Answer:
[tex]\dfrac{5x^2+20x+16}{x(x+1)^2} & \equiv \dfrac{16}{x}-\dfrac{11}{(x+1)}-\dfrac{1}{(x+1)^2}[/tex]
Step-by-step explanation:
When writing an algebraic fraction as an identity, if its denominator has repeated linear factors, the power of the repeated factor indicates how many times that factor should appear in the partial fractions.
Write out the fraction as an identity:
[tex]\begin{aligned}\dfrac{5x^2+20x+16}{x(x+1)^2} & \equiv \dfrac{A}{x}+\dfrac{B}{(x+1)}+\dfrac{C}{(x+1)^2}\\\\\implies \dfrac{5x^2+20x+16}{x(x+1)^2} & \equiv \dfrac{A(x+1)^2}{x(x+1)^2}+\dfrac{Bx(x+1)}{x(x+1)^2}+\dfrac{Cx}{x(x+1)^2}\\\\\implies 5x^2+20x+16 & \equiv A(x+1)^2+Bx(x+1)+Cx\end{aligned}[/tex]
Calculate the values of A and C using substitution:
[tex]\begin{aligned}\textsf{When }x=0 \implies 16 & =A(1)+B(0)+C(0) \implies A=16\\\\\textsf{When }x=-1 \implies 1 & =A(0)+B(0)+C(-1) \implies C=-1\end{aligned}[/tex]
Substitute the found values of A and C and expand the identity:
[tex]\begin{aligned}\implies 5x^2+20x+16 & \equiv 16(x+1)^2+Bx(x+1)-x\\& \equiv 16x^2+32x+16+Bx^2+Bx-x\\& \equiv (16+B)x^2+(31+B)x+16\\\end{aligned}[/tex]
Compare the coefficients of the x² term to find B:
[tex]x^2 \textsf{ term } \implies 16+B =5\implies B & =-11[/tex]
Replace the found values of A, B and C in the original identity:
[tex]\dfrac{5x^2+20x+16}{x(x+1)^2} & \equiv \dfrac{16}{x}-\dfrac{11}{(x+1)}-\dfrac{1}{(x+1)^2}[/tex]
Learn more about partial fractions here:
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