A factorization of 
is 
.
<h3>What are the properties of roots of a polynomial?</h3>
- The maximum number of roots of a polynomial of degree
is . - For a polynomial with real coefficients, the roots can be real or complex.
- The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if
is a root, then 
is also a root.
If the roots of the polynomial 
are 
, then it can be factorized as 
.
Here, we are to find a factorization of 
. Also, given that 
and 
are roots of the polynomial.
Since is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.
Hence, 
and 
are also roots of the given polynomial.
Thus, all the four roots of the polynomial , are: 
.
So, the polynomial can be factorized as follows:
![\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)](https://tex.z-dn.net/?f=%5C%7Bx-%28-2%2Bi%5Csqrt%7B7%7D%29%5C%7D%5C%7Bx-%28-2-i%5Csqrt%7B7%7D%29%5C%7D%5C%7Bx-%281-i%5Csqrt%7B3%7D%29%5C%7D%5C%7Bx-%281%2Bi%5Csqrt%7B3%7D%29%5C%7D%5C%5C%3D%28x%2B2-i%5Csqrt%7B7%7D%29%28x%2B2%2Bi%5Csqrt%7B7%7D%29%28x-1%2Bi%5Csqrt%7B3%7D%29%28x-1-i%5Csqrt%7B3%7D%29%5C%5C%3D%5C%7B%28x%2B2%29%5E2%2B7%5C%7D%5C%7B%28x-1%29%5E2%2B3%5C%7D%5Chspace%7B1cm%7D%20%5B%5Cbecause%20%28a%2Bb%29%28a-b%29%3Da%5E2-b%5E2%5D%5C%5C%3D%28x%5E2%2B4x%2B4%2B7%29%28x%5E2-2x%2B1%2B3%29%5C%5C%3D%28x%5E2%2B4x%2B11%29%28x%5E2-2x%2B4%29)
Therefore, a factorization of is .
To know more about factorization, refer: brainly.com/question/25829061
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It opens downwards so it looks like this “n”. That is because a in the formula is a negative number in this situation
90 < [( n + n + 2 + n + 4) / 2] < 105
90 < (3n + 6) / 2 < 105
3n + 6 > 180 and 3n + 6 < 210
n > 58 , n < 68
58 < n < 68 answer
I believe the number is 406,152
