Answer:
Zeroes : 1, 4 and -5.
Potential roots: [tex]\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20[/tex].
Step-by-step explanation:
The given equation is
[tex]x^3-21x=-20[/tex]
It can be written as
[tex]x^3+0x^2-21x+20=0[/tex]
Splitting the middle terms, we get
[tex]x^3-x^2+x^2-x-20x+20=0[/tex]
[tex]x^2(x-1)+x(x-1)-20(x-1)=0[/tex]
[tex](x-1)(x^2+x-20)=0[/tex]
Splitting the middle terms, we get
[tex](x-1)(x^2+5x-4x-20)=0[/tex]
[tex](x-1)(x(x+5)-4(x+5))=0[/tex]
[tex](x-1)(x+5)(x-4)=0[/tex]
Using zero product property, we get
[tex]x-1=0\Rightarrow x=1[/tex]
[tex]x-4=0\Rightarrow x=4[/tex]
[tex]x+5=0\Rightarrow x=-5[/tex]
Therefore, the zeroes of the equation are 1, 4 and -5.
According to rational root theorem, the potential root of the polynomial are
[tex]x=\dfrac{\text{Factor of constant}}{\text{Factor of leading coefficient}}[/tex]
Constant = 20
Factors of constant ±1, ±2, ±4, ±5, ±10, ±20.
Leading coefficient= 1
Factors of leading coefficient ±1.
Therefore, the potential root of the polynomial are [tex]\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20[/tex].
