praisemickey06
Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Someone pls help me ASAP! Express the following as the sum of its partial fraction: 2x+4/x(x+2)(x-4)
[tex] \frac{2x + 4}{(x + 2)(x - 4)} [/tex]
​

Sagot :

semsee45

Answer:

[tex]\dfrac{2x+4}{x(x+2)(x-4)} \equiv \dfrac{1}{2(x-4)}-\dfrac{1}{2x}[/tex]

Step-by-step explanation:

Partial Fractions

Write out the expression as an identity:

[tex]\begin{aligned}\dfrac{2x+4}{x(x+2)(x-4)} & \equiv \dfrac{A}{x}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-4)}\\\\\implies \dfrac{x(2x+4)(x+2)(x-4)}{x(x+2)(x-4)} & \equiv \dfrac{Ax(x+2)(x-4)}{x}+\dfrac{Bx(x+2)(x-4)}{(x+2)}+\dfrac{Cx(x+2)(x-4)}{(x-4)}\\\\\implies 2x+4 & \equiv A(x+2)(x+4)+ Bx(x-4)+Cx(x+2)\end{aligned}[/tex]

Calculate the values of A, B and C using substitution:

[tex]\begin{aligned}2x+4 & = A(x+2)(x-4)+Bx(x-4)+Cx(x+2)\\\\x=4 \implies 12 & = A(0)+B(0)+C(24)\implies C=\dfrac{1}{2}\\\\x=-2 \implies 0 & = A(0)+B(12)+C(0) \implies B=0\\\\ x=0 \implies 4 & = A(-8)+B(0)+C(0) \implies A=-\dfrac{1}{2}\end{aligned}[/tex]

Replace A, B and C in the original identity:

[tex]\begin{aligned}\dfrac{2x+4}{x(x+2)(x-4)} & \equiv \dfrac{A}{x}+ \dfrac{B}{(x+2)}+\dfrac{C}{(x-4)}\\\\& \equiv -\dfrac{1}{2x}+\dfrac{1}{2(x-4)}\\\\\implies \dfrac{2x+4}{x(x+2)(x-4)}& \equiv \dfrac{1}{2(x-4)}-\dfrac{1}{2x}\end{aligned}[/tex]

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.