Math / Algebra

Question $\begin{array}{l} f(x)=2 x^{2}-5 x-3 \\ g(x)=2 x^{2}+5 x+2 \end{array}$
Find: $\left(\frac{f}{g}\right)(x)$ $\frac{x+3}{x+2}$ $\frac{x-3}{x+2}$ $x^{2}-1 x-\frac{3}{2}$ $\frac{2 x^{2}-5 x-3}{2 x^{2}+5 x+2}$

Studdy Solution

STEP 1

What is this asking? We're asked to divide the function $f(x)$ by the function $g(x)$ , creating a brand new function! Watch out! Don't forget to factor and simplify whenever possible.
Sometimes, things magically disappear!

STEP 2

1. Define the functions
2. Divide f(x) by g(x)
3. Factor the numerator
4. Factor the denominator
5. Simplify the expression

STEP 3

We're given two fantastic functions, $f(x)$ and $g(x)$ .
Let's write them down clearly: $f(x) = 2x^2 - 5x - 3$ $g(x) = 2x^2 + 5x + 2$ This is our starting point.
Woohoo!

STEP 4

Now, let's create our new function, which is $\frac{f(x)}{g(x)}$ .
It's as simple as putting $f(x)$ on top and $g(x)$ on the bottom.
Like a mathematical sandwich! $\frac{f(x)}{g(x)} = \frac{2x^2 - 5x - 3}{2x^2 + 5x + 2}$

STEP 5

Let's factor the top part of our fraction, the numerator.
We're looking for two numbers that multiply to $(2 \cdot -3) = \mathbf{-6}$ and add up to $\mathbf{-5}$ .
Those magical numbers are $\mathbf{-6}$ and $\mathbf{1}$ .

STEP 6

Rewriting our numerator, we get: $2x^2 - 6x + 1x - 3$

STEP 7

Now, we can factor by grouping: $2x(x - 3) + 1(x - 3)$ $(2x + 1)(x - 3)$ Look at that, factored!

STEP 8

Time to factor the bottom part of our fraction, the denominator.
We need two numbers that multiply to $(2 \cdot 2) = \mathbf{4}$ and add up to $\mathbf{5}$ .
Those numbers are $\mathbf{4}$ and $\mathbf{1}$ .

STEP 9

Rewriting our denominator, we get: $2x^2 + 4x + 1x + 2$

STEP 10

Factoring by grouping gives us: $2x(x + 2) + 1(x + 2)$ $(2x + 1)(x + 2)$ Boom! Factored!

STEP 11

Now, let's put our factored numerator and denominator back into the fraction: $\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)}$

STEP 12

Notice that $(2x + 1)$ appears on both the top and the bottom.
We can divide both by $(2x+1)$ to simplify!
We're essentially multiplying by $\frac{1/(2x+1)}{1/(2x+1)}$ , which is the same as multiplying by one! $\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)} \cdot \frac{1/(2x+1)}{1/(2x+1)} = \frac{x-3}{x+2}$

STEP 13

Our final simplified function is $\frac{x-3}{x+2}$ .
Fantastic!

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Question $\begin{array}{l} f(x)=2 x^{2}-5 x-3 \\ g(x)=2 x^{2}+5 x+2 \end{array}$
Find: $\left(\frac{f}{g}\right)(x)$ $\frac{x+3}{x+2}$ $\frac{x-3}{x+2}$ $x^{2}-1 x-\frac{3}{2}$ $\frac{2 x^{2}-5 x-3}{2 x^{2}+5 x+2}$

Studdy Solution

STEP 1

What is this asking? We're asked to divide the function $f(x)$ by the function $g(x)$ , creating a brand new function! Watch out! Don't forget to factor and simplify whenever possible.
Sometimes, things magically disappear!

STEP 2

1. Define the functions
2. Divide f(x) by g(x)
3. Factor the numerator
4. Factor the denominator
5. Simplify the expression

STEP 3

We're given two fantastic functions, $f(x)$ and $g(x)$ .
Let's write them down clearly: $f(x) = 2x^2 - 5x - 3$ $g(x) = 2x^2 + 5x + 2$ This is our starting point.
Woohoo!

STEP 4

Now, let's create our new function, which is $\frac{f(x)}{g(x)}$ .
It's as simple as putting $f(x)$ on top and $g(x)$ on the bottom.
Like a mathematical sandwich! $\frac{f(x)}{g(x)} = \frac{2x^2 - 5x - 3}{2x^2 + 5x + 2}$

STEP 5

Let's factor the top part of our fraction, the numerator.
We're looking for two numbers that multiply to $(2 \cdot -3) = \mathbf{-6}$ and add up to $\mathbf{-5}$ .
Those magical numbers are $\mathbf{-6}$ and $\mathbf{1}$ .

STEP 6

Rewriting our numerator, we get: $2x^2 - 6x + 1x - 3$

STEP 7

Now, we can factor by grouping: $2x(x - 3) + 1(x - 3)$ $(2x + 1)(x - 3)$ Look at that, factored!

