -3)
Multiply x^2 by everything in the first term
x^2 (x^2 -7x+10) =x^4 - 7x^3 + 10x^2
Multiply -3 by everything in the first term
-3 (x^2 -7x+ 10) = -3x^2 +21x -30
Add these results together
x^4 - 7x^3 + 10x^2
-3x^2 +21x -30
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x^4 - 7x^3 +7x^2 +21x-30