Answer :
If a polynomial is divisible by [tex]x-a[/tex], then [tex]a[/tex] is a root of this polynomial.
Here (x-a) is equal to x+2:
[tex]x-a=x+2 \\ -a=2 \\ a=-2[/tex]
So -2 is a root of the polynomial. That means x³-6x²+kx+10=0 for x=-2. Now substitute -2 for x in the equation and solve for k:
[tex](-2)^3-6 \times (-2)^2+ k \times (-2)+10=0 \\ -8-6 \times 4-2k+10=0 \\ -8-24-2k+10=0 \\ -2k-22=0 \\ -2k=22 \\ \boxed{k=-11}[/tex]
Here (x-a) is equal to x+2:
[tex]x-a=x+2 \\ -a=2 \\ a=-2[/tex]
So -2 is a root of the polynomial. That means x³-6x²+kx+10=0 for x=-2. Now substitute -2 for x in the equation and solve for k:
[tex](-2)^3-6 \times (-2)^2+ k \times (-2)+10=0 \\ -8-6 \times 4-2k+10=0 \\ -8-24-2k+10=0 \\ -2k-22=0 \\ -2k=22 \\ \boxed{k=-11}[/tex]
