Given: (2x^3+3x^2-15x+5)/(x^2+3x-4)2x3+3x2−15x+5x2+3x−4
We cannot do partial fraction expansion until we do the implied division so that the power of the numerator is less than the power of the denominator. Then, we do partial fraction expansion on the remainder.
Do polynomial long division:
color(white)( (x^2+3x-4)/color(black)(x^2+3x-4))color(white)((2x^3+3x^2-15x+5))/(")" color(white)(x)2x^3+3x^2-15x+5)x2+3x−4x2+3x−4(2x3+3x2−15x+5))x2x3+3x2−15x+5
color(white)( (x^2+3x-4)/color(black)(x^2+3x-4))(2xcolor(white)(3x^2-15x+5))/(")" color(white)(x)2x^3+3x^2-15x+5)x2+3x−4x2+3x−42x3x2−15x+5)x2x3+3x2−15x+5
color(white)(....................)ul(-2x^3-6x^2+8x)
color(white)(.............................)-3x^2-7x+5
color(white)( (x^2+3x-4)/color(black)(x^2+3x-4))(2x-3color(white)(15x+5))/(")" color(white)(x)2x^3+3x^2-15x+5)
color(white)(....................)ul(-2x^3-6x^2+8x)
color(white)(.............................)-3x^2-7x+5
color(white)(..................................)ul(3x^2+9x-12)
color(white)(.............................................)2x-7
This means that:
(2x^3+3x^2-15x+5)/(x^2+3x-4) = 2x-3+(2x-7)/(x^2+3x-4)
We perform partial fraction expansion on the last term:
(2x-7)/(x^2+3x-4) = A/(x+4)+ B/(x-1)
Multiply both sides by the numerator:
2x-7 = A(x-1)+ B(x+4)
Let x = -4:
2(-4)-7 = A(-4-1)+ B(-4+4)
-15 = A(-5)+ B(0)
A = 3
Let x = 1:
2(1)-7 = A(1-1)+ B(1+4)
#-5 = A(0)+B(5)
B = -1
(2x^3+3x^2-15x+5)/(x^2+3x-4) = 2x-3+3/(x+4)-1/(x-1)
I double checked using WolframAlpha .
