given the equation fxx32x25x4 please answer what is the degr

given the equation f(x)=x^3+2x^2+5x+4 ..............please answer what is the degree??.......Also answer how many zeroes and how you found that out so I can learn it (including how many negative and positive zeroes).....also list the number of sign changes....also what are the possible real zeroes and explain how you figured that out or show work please.

Solution

f(x)=x^3+2x^2+5x+4

Degree of polynomial = 3

The factor of the leading coefficient (1) is 1 .The factors of the constant term (4) are 1 2 2 4 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±2/1, ±4/1

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(1)=0.

So, we have (x^3+2x^2+5x+4 )/(x+1) = x^2 +x +4

On solving we get : x= -1 , -1/2 + sqrt(15i)/2 , -1/2 - sqrt(15i)/2

Now graph sign change at x= -2; f(-2) = -8 +8 -5 +4 = -1 ---- (-ve)

at x= 2; f(2) = 8 +8 +8 +4 ---- (+ve)

One sign change: