Answer:
A = 1, B = 7, C = 4
Step-by-step explanation:
[tex] 2x^2 +5x-C= 2(x - 2)(Ax+1)+Bx\\
2x^2 +5x-C= (2x - 4)(Ax+1)+Bx\\
2x^2 +5x-C= 2Ax^2+2x - 4Ax-4+Bx\\
2x^2 +5x-C= 2Ax^2 + (2-4A+B)x-4\\[/tex]
Equating like terms on both sides:
[tex] 2x^2 = 2Ax^2\\
\frac{2x^2}{2x^2} = A\\
\huge \purple {\boxed {\implies A = 1}} \\\\
5x = (2-4A+B)x\\
\frac{5x}{x} =2-4A + B\\
2-4A + B = 5....(1)\\
[Plug \: A = 1\: in\: equation \: (1)] \\
2-4\times 1 + B = 5\\
2-4 +B = 5\\
-2 + B = 5\\
B=5+2\\
\huge \purple {\boxed {\implies B = 7}} \\\\
-C = - 4\\
\huge \purple {\boxed {\implies C = 4}}[/tex]
