6x³y + 6xy − 12x² − 12 = 6(xy-2) * (x²+1)
(x²+1) is a factor of 6x³y + 6xy − 12x² − 12
If 2 + 5i is a zero, then by the complex conjugate root theorem, we must have its conjugate as a zero to have a polynomial containing real coefficients. Therefore, the zeros are -3, 2 + 5i, and 2 - 5i. We have three zeros so this is a degree 3 polynomial (n = 3).
f(x) has the equation
f(x) = (x+3)(x - (2 + 5i))(x - (2 - 5i))
If we expand this polynomial out, we get the simplest standard form
f(x) = x^3-x^2+17x+87
Therefore the answer to this question is f(x) = x^3-x^2+17x+87
Be right there! Let's do this!
Answer:
See explanation
Step-by-step explanation:
Assuming the given inequality is 
Then the corresponding linear equation is 
When x=0, we have 
When y=0, we have 
The T-table is:
<u>x | y</u>
0 | 4
6 | 12
We plot this points and draw a solid straight line as shown in the attachment.
Now let us test the origin: (0,0) by plugging x=0 and y=0 into the inequality.
....This is true so we shade the lower half plane as shown in the attachment.
h=9
A=((a+b)/2)h
234=((32+20)/2)h
234=(52/2)h
234=26h
9=h
