Rational Zeros of A Polynomial

[nycmathguy]

Section 2.5
Question 96

Can you work out 96 as a guide for me to do the rest? This is the final Section 2.5 question for us.

 

[MathLover1]

f(x)=x^3-(3/2)x^2-(23/2)x+6....factor out 1/2

f(x)=(1/2)(2x^3-3x^2-23x+12)

factor (2x^3-3x^2-23x+12)
write
-3x^2 as -x^2-2x^2
and -23x as x-24x

2x^3-x^2-2x^2+x-24x+12....group

(2x^3-x^2)-(2x^2-x)-(24x-12)

x^2(2x-1)-x(2x-1)-12(2x-1)

(2x - 1) (x^2 - x - 12)

now factor (x^2 - x - 12)
x^2 - x - 12
x^2 +3x- 4x - 12
(x^2 +3x)- (4x +12)
x(x +3)- 4(x +3)
(x + 3) (x - 4)

then you have
f(x)=(2x - 1) (x + 3) (x - 4)

 

[nycmathguy]

f(x)=x^3-(3/2)x^2-(23/2)x+6....factor out 1/2

f(x)=(1/2)(2x^3-3x^2-23x+12)

factor (2x^3-3x^2-23x+12)
write
-3x^2 as -x^2-2x^2
and -23x as x-24x

2x^3-x^2-2x^2+x-24x+12....group

(2x^3-x^2)-(2x^2-x)-(24x-12)

x^2(2x-1)-x(2x-1)-12(2x-1)

(2x - 1) (x^2 - x - 12)

now factor (x^2 - x - 12)
x^2 - x - 12
x^2 +3x- 4x - 12
(x^2 +3x)- (4x +12)
x(x +3)- 4(x +3)
(x + 3) (x - 4)

then you have
f(x)=(2x - 1) (x + 3) (x - 4)

I can take it from here. Thank you.