Expand the right side.
[tex]A(x+2)(x-4) = A(x^2 - 2x - 8) = Ax^2 - 2Ax - 8A[/tex]
[tex]B(x-4)(x-1) = B(x^2-5x+4) = Bx^2-5Bx+4B[/tex]
[tex]C(x-1)(x+2) = C(x^2+x-2) = Cx^2 + Cx - 2C[/tex]
Then
[tex]4x^2 + 4x - 26 = (A + B + C)x^2 + (-2A - 5B + C)x + (-8A + 4B - 2C)[/tex]
Two polynomials are equal if the coefficients on like-power terms are equal, so
[tex]\begin{cases} A + B + C = 4 \\ -2A - 5B + C = 4 \\ -8A + 4B - 2C = -26 \end{cases}[/tex]
Solve the system of equations.
Eliminating [tex]C[/tex], we have
[tex](A + B + C) - (-2A - 5B + C) = 4 - 4 \\\\ ~~~~ \implies 3A + 6B = 0 \\\\ ~~~~ \implies A + 2B = 0[/tex]
[tex]2(A + B + C) + (-8A + 4B - 2C) = 8 - 26 \\\\ ~~~~ \implies -6A + 6B = -18 \\\\ ~~~~ \implies -A + B = -3[/tex]
Eliminating [tex]B[/tex],
[tex](A + 2B) - 2(-A + B) = 0 + 6 \\\\ ~~~~ \implies 3A = 6 \\\\ ~~~~ \implies \boxed{A = 2}[/tex]
Solve for [tex]B[/tex] and [tex]C[/tex].
[tex]A + 2B = 0 \\\\ ~~~~ \implies B = -\dfrac A2 \\\\ ~~~~ \implies \boxed{B = -1}[/tex]
[tex]A + B + C = 4 \\\\ ~~~~ \implies C = 4 - A - B \\\\ ~~~~ \implies \boxed{C = 3}[/tex]
