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
Answer:
{x,y} = {5,6}
Step-by-step explanation:
// Solve equation [1] for the variable x
[1] x = y - 1
// Plug this in for variable x in equation [2]
[2] 2•(y -1) - y = 4
[2] y = 6
// Solve equation [2] for the variable y
[2] y = 6
// By now we know this much :
x = y-1
y = 6
// Use the y value to solve for x
x = (6)-1 = 5
Answer: 912
===================================
Work Shown:
The starting term is a1 = 3. The common difference is d = 5 (since we add 5 to each term to get the next term). The nth term formula is
an = a1+d(n-1)
an = 3+5(n-1)
an = 3+5n-5
an = 5n-2
Plug n = 19 into the formula to find the 19th term
an = 5n-2
a19 = 5*19-2
a19 = 95-2
a19 = 93
Add the first and nineteenth terms (a1 = 3 and a19 = 93) to get a1+a19 = 3+93 = 96
Multiply this by n/2 = 19/2 = 9.5 to get the final answer
96*9.5 = 912
I used the formula
Sn = (n/2)*(a1 + an)
where you add the first term (a1) to the nth term (an), then multiply by n/2
-----------------
As a check, here are the 19 terms listed out and added up. We get 912 like expected.
3+8+13+18 +23+28+33+38 +43+48+53+58 +63+68+73+78 +83+88+93 = 912
There are 19 values being added up in that equation above. I used spaces to help group the values (groups of four; except the last group which is 3 values) so it's a bit more readable.
Answer:
12.52 in.
Step-by-step explanation:
Area of square(A)=157in^2
length of square (l)=?
we know,
A=l^2
or,sqrt 157=l
12.52 in=l
The measurement of <ABC is 50 degrees
