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(a) If (x + 2) is a factor of g(x) = 2x\(^3\) +11x\(^2\) - x - 30, find the zeros of ...

(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\(^{y + 1}\) = -19 simultaneously.
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    Correct Answer: Option
    Explanation:
    (a) Using long division
    x + 2 2x\(^2\) + 7x - 15
    2x\(^3\) + 11x^2 - x -30- (2x\(^3\) + 4x\(^2\))\(\overline{7^2 - x - 30}\)- (7x\(^2\) + 14x) - 3\(\overline{-15x - 30}\)- 15x - 30\(\overline{0}\)




    (x + 2)(2x\(^2\) + 7x - 15)
    (x + 2)(2x - 3)(x - 5)
    zeros of g (x) are;
    x + 2 =0, 2x - 3 = 0
    x + 5 = 0, x = -2
    x = \(\frac{5}{2}\) and x = -5

    (b) 3(2^\(^x\)) + 3\(^{y - 2}\) = 25 .......x1
    2\(^x\) - 3\(^{y + 1}\) = -19 .........x3
    3(2\(^x\)) + 3\(^{y - 2}\) = 25
    3\(^{y - 2}\) + 3(3\(^{y + 1}\)) = 82
    \(\frac{3^y}{9} + \frac{3^y}{1}\) = 82
    \(\frac{3^y + 9(3^y)}{9}\) = 82
    3\(^y\) + 9(3\(^y)\)) = 82 x 9
    let p = 3y
    p + qp = 738
    \(\frac{10p}{10} = \frac{738}{10}\)
    p = 73.8, 3\(^y\) = 73.8
    log3\(^y\) = log 73.8
    \(\frac{y log 3}{log 3} = \frac{log 73.8}{\log 3}\)
    y = 3.9
    from
    2\(^x\) + 3\(^{y + 1}\) = - 19
    2\(^x\) + 3\(^{3.9 + 1}\) = - 19
    2\(^x\) + 3\(^{4 - 9}\) = - 19
    2\(^x\) - 217.7 = - 19
    2\(^x\) = 198.71
    log 2\(^x\) = log 198.71
    \(\frac{x \log 2}{log 2} = \frac{log 198.71}{log 2}\)
    x = 7.6

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