Math / Algebra

QuestionFind the partial fraction decomposition of the following rational expression. $\begin{array}{l} \frac{12}{x\left(x^{2}+4\right)} \\ \frac{12}{x\left(x^{2}+4\right)}= \end{array}$ $\square$ (Use integers or fractions for any numbers in the expression.)

Studdy Solution

STEP 1

What is this asking? Rewrite this big fraction as the sum of smaller, simpler fractions! Watch out! Remember that $x^2 + 4$ can't be factored further using real numbers, so we'll need to handle it differently than if it were $x^2 - 4$ .

STEP 2

1. Set up the decomposition.
2. Multiply and simplify.
3. Solve for the unknowns.
4. Write the final decomposition.

STEP 3

We've got a factor of $x$ and a factor of $x^2 + 4$ , both in the denominator.
Since $x$ is just a regular linear factor, it gets a constant, say $A$ , on top.
So, we have a fraction $\frac{A}{x}$ .

STEP 4

Now, $x^2 + 4$ is a quadratic factor that can't be factored further.
That means it gets a linear term, say $Bx + C$ , on top.
So, we have another fraction $\frac{Bx + C}{x^2 + 4}$ .

STEP 5

Putting it all together, our partial fraction decomposition looks like this: $\frac{12}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$

STEP 6

To get rid of those pesky denominators, let's multiply both sides of our equation by $x(x^2 + 4)$ .
This gives us: $12 = A(x^2 + 4) + (Bx + C)x$

STEP 7

Now, let's expand and group like terms: $12 = Ax^2 + 4A + Bx^2 + Cx$ $12 = (A + B)x^2 + Cx + 4A$

STEP 8

Now comes the fun part!
We're going to match up the coefficients on both sides of the equation.
Notice that the left side has no $x^2$ term, so the coefficient of $x^2$ is zero.
This means $A + B = 0$ .

STEP 9

The left side also has no $x$ term, so $C = 0$ .
That simplifies things nicely!

STEP 10

Finally, the constant term on the left side is 12, so $4A = 12$ .
Dividing both sides by 4, we get $A = 3$ .

STEP 11

Since $A + B = 0$ and $A = 3$ , we have $3 + B = 0$ , which means $B = -3$ .

STEP 12

We found $A = 3$ , $B = -3$ , and $C = 0$ .
Plugging these values back into our decomposition setup, we get: $\frac{12}{x(x^2 + 4)} = \frac{3}{x} + \frac{-3x + 0}{x^2 + 4}$ which simplifies to: $\frac{12}{x(x^2 + 4)} = \frac{3}{x} - \frac{3x}{x^2 + 4}$

STEP 13

Our final answer is $\frac{3}{x} - \frac{3x}{x^2 + 4}$ .

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QuestionFind the partial fraction decomposition of the following rational expression. $\begin{array}{l} \frac{12}{x\left(x^{2}+4\right)} \\ \frac{12}{x\left(x^{2}+4\right)}= \end{array}$ $\square$ (Use integers or fractions for any numbers in the expression.)

Studdy Solution

STEP 1

What is this asking? Rewrite this big fraction as the sum of smaller, simpler fractions! Watch out! Remember that $x^2 + 4$ can't be factored further using real numbers, so we'll need to handle it differently than if it were $x^2 - 4$ .

STEP 2

1. Set up the decomposition.
2. Multiply and simplify.
3. Solve for the unknowns.
4. Write the final decomposition.

STEP 3

We've got a factor of $x$ and a factor of $x^2 + 4$ , both in the denominator.
Since $x$ is just a regular linear factor, it gets a constant, say $A$ , on top.
So, we have a fraction $\frac{A}{x}$ .

STEP 4

Now, $x^2 + 4$ is a quadratic factor that can't be factored further.
That means it gets a linear term, say $Bx + C$ , on top.
So, we have another fraction $\frac{Bx + C}{x^2 + 4}$ .

STEP 5

Putting it all together, our partial fraction decomposition looks like this: $\frac{12}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$

STEP 6

To get rid of those pesky denominators, let's multiply both sides of our equation by $x(x^2 + 4)$ .
This gives us: $12 = A(x^2 + 4) + (Bx + C)x$

STEP 7

Now, let's expand and group like terms: $12 = Ax^2 + 4A + Bx^2 + Cx$ $12 = (A + B)x^2 + Cx + 4A$

STEP 8

Now comes the fun part!
We're going to match up the coefficients on both sides of the equation.
Notice that the left side has no $x^2$ term, so the coefficient of $x^2$ is zero.
This means $A + B = 0$ .

STEP 9

The left side also has no $x$ term, so $C = 0$ .
That simplifies things nicely!

STEP 10

Finally, the constant term on the left side is 12, so $4A = 12$ .
Dividing both sides by 4, we get $A = 3$ .

STEP 11

Since $A + B = 0$ and $A = 3$ , we have $3 + B = 0$ , which means $B = -3$ .

