Find All Real Solutions

[nycmathguy]

Section 2.5
Question 32

Possible Rational Zeros:

±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42

With so many possibilities, it's best to graph the equation.

Solutions:

x = -3, x = -2

This means (x + 2) and (x + 3) are factors of the polynomial.

After applying synthetic division, I found the following trinomial as a third factor: x^2 + 3x -7.

Applying the quadratic formula, I found two extra solutions:

x = (-3 -sqrt{37})/2

x = (-3 + sqrt{37})/2

Note:

The original equation can be expressed as
(x + 2)(x + 3)(x^2 + 3x - 7) = 0.

You say?

 

[MathLover1]

correct

but you can use last two zeros too and expressed the equation as

(x + 2)(x + 3)(x-(-3 -sqrt(37))/2))(x-(-3 + sqrt(37)/2)) = 0

 

[nycmathguy]

correct

but you can use last two zeros too and expressed the equation as

(x + 2)(x + 3)(x-(-3 -sqrt(37))/2))(x-(-3 + sqrt(37)/2)) = 0

Thanks for letting me know. Section 2.5 is very detailed. I like it but very tedious.