Answers:
15) x = 8,
![(1)/(3), - (2)/(5)](https://img.qammunity.org/2019/formulas/mathematics/high-school/k8ls1hfiwg48rywus6gygtfp8i9xwjnhhb.png)
16) x = 9,
,
![\frac{5 -√(17)} {2}](https://img.qammunity.org/2019/formulas/mathematics/high-school/uzo0zo7eoot86qrij4ahbk05au9zozesyo.png)
17) x = 6,
,
![-\frac{3-i√(11)} {10}](https://img.qammunity.org/2019/formulas/mathematics/high-school/nyoxjz0jry0mpvebul1z4usdgkm1mfqxmh.png)
Explanation:
15x³ - 119x² - 10x + 16 = 0
![(p)/(q): (16)/(15):+/- (1*2*4*8*16)/(1*3*5*15)](https://img.qammunity.org/2019/formulas/mathematics/high-school/jlcpdwnflso7u98jvlakinjm6bwtr1gzbh.png)
So, the possible rational roots are: +/-
![1, 2, 4, 8, 16,(1)/(3),(2)/(3),(4)/(3),(8)/(3),(16)/(3),(1)/(5),(2)/(5),(4)/(5),(8)/(5),(16)/(5),(1)/(15),(2)/(15),(4)/(15),(8)/(15),(16)/(15)](https://img.qammunity.org/2019/formulas/mathematics/high-school/yywd23cb1dlg7ekttr0p1f1bkc0x5jgnm7.png)
Use synthetic division with each one until you find a remainder of zero. I am not going to go through each one because it is too time consuming, however, the first one that works is x = 8
(x - 8)(15x² + x - 2)
Next, factor 15x² + x - 2 using any method
(x - 8)(3x - 1)(5x + 2)
Now, solve for x.
x = 8, x =
, x =
![-(2)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zt8cp8zhia8y1cxrxo1xrrhk5vy9k4nykh.png)
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For #16 & 17, follow the same process as above.