Hi, can someone please help me find all the zeros in order using synthetic division and quadratic formula for equation:
P(x)= 2x⁹ - x⁸ - 16x⁷ + 17x⁶ - 6x⁵ - 5x⁴+ 68x³ - 77x² + 18
Try 1 as a zero
1|2 -1 -16 17 -6 -5 68 -77 0 18
| 2 1 -15 2 -4 -9 59 -18 -18
2 1 -15 2 -4 -9 59 -18 -18 0
That factors P(x) as
P(x) = (x - 1)(2x⁸ + x⁷ - 15x⁶ + 2x⁵ - 4x⁴- 9x³ - 59x² - 18x - 18)
Try 1 as a zero again
1|2 1 -15 2 -4 -9 59 -18 -18
| 2 3 -12 -10 -14 -23 36 18
2 3 -12 -10 -14 -23 36 18 0
That factors P(x) as
P(x) = (x - 1)(x - 1)(2x⁷ + 3x⁶ - 12x⁵ - 10x⁴- 14x³ - 23x² + 36x + 18)
Try 1 as a zero again
1|2 3 -12 -10 -14 -23 36 18
| 2 5 -7 -17 -31 -54 -18
2 5 -7 -17 -31 -54 -18 0
That factors P(x) as
P(x) = (x - 1)(x - 1)(x - 1)(2x⁶ + 5x⁵ - 7x⁴- 17x³ - 31x² - 54x - 18)
Try -3 as a zero
-3|2 5 -7 -17 -31 -54 -18
| -6 3 12 15 48 18
2 -1 -4 -5 -16 -6 0
That factors P(x) as
P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(2x⁵ - x⁴- 4x³ - 5x² - 16x - 6)
Try -3/2 as a zero
-3/2|2 -1 -4 -5 -16 -6
| -3 6 -3 12 6
2 -4 2 -8 -4 0
That factors P(x) as
P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(2x⁴- 4x³ + 2x² - 8x - 4)
We can factor 2 out of that last parentheses:
P(x) = (x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(2)(x⁴- 2x³ + x² - 4x - 2)
P(x) = 2(x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(x⁴- 2x³ + x² - 4x - 2)
Now we must factor the last polynomial:
x⁴- 2x³ + x² - 4x - 2
Rearrange as
x⁴+ x² - 2 - 2x³ - 4x
Factor -2x out of the last two terms:
(x⁴+ x² - 2) - 2x(x² + 2)
Factor the first parentheses:
(x² + 2)(x² - 1) - 2x(x² + 2)
Factor out (x² + 2)
(x² + 2)[(x² - 1) - 2x]
(x² + 2)(x² - 1 - 2x)
(x² + 2)(x² - 2x - 1)
Now we have factored P(x) as
P(x) = 2(x - 1)(x - 1)(x - 1)(x + 3)(x + 3/2)(x² + 2)(x² - 2x - 1)
P(x) = 2(x - 1)³(x + 3)(x + 3/2)(x² + 2)(x² - 2x - 1)
We set x² + 2 = 0
x² = -2 _
x = ±i√2
We set x² - 2x - 1 = 0
_
x = 1 ± √2
So, 1 is a zero of multiplicity 3,
3/2 is a zero
_ _
i√2 and -i√2 are zeros
_ _
1 + √2 and 1 - √2 are zeros.
That's really a humdinger of a problem!
Edwin