Question

Identify all of the real roots of
x^3 − 3x^2 − 33x + 35 = 0.

Answer 1

Answer:

$\text{1, -5, 7}$

Step-by-step explanation:

$\text{Here, we know that the factors of 35 are 1, 3, 5 and 7.}\\\text{So we will test values of }x\text{ as }\pm1,\ \pm3,\ \pm5\text{ and }\pm7\text{ one by one until we find }\\\text{one root of the equation.}$

$\text{When }x=1,\ x^3-3x^2-33x+35=1^3-3(1)^2-33(1)+35=0\\\therefore\ (x-1)\text{ is a factor.}$

$\text{Now we divide the terms except for the constant of the expression }\\x^3-3x^2-33x+35=0\text{ in such a way that each pair has }(x-1)\text{ as a common}\\\text{factor.}$

$x^3-3x^2-33x+35=0\\\text{or, }x^3-x^2-2x^2+2x-35x+35x=0\\\text{or, }x^2(x-1)-2x(x-1)-35(x-1)=0\\\text{or, }(x-1)(x^2-2x-35)=0\\\text{or, }(x-1)\{x^2-7x+5x-35\}=0\\\text{or, }(x-1)\{x(x-7)+5(x-7)\}=0\\\text{or, }(x-1)(x+5)(x-7)=0\\\therefore\ x=1,\ -5\text{ or }7$