Question 83034: Solution to the problem: x^3-5x^2-7x+51
answer choices:
4+i, 4-i
4i, -4i
4, -4
4+i, 4-i
Answer by Edwin McCravy(20063)   (Show Source):
You can put this solution on YOUR website!
Solution to the problem: x^3-5x^2-7x+51
answer choices:
4+i, 4-i
4i, -4i
4, -4
4+i, 4-i
-----------------------------------------------------
x³ - 5x² - 7x + 51
The posible rational zeros are ± the factors of 51.
They are ±1, ±3, ±17, ±51
We'll start trying them with synthetic division.
We'll try 1, and it'll leave a remainder 40 (no good!)
Then we'll try -1 and it'll leave a remainder of 52,
(also no good!), then we'll try -3, and get this:
-3| 1 -5 -7 51
| -3 24 -51
1 -8 17 0
Whoopee! A zero remainder!
So the cubic polynomial
x³ - 5x² - 7x + 51
factors as
(x + 3)(x² - 8x + 17)
Now we can find the zeros of the polynomial
x² - 8x + 17
by setting it = 0
x² - 8x + 17 = 0
and solving it by the quadratic formula:
______
-b ± Öb²-4ac
x = ——————————————
2a
where a = 1; b = -8; c = 17
______________
-(-8) ± Ö(-8)²-4(1)(17)
x = —————————————————————————
2(1)
_____
8 ± Ö64-68
x = ————————————
2
__
8 ± Ö-4
x = —————————
2
_
8 ± iÖ4
x = —————————
2
8 ± 21
x = —————————
2
8 2i
x = ——— ± —————
2 2
x = 4 ± i
So the zeros are 4 + i and 4 - 1.
Edwin
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