Answer:
(t-8)+10
Step-by-step explanation:
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
#SPJ9
Hello!
if x = -3, just substitute it for x (replace the x with -3)
2(-3) + 6 = 0
2 * -3 = -6
-6 + 6 = 0
I hope this helps, and have a nice day!
Answer:
= 7
Step-by-step explanation:
get it right! hope you get it right!\
Answer:
6
Step-by-step explanation:
subtract 2 from2 equal zero
add 2and 4 together get six
zero divided by any non-zero number gives zero
multiply 3 times 4 gets 12
0+12/2
divide 12 by 2 equal six
0+6
zero 0 and 6 to get 6
6
