-2x^3-8x^2+6x+36−2x3−8x2+6x+36
=-2(x^3+4x^2-3x-18)=−2(x3+4x2−3x−18)
Let f(x) = x^3+4x^2-3x-18f(x)=x3+4x2−3x−18
By the rational roots theorem, any rational root of f(x) = 0f(x)=0 of the form p/qpq in lowest terms must be such that pp is a divisor of 1818 and qq is a divisor of 11. That is:
x = +-1x=±1, +-2±2, +-3±3, +-6±6, +-9±9 or +-18±18
Trying the first few:
f(1) = 1+4-3-18 = -16f(1)=1+4−3−18=−16
f(-1) = -1+4+3-18 = -12f(−1)=−1+4+3−18=−12
f(2) = 8+16-6-18 = 0f(2)=8+16−6−18=0
So x=2x=2 is a root of f(x) = 0f(x)=0 and (x-2)(x−2) is a factor of f(x)f(x)
x^3+4x^2-3x-18 = (x-2)(x^2+6x+9)x3+4x2−3x−18=(x−2)(x2+6x+9)
x^2+6x+9x2+6x+9 is recognisable as a perfect square trinomial:
It is of the form a^2+2ab+b^2 = (a+b)^2a2+2ab+b2=(a+b)2 with a=xa=x and b=3b=3, so
(x^2+6x+9) = (x+3)^2(x2+6x+9)=(x+3)2
Putting this all together we get:
-2x^3-8x^2+6x+36 = -2(x-2)(x+3)(x+3)−2x3−8x2+6x+36=−2(x−2)(x+3)(x+3)
So -2x^3-8x^2+6x+36=0−2x3−8x2+6x+36=0 has one root x=2x=2 and one repeated root x=-3x=−3 with multiplicity 22.
