(a) If (x + 2) is a factor of g(x) = 2x\(^3\) +11x\(^2\) - x - 30, find the zeros of g(x). (b) Solve 3(2\(^x\)) +3\(^{y - 2}\) = 25 and 2x - 3\(^{y + 1}\) =...

Answer Details

(a) If (x + 2) is a factor of g(x), it means that (x + 2) is a root of the polynomial g(x). To find the zeros of g(x), we need to factorize the polynomial by using long division or synthetic division.

Using synthetic division, we get:

-2 | 2 11 -1 -30
   |-----------
   | 2 -6 11 | 0

Therefore, g(x) can be factorized as g(x) = 2x^2 - 6x + 11. To find the zeros of g(x), we need to solve the equation 2x^2 - 6x + 11 = 0.

Using the quadratic formula, we get:

x = \(\frac{6 \pm \sqrt{6^2 - 4(2)(11)}}{4}\)

x = \(\frac{3}{2} \pm \frac{\sqrt{23}}{2}\)

Therefore, the zeros of g(x) are \(\frac{3}{2} + \frac{\sqrt{23}}{2}\), and \(\frac{3}{2} - \frac{\sqrt{23}}{2}\).

(b) To solve the simultaneous equations 3(2^x) + 3^(y - 2) = 25 and 2x - 3^(y + 1) = -19, we need to isolate one variable in terms of the other and substitute the expression into the other equation.

From the second equation, we have:

2x = 3^(y + 1) - 19

Substituting this expression into the first equation, we get:

3(2^x) + 3^(y - 2) = 25

3(2^x) + 3^(y - 2) = 3^2

Dividing both sides by 3^2, we get:

2^x + 3^(y - 2 - 2) = 1

2^x + 3^(y - 4) = 1

Substituting the expression for 2x from the second equation, we get:

2^x + 3^(y - 4) = 1 + 3^(y + 1) - 19

2^x + 3^(y - 4) = 3^(y + 1) - 18

Rearranging this equation, we get:

2^x = 3^(y + 1) - 3^(y - 4) - 18

Using logarithms, we can solve for x:

x = log₂(3^(y + 1) - 3^(y - 4) - 18)

Substituting this expression for x into the second equation, we get:

2log₂(3^(y + 1) - 3^(y - 4) - 18) - 3^(y + 1) = -19

Simplifying this