How do you express $(4x^2 - 7x - 12) / ((x) (x+2) (x-3))$ in partial fractions?

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Answer 1 · 2016-12-16T06:52:35

The answer is $=2/(x)+(9/5)/(x+2)+(1/5)/(x-3)$

Let's do the decomposition into partial fractions

$(4x^2-7x-12)/(x(x+2)(x-3))=A/(x)+B/(x+2)+C/(x-3)$

$=(A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2))/(x(x+2)(x-3))$

Therefore,

$4x^2-7x-12=A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2)$

Let $x=0$ , $=>$ , $-12=-6A$ , $=>$ , $A=2$

Let $x=-2$ , $=>$ , $18=10B$ , $=>$ , $B=9/5$

Let $x=3$ , $=>$ , $3=15C$ , $=>$ , $C=1/5$

So,

$(4x^2-7x-12)/(x(x+2)(x-3))=2/(x)+(9/5)/(x+2)+(1/5)/(x-3)$

Answer 2 · 2016-12-16T06:53:04

$(4x^2-7x-12)/(x(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))$

Let $(4x^2-7x-12)/(x(x+2)(x-3))hArrA/x+B/(x+2)+C/(x-3)$

Hence $(4x^2-7x-12)/(x(x+2)(x-3))hArr(A(x+2)(x-3)+Bx(x-3)+Cx(x+2))/(x(x+2)(x-3))$

i.e. $4x^2-7x-12hArrA(x+2)(x-3)+Bx(x-3)+Cx(x+2)$

Now putting $x=0$ , we get $-6A=-12$ or $A=2$

putting $x=3$ , we get $15C=3$ or $C=1/5$

and putting $x=-2$ , we get $10B=16+14-12=18$ or $B=9/5$

Hence $(4x^2-7x-12)/(x(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))$

How do you express (4x^2 - 7x - 12) / ((x) (x+2) (x-3))(4x^2 - 7x - 12) / ((x) (x+2) (x-3)) in partial fractions?