# Thread: Equation with x,A,B,C need A,B,C

1. ## Equation with x,A,B,C need A,B,C

This is for integration of rational functions but there's a part I don't understand that is pure algebra.

2x + 4 = Ax(x-2) + B(x - 2) + Cx^2

In the book they use x = 0 and x = 2 to get B = -2 and C = 2, respectively. This makes perfect sense to me but they get A = -2 by

equating the coefficients of x^2 on the two sides

I really don't know what they mean by this. I know what a coefficient is but I still do not follow. Can anyone explain or show how they would do this. My book assumes a ton of knowledge from classes I took like 10 years ago

2. Originally Posted by TYTY
2x + 4 = Ax(x-2) + B(x - 2) + Cx^2

In the book they use x = 0 and x = 2 to get B = -2 and C = 2, respectively. This makes perfect sense to me but they get A = -2 by

equating the coefficients of x^2 on the two sides

I really don't know what they mean by this. I know what a coefficient is but I still do not follow. Can anyone explain or show how they would do this. My book assumes a ton of knowledge from classes I took like 10 years ago

If you expand the right-hand side you get

$2x+4=Ax^2-2Ax + Bx - 2B + Cx^2$

$\Rightarrow2x+4=(A+C)x^2+(-2A + B)x - 2B$

The coefficient of $x^2$ on the left-hand side is 0 (if you wanted, you could write $0x^2$ on the left), and on the right-hand side it is $A+C.$ So, $A+C=0,$ and since $C=2,$ we get $A=-2.$

3. Originally Posted by Reckoner

If you expand the right-hand side you get

$2x+4=Ax^2-2Ax + Bx - 2B + Cx^2$

$\Rightarrow2x+4=(A+C)x^2+(-2A + B)x - 2B$

The coefficient of $x^2$ on the left-hand side is 0 (if you wanted, you could write $0x^2$ on the left), and on the right-hand side it is $A+C.$ So, $A+C=0,$ and since $C=2,$ we get $A=-2.$
Yeah I understand everything you said but I don't understand why the coefficient of x^2 is relevant, e.g. why can we ignore everything but the coefficient of x^2 on both sides? There is something I am completely ignorant of here.

4. Originally Posted by TYTY
Yeah I understand everything you said but I don't understand why the coefficient of x^2 is relevant, e.g. why can we ignore everything but the coefficient of x^2 on both sides? There is something I am completely ignorant of here.
There is NO $x^2$term on the left hand side of the equation, so your factors $(A + C)x^2$ must be equal to 0. You aren't ignoring everything, you are placing values with everything, but in terms of the coefficient for your like terms, there isn't a $x^2$ term to equate with, that is why it is zero. The other two groups of like terms on the right side DO have terms on the left and so you put those equal to the coefficients of those like terms.

5. Originally Posted by TYTY
Yeah I understand everything you said but I don't understand why the coefficient of x^2 is relevant, e.g. why can we ignore everything but the coefficient of x^2 on both sides? There is something I am completely ignorant of here.
If you have two polynomial functions,

$f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$

and

$g(x)=b_0+b_1x+b_2x^2+\cdots+b_nx^n,$

$f$ and $g$ are equal if and only if $a_i=b_i$ for $0\leq i\leq n.$

So if the $x^m$ term in the polynomial on the left has a coefficient of $t$, the $x^m$ term in the polynomial on the right must also have a coefficient of $t$.