Solve the polynomial by finding all real roots
G(x)= x^3 + 3x^2 +9x+ 27
I got x^2 (x+3) +3(x+9) so far
Thank you
Recall that
a^2 - b^2 = (a-b)(a+b)
a^3-b^3 = (a-b)(a^2+ab+b^2)
a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3)
So,
x^3 + 3x^2 +9x+ 27 = (x^4-3^4)/(x-1)
= (x-3)(x+3)(x^2+9)/(x-3)
= (x+3)(x^2+9)
So x = -3 is the only real root
Or,
Descartes' Rule of Signs says there are no positive roots
The Rational Root Theorem says that the roots must be among -1,-3,-9,-27
So, trying -3, that works, meaning (x+3) is a factor
Divide by (x+3) to get (x^2+9)