Answer:
Step-by-step explanation:
As it is a third order polynomial, we expect no more than three x axis touches or three factors.
factors have a form of (ax¹ + bx⁰)
As 27 is negative, we will have only one factor with a negative x⁰ value meaning only one crossing point is positive. The other two must be negative and have positive x⁰ values
as the x³ term has a multiplier of 2, the likely multipliers for the x¹ terms are 2, 1, 1
p(x) = ( ?x + ?)(?x + ?)(?x - ?)
Let's guess that a factor is (x + 3)
2x² - 3x - 9
x + 3 | 2x³ + 3x² - 18x - 27
- (2x³ + 6x²)
-3x² - 18x
- (-3x² - 9x )
-9x - 27
- (-9x - 27)
0
so (x + 3) is a factor and x = -3 is a zero
We are left with a quadratic which we can use the formula on
2x² - 3x - 9
x = (3 ±√(3² - 4(2)(-9))) / (2(2))
x = (3 ±√(81)) / 4
x = (3 ± 9) / 4
x = 12/4 = 3
or
x = -6/4 = -1.5
so the zeros are at -3, -1.5, and 3