STEP 12

We found $A = 3$ , $B = -3$ , and $C = 0$ .
Plugging these values back into our decomposition setup, we get: $\frac{12}{x(x^2 + 4)} = \frac{3}{x} + \frac{-3x + 0}{x^2 + 4}$ which simplifies to: $\frac{12}{x(x^2 + 4)} = \frac{3}{x} - \frac{3x}{x^2 + 4}$

STEP 13

Our final answer is $\frac{3}{x} - \frac{3x}{x^2 + 4}$ .

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Studdy Solution

STEP 1

What is this asking? Rewrite this big fraction as the sum of smaller, simpler fractions! Watch out! Remember that x2+4x^2 + 4x2+4 can't be factored further using real numbers, so we'll need to handle it differently than if it were x2−4x^2 - 4x2−4.

STEP 2

1. Set up the decomposition.2. Multiply and simplify.3. Solve for the unknowns.4. Write the final decomposition.

STEP 3

We've got a factor of xxx and a factor of x2+4x^2 + 4x2+4, both in the **denominator**.Since xxx is just a regular **linear factor**, it gets a constant, say AAA, on top.So, we have a fraction Ax\frac{A}{x}xA​.

STEP 4

Now, x2+4x^2 + 4x2+4 is a **quadratic factor** that can't be factored further.That means it gets a linear term, say Bx+CBx + CBx+C, on top.So, we have another fraction Bx+Cx2+4\frac{Bx + C}{x^2 + 4}x2+4Bx+C​.

STEP 5

Putting it all together, our **partial fraction decomposition** looks like this: 12x(x2+4)=Ax+Bx+Cx2+4 \frac{12}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4} x(x2+4)12​=xA​+x2+4Bx+C​

STEP 6

To get rid of those pesky **denominators**, let's multiply both sides of our equation by x(x2+4)x(x^2 + 4)x(x2+4).This gives us: 12=A(x2+4)+(Bx+C)x 12 = A(x^2 + 4) + (Bx + C)x 12=A(x2+4)+(Bx+C)x

STEP 7

Now, let's **expand** and **group like terms**: 12=Ax2+4A+Bx2+Cx 12 = Ax^2 + 4A + Bx^2 + Cx 12=Ax2+4A+Bx2+Cx 12=(A+B)x2+Cx+4A 12 = (A + B)x^2 + Cx + 4A 12=(A+B)x2+Cx+4A

STEP 8

Now comes the fun part!We're going to match up the **coefficients** on both sides of the equation.Notice that the left side has no x2x^2x2 term, so the coefficient of x2x^2x2 is **zero**.This means A+B=0A + B = 0A+B=0.

STEP 9

The left side also has no xxx term, so C=0C = 0C=0.That simplifies things nicely!

STEP 10

Finally, the **constant term** on the left side is **12**, so 4A=124A = 124A=12.Dividing both sides by **4**, we get A=3A = 3A=3.

STEP 11

Since A+B=0A + B = 0A+B=0 and A=3A = 3A=3, we have 3+B=03 + B = 03+B=0, which means B=−3B = -3B=−3.

STEP 12

STEP 13

Our **final answer** is 3x−3xx2+4\frac{3}{x} - \frac{3x}{x^2 + 4}x3​−x2+43x​.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? Rewrite this big fraction as the sum of smaller, simpler fractions! Watch out! Remember that $x^2 + 4$ can't be factored further using real numbers, so we'll need to handle it differently than if it were $x^2 - 4$ .

1. Set up the decomposition.
2. Multiply and simplify.
3. Solve for the unknowns.
4. Write the final decomposition.

We've got a factor of $x$ and a factor of $x^2 + 4$ , both in the denominator.
Since $x$ is just a regular linear factor, it gets a constant, say $A$ , on top.
So, we have a fraction $\frac{A}{x}$ .

Now, $x^2 + 4$ is a quadratic factor that can't be factored further.
That means it gets a linear term, say $Bx + C$ , on top.
So, we have another fraction $\frac{Bx + C}{x^2 + 4}$ .

Putting it all together, our partial fraction decomposition looks like this: $\frac{12}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$

To get rid of those pesky denominators, let's multiply both sides of our equation by $x(x^2 + 4)$ .
This gives us: $12 = A(x^2 + 4) + (Bx + C)x$

Now, let's expand and group like terms: $12 = Ax^2 + 4A + Bx^2 + Cx$ $12 = (A + B)x^2 + Cx + 4A$

Now comes the fun part!
We're going to match up the coefficients on both sides of the equation.
Notice that the left side has no $x^2$ term, so the coefficient of $x^2$ is zero.
This means $A + B = 0$ .

The left side also has no $x$ term, so $C = 0$ .
That simplifies things nicely!

Finally, the constant term on the left side is 12, so $4A = 12$ .
Dividing both sides by 4, we get $A = 3$ .

Since $A + B = 0$ and $A = 3$ , we have $3 + B = 0$ , which means $B = -3$ .

Our final answer is $\frac{3}{x} - \frac{3x}{x^2 + 4}$ .