Asked by damilous
                EXPRESS 15x^2-x+2\:=\frac{a}{\left(x-5\right)}+\frac{\left(bx+c\right)}{\left(3x^2+4x-2\right)} IN PARTIAL FRACTION
            
            
        Answers
                    Answered by
            damilous
            
    EXPRESS 15x^2-x+2\(x-5)(3x^2+4x-2) IN PARTIAL FRACTION
    
                    Answered by
            Reiny
            
    Write it this way:
(15x^2-x+2)/( (x-5)(3x^2+4x-2) )
assume:
(15x^2-x+2)/( (x-5)(3x^2+4x-2) ) = A/(x-5) + (Bx+C)/(3x^2+4x-2)
= ( A(3x^2+4x-2) + (Bx+C)(x-5) )/ ( (x-5)(3x^2+4x-2) )
then:
A(3x^2+4x-2) + (Bx+C)(x-5) = 15x^2-x+2
let x=5 ----> 93A = 372 , A = 4
So you have
4(3x^2+4x-2) + (Bx+C)(x-5) = 15x^2-x+2
let x = 1
20 + (B+C)(-4) = 16
B+C = 1 , B = 1 - C
let x = 0
-8 + C(-5) = 2
C = -2
then B = 3
finally:
(15x^2-x+2)/( (x-5)(3x^2+4x-2) ) = 4/(x-5) + (3x - 2)/(3x^2+4x-2)
    
(15x^2-x+2)/( (x-5)(3x^2+4x-2) )
assume:
(15x^2-x+2)/( (x-5)(3x^2+4x-2) ) = A/(x-5) + (Bx+C)/(3x^2+4x-2)
= ( A(3x^2+4x-2) + (Bx+C)(x-5) )/ ( (x-5)(3x^2+4x-2) )
then:
A(3x^2+4x-2) + (Bx+C)(x-5) = 15x^2-x+2
let x=5 ----> 93A = 372 , A = 4
So you have
4(3x^2+4x-2) + (Bx+C)(x-5) = 15x^2-x+2
let x = 1
20 + (B+C)(-4) = 16
B+C = 1 , B = 1 - C
let x = 0
-8 + C(-5) = 2
C = -2
then B = 3
finally:
(15x^2-x+2)/( (x-5)(3x^2+4x-2) ) = 4/(x-5) + (3x - 2)/(3x^2+4x-2)
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