Start by distributing the 6 through the parenthses on the left side.
So we have 6x + 24 = 30.
Solving from here, I would subtract 24 from both sides to get 6x = 6.
Finally, dividing both sides of the equation by 6, we have x = 1.
Although the question is incomplete, I have seen it before.
The equations are:
3a + 6b = 12
-3a + 6b = -12
Adding the equations, we get:
12b = 0
And b = 0
I believe it’s
D.51is it B ??
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