Respuesta :
Answer:
[tex]\boxed{\textsf{ The possible rational roots of equation are \textbf{ 2 , (-1) and (-3)}.}}[/tex]
Step-by-step explanation:
A cubic polynomial is given to us . And we need to find the rational roots . Rational roots means the roots which really exits and not imaginary . The given polynomial is :-
[tex]\sf\implies p(x)= x^3+2x^2-5x-6 [/tex]
On equating it with 0 , it becomes cubic equation. That is ,
[tex]\sf\implies x^3+2x^2-5x-6=0 [/tex]
Here we will factorise out the given equation and then equate it with 0 to find the roots . Here the constant term is 6 . So the possible factors is 6 is :-
[tex]\sf\implies Factors\ of \ 6 \ = \ \pm 1 , \pm 2 , \ \pm 3 , \ \pm 6. [/tex]
So put them one by one checking out the factors .
Put x = 2 :-
[tex]\sf\implies p(x)= x^3+2x^2-5x-6\\\\\sf\implies p(2)= 2^3 +2(2)^2 -5(2)-6\\\\\sf\implies p(2)= 8 +8 -10 -6 \\\\\sf\implies p(2)=16-16\\\\\sf\implies \boxed{\sf p(2)= 0 }[/tex]
This implies that (x-2) is a factor of p(x) . Now let's divide the polynomial by (x-2) .
[tex]\rule{200}2[/tex]
x-2 ) x³ + 2x² - 5x -6 ( x² +4x +3
⠀⠀⠀ (-) x³(+) -2x²
_____________
⠀⠀⠀+4x² -5x - 6
⠀⠀⠀ (-) 4x² (+) -8x
_____________
⠀⠀⠀ 3x -6
⠀⠀⠀(+) 3x(+) - 6
_______________
⠀⠀⠀⠀ 0
Hence we can write , x³ +2x² -5x -6 as ,
[tex]\sf\implies x^3+2x^2-5x-6 = (x-2)(x^2 +4x+3) [/tex]
Equating it with 0 .
[tex]\sf\implies x^3+2x^2-5x-6 = 0\\\\\sf\implies (x-2)(x^2 +4x+3) = 0 \\\\\sf\implies (x-2)(x^2+3x+x+3)=0 \\\\\sf\implies (x-2)[ x(x+3)+1(x+3)]=0 \\\\\sf\implies (x-2)(x+1)(x+3)=0 \\\\\sf\implies\boxed{\pink{\frak { x = 2 , (-1) , (-3) }}}[/tex]
