How do you use the rational root theorem to find the roots of f(x)=x^3+x^2-8x-6 f(x)=x3+x2−8x−6?
1 Answer
The zeros of
Explanation:
Given:
f(x) = x^3+x^2-8x-6f(x)=x3+x2−8x−6
The rational root theorem tells us that any rational zero of
That means that the only possible rational zeros are:
+-1, +-2, +-3, +-6±1,±2,±3,±6
Trying each of these in turn, we eventually find:
f(-3) = (color(blue)(-3))^3+(color(blue)(-3))^2-8(color(blue)(-3))-6 = -27+9+24-6 = 0f(−3)=(−3)3+(−3)2−8(−3)−6=−27+9+24−6=0
So
x^3+x^2-8x-6 = (x+3)(x^2-2x-2)x3+x2−8x−6=(x+3)(x2−2x−2)
We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:
A^2-B^2 = (A-B)(A+B)A2−B2=(A−B)(A+B)
with
x^2-2x-2 = x^2-2x+1-3x2−2x−2=x2−2x+1−3
color(white)(x^2-2x-2) = (x-1)^2-(sqrt(3))^2x2−2x−2=(x−1)2−(√3)2
color(white)(x^2-2x-2) = ((x-1)-sqrt(3))((x-1)+sqrt(3))x2−2x−2=((x−1)−√3)((x−1)+√3)
color(white)(x^2-2x-2) = (x-1-sqrt(3))(x-1+sqrt(3))x2−2x−2=(x−1−√3)(x−1+√3)
Hence the other two zeros are:
x = 1+-sqrt(3)x=1±√3
