Factor Theorem...4

[nycmathguy]

Section 2.3
Question 60

x^3 + 2x^2 - 2x - 4 = 0 given x = sqrt{2}.

If x = sqrt{2} leads to 0, one factor of the polynomial is x - sqrt{2}.

Using synthetic division, I came up with
x^2 + (2 + 2sqrt{2}) + 2sqrt{2}.

Factoring I got (x + 2)(x + sqrt{2}).

List zeros:

x = -2, x = -sqrt{2}

You say?

 

[MathLover1]

you forgot x
x^2 + (2 + sqrt(2))x + 2sqrt(2).........factor
x^2 + 2x + 2sqrt(2)x + 2sqrt(2)
(x^2 + 2sqrt(2)x )+ (2x + 2sqrt(2))
x(x + 2sqrt(2) )+ 2(x + sqrt(2))
(x+ 2)(x + sqrt(2))

=> x^3 + 2x^2 - 2x - 4 = (x - sqrt(2))(x+ 2)(x + sqrt(2))

List zeros:
x =sqrt(2)
x=- 2
x =-sqrt(2)

 

[nycmathguy]

you forgot x
x^2 + (2 + sqrt(2))x + 2sqrt(2).........factor
x^2 + 2x + 2sqrt(2)x + 2sqrt(2)
(x^2 + 2sqrt(2)x )+ (2x + 2sqrt(2))
x(x + 2sqrt(2) )+ 2(x + sqrt(2))
(x+ 2)(x + sqrt(2))

=> x^3 + 2x^2 - 2x - 4 = (x - sqrt(2))(x+ 2)(x + sqrt(2))

List zeros:
x =sqrt(2)
x=- 2
x =-sqrt(2)

Not only did I forget x, I also forgot x = sqrt{2} as a third zero of the polynomial. I may have to read Ron Larson's chapter notes again. I started Section 2.3 in the middle of the moving process from Queens to Brooklyn.

This is not an excuse. I have been very stressed out lately. I just found out that a very painful procedure called prostate biopsy is needed before the year ends. This means that there's a good chance I may have prostate cancer. Prostate cancer is > math.