Since (x + 1), (x - 4) are factors of
[tex]x^3+ax^2+bx+c[/tex]
That means if x = -1 make the polynomial equal to 0, and x = 4 also make it = 0
Then we can make 2 equations from these informations
[tex]\begin{gathered} x=-1 \\ (-1)^3+a(-1)^2+b(-1)+c=0 \\ -1+a-b+c=0 \\ a-b+c=1\rightarrow(1) \end{gathered}[/tex][tex]\begin{gathered} x=4 \\ (4)^3+a(4)^2+b(4)+c=0 \\ 64+16a+4b+c=0 \\ 16a+4b+c=-64\rightarrow(2) \end{gathered}[/tex]
Since when dividing it by x + 2 the remainder is 6
That means when substitute x by -2 the answer will be 6
[tex]\begin{gathered} (-2)^3+a(-2)^2+b(-2)+c=6 \\ -8+4a-2b+c=6 \\ 4a-2b+c=6+8 \\ 4a-2b+c=14\rightarrow(3) \end{gathered}[/tex]
Now, we have a system of equations to solve it to find a, b, c
We will Subtract equation (1) from equation (2) to eliminate c
[tex]\begin{gathered} (16a-a)+(4b--b)+(c-c)=(-64-1) \\ 15a+5b=-65 \\ \frac{15a}{5}+\frac{5b}{5}=\frac{-65}{5} \\ 3a+b=-13\rightarrow(4) \end{gathered}[/tex]
Subtract equation (1) from equation (3) to eliminate c
[tex]\begin{gathered} (4a-a)+(-2b--b)=(14-1) \\ 3a-b=13\rightarrow(5) \end{gathered}[/tex]
Subtract equation (5) from equation (4) to eliminate a
[tex]\begin{gathered} (3a-3a)+(b--b)=(-13-13) \\ 0+2b=-26 \\ 2b=-26 \\ \frac{2b}{2}=\frac{-26}{2} \\ b=-13 \end{gathered}[/tex]
Substitute the value of b in equation (4) to find a
[tex]\begin{gathered} 3a+(-13)=-13 \\ 3a-13+13=-13+13 \\ 3a=0 \\ a=0 \end{gathered}[/tex]
Substitute the values of a and b in equation (1) to find c
[tex]\begin{gathered} 0-(-13)+c=1 \\ 0+13+c=1 \\ 13+c=1 \\ 13-13+c=1-13 \\ c=-12 \end{gathered}[/tex]
The values of a, b, c are
a = 0
b = -13
c = -12