STEP 8

Time to factor the bottom part of our fraction, the denominator.
We need two numbers that multiply to $(2 \cdot 2) = \mathbf{4}$ and add up to $\mathbf{5}$ .
Those numbers are $\mathbf{4}$ and $\mathbf{1}$ .

STEP 9

Rewriting our denominator, we get: $2x^2 + 4x + 1x + 2$

STEP 10

Factoring by grouping gives us: $2x(x + 2) + 1(x + 2)$ $(2x + 1)(x + 2)$ Boom! Factored!

STEP 11

Now, let's put our factored numerator and denominator back into the fraction: $\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)}$

STEP 12

STEP 13

Our final simplified function is $\frac{x-3}{x+2}$ .
Fantastic!

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Studdy Solution

STEP 1

What is this asking? We're asked to divide the function f(x)f(x)f(x) by the function g(x)g(x)g(x), creating a brand new function! Watch out! Don't forget to factor and simplify whenever possible.Sometimes, things magically disappear!

STEP 2

1. Define the functions2. Divide *f(x)* by *g(x)*3. Factor the numerator4. Factor the denominator5. Simplify the expression

STEP 3

We're given two fantastic functions, f(x)f(x)f(x) and g(x)g(x)g(x).Let's write them down clearly: f(x)=2x2−5x−3f(x) = 2x^2 - 5x - 3f(x)=2x2−5x−3 g(x)=2x2+5x+2g(x) = 2x^2 + 5x + 2g(x)=2x2+5x+2This is our starting point.Woohoo!

STEP 4

STEP 5

Let's factor the top part of our fraction, the **numerator**.We're looking for two numbers that multiply to (2⋅−3)=−6(2 \cdot -3) = \mathbf{-6}(2⋅−3)=−6 and add up to −5\mathbf{-5}−5.Those magical numbers are −6\mathbf{-6}−6 and 1\mathbf{1}1.

STEP 6

Rewriting our numerator, we get: 2x2−6x+1x−32x^2 - 6x + 1x - 32x2−6x+1x−3

STEP 7

Now, we can factor by grouping: 2x(x−3)+1(x−3)2x(x - 3) + 1(x - 3)2x(x−3)+1(x−3) (2x+1)(x−3)(2x + 1)(x - 3)(2x+1)(x−3)Look at that, factored!

STEP 8

Time to factor the bottom part of our fraction, the **denominator**.We need two numbers that multiply to (2⋅2)=4(2 \cdot 2) = \mathbf{4}(2⋅2)=4 and add up to 5\mathbf{5}5.Those numbers are 4\mathbf{4}4 and 1\mathbf{1}1.

STEP 9

Rewriting our denominator, we get: 2x2+4x+1x+22x^2 + 4x + 1x + 22x2+4x+1x+2

STEP 10

Factoring by grouping gives us: 2x(x+2)+1(x+2)2x(x + 2) + 1(x + 2)2x(x+2)+1(x+2) (2x+1)(x+2)(2x + 1)(x + 2)(2x+1)(x+2)Boom! Factored!

STEP 11

Now, let's put our factored numerator and denominator back into the fraction: (2x+1)(x−3)(2x+1)(x+2)\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)}(2x+1)(x+2)(2x+1)(x−3)​

STEP 12

STEP 13

Our final simplified function is x−3x+2\frac{x-3}{x+2}x+2x−3​.Fantastic!

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Studdy solves anything!

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? We're asked to divide the function $f(x)$ by the function $g(x)$ , creating a brand new function! Watch out! Don't forget to factor and simplify whenever possible.
Sometimes, things magically disappear!

1. Define the functions
2. Divide f(x) by g(x)
3. Factor the numerator
4. Factor the denominator
5. Simplify the expression

We're given two fantastic functions, $f(x)$ and $g(x)$ .
Let's write them down clearly: $f(x) = 2x^2 - 5x - 3$ $g(x) = 2x^2 + 5x + 2$ This is our starting point.
Woohoo!

Let's factor the top part of our fraction, the numerator.
We're looking for two numbers that multiply to $(2 \cdot -3) = \mathbf{-6}$ and add up to $\mathbf{-5}$ .
Those magical numbers are $\mathbf{-6}$ and $\mathbf{1}$ .

Rewriting our numerator, we get: $2x^2 - 6x + 1x - 3$

Now, we can factor by grouping: $2x(x - 3) + 1(x - 3)$ $(2x + 1)(x - 3)$ Look at that, factored!

Time to factor the bottom part of our fraction, the denominator.
We need two numbers that multiply to $(2 \cdot 2) = \mathbf{4}$ and add up to $\mathbf{5}$ .
Those numbers are $\mathbf{4}$ and $\mathbf{1}$ .

Rewriting our denominator, we get: $2x^2 + 4x + 1x + 2$

Factoring by grouping gives us: $2x(x + 2) + 1(x + 2)$ $(2x + 1)(x + 2)$ Boom! Factored!

Now, let's put our factored numerator and denominator back into the fraction: $\frac{(2x + 1)(x - 3)}{(2x + 1)(x + 2)}$

Our final simplified function is $\frac{x-3}{x+2}$ .
Fantastic!