praisemickey06
Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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]

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.