(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.
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