The number of real roots of equation X3 + 6x2 + 9x + 4 = 0 is () and the answer is 2 x followed by several powers -3 and - 1 do wrong friends, please look through the problem, this is the derivative part of the process, let its derivative = 0, and then find - 1, - 3 did not understand what the real root refers to

The number of real roots of equation X3 + 6x2 + 9x + 4 = 0 is () and the answer is 2 x followed by several powers -3 and - 1 do wrong friends, please look through the problem, this is the derivative part of the process, let its derivative = 0, and then find - 1, - 3 did not understand what the real root refers to

x^3+6x^2+9x+4
=(x+1)(x^2+5x+4)
=(x+1)(x+1)(x+4)
So x = - 1, x = - 4 are the roots of the equation
So there are two roots
So you're in high school
That's easy to say
f'(x)=3x^2+12x+9=3(x+3)(x+1)
So when x = - 1, - 3, the slope of F (x) is 0
That is, x = - 1 and x = - 3 are maxima
F (x) is an increasing function at (- infinity, - 3], a decreasing function at [- 3, - 1], and an increasing function at [- 1, + infinity)
And f (- 1) = 0
f(-3)=4>0
So f (x) has a root on (- infinity, - 3), no root on [- 3, - 1), and a root on [1, + infinity]
So there are two roots