In O level Additional Mathematics, there is a small section on Identity inside the topic of Factor & Remainder Theorem. Today I am going to share with you the 2 different approaches to solve this kind of questions.

**Substitution Method **(My preferred method)

I am going to use the question below to show you the step by step solutions of both methods.

**Given that [pmath]3x^2+x-2=A(x-1)(x+2)+B(x-1)+C[/pmath] for all values of ***x*, find the value of A, of B and of C.

**Let ***x* = 1,

[pmath]3+1-2 = C[/pmath]

[pmath]C=2[/pmath]

**Let ***x* = -2,

[pmath]3(4)-2-2 = B(-3) + 2[/pmath]

[pmath]3(4)-2-2 = B(-3) + 2[/pmath]

[pmath]B= -2[/pmath]

**Let ***x* = 0,

[pmath]-2 = -2A+ 2 + 2[/pmath]

[pmath]A = 3[/pmath]

Thus A = 3, B = -2 and C = 2

**Concept behind the Subsitution method:** The value of *x* choosen will cause one or more of the unknowns to be "cancel off", leaving just 1 unknown left. For example, when I choose *x* = 1 in the first subsituition, A & B are eliminated, allowing me to find 'C'.

**Comparing Coefficients Method**

**Given that [pmath]3x^2+x-2=A(x-1)(x+2)+B(x-1)+C[/pmath] for all values of ***x*, find the value of A, of B and of C.

By comparing coefficient of [pmath]x^2[/pmath]:

LHS: 3 = A => A = 3

By comparing coefficient of [pmath]x[/pmath]:

LHS: 1 = 2A - A + B => B = -2

By comparing coefficient of [pmath]x^0[/pmath]:

LHS: -2 = -2A - B + C => C = 2

Thus A = 3, B = -2 and C = 2

**Concept behind the Comparing Coefficient method:** Expansion is usually required on one side of the equation. It takes up time. The reason for the insignificant working shown is due to the fact that the expansion is done mentally instead of written. This method is highly recommended if there is more than 1 unknown other than *x *on the left hand side of the equation. For example, there's an unknown 'D' on the left hand side of the equation.

**Which method do you usually use? And which method does your school teach you? Leave me your answer in the comment section below. **

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