Answer: The answer is (b) [tex]x^3-x^2-26x-24.[/tex]
Step-by-step explanation: We are given four polynomials and we are to check which one of them has roots -4, -1 and 6. Obviously, if putting these three values of 'x' in a polynomials yields 0, then that particular value will be a root of that polynomial.
Let us denote the polynomials as follows -
[tex]P(x)=x^3-9x^2-22x+24,\\\\Q(x)=x^3-x^2-26x-24,\\\\R(x)=x^3+x^2-26x+24\\\\\textup{and}\\\\S(x)=x^3+9x^2+14x-24.[/tex]
Let us check for x = -1 first. So, substituting x = -1 in all the four polynomils, we get
[tex]P(-1)=36\neq 0,~~Q(-1)=0,~~R(x)=50\neq 0~~\textup{and}~~S(x)=-30\neq 0.[/tex]
Therefore, only possibility is Q(x).
If we put x = -4 and x = 6 in Q(x), we find that
[tex]Q(-4)=0~~\textup{and}~~Q(6)=0.[/tex]
Thus, the correct option is (b) [tex]x^3-x^2-26x-24.[/tex]
