Given the functions h(x)=2x3−4h\left(x\right)=2x^3-4h(x)=2x3−4 and k(x)=x2+2k\left(x\right)=x^2+2k(x)=x2+2, find and fully simplify h⋅k(x)h\cdot k\left(x\right)h⋅k(x)

h ⋅ k ( x ) = 2 ( x 5 + 2 x 3 − 2 x 2 − 4 ) h\cdot k\left(x\right)=2\left(x^5+2x^3-2x^2-4\right) h ⋅ k ( x ) = 2 ( x 5 + 2 x 3 − 2 x 2 − 4 )

Given the functions h ( x ) = 2 x 3 − 4 h\left(x\right)=2x^3-4 h ( x ) = 2 x 3 − 4 and k ( x ) = x 2 + 2 k\left(x\right)=x^2+2 k ( x ) = x 2 + 2 , find and fully simplify h ⋅ k ( x ) h\cdot k\left(x\right) h ⋅ k ( x )

Okay. So let's now see if we can solve a practice problem here. So in this problem, we are given 2 functions. We're given h of x is equal to 2x cubed minus 4, and k of x is equal to x squared plus 2. We're asked to find and fully simplify h multiplied by k of x. So to find this product here, what we're basically just doing is multiplying these two functions together and simplifying our answer. So in this case, we'll have h of x, which is 2x cubed minus 4 multiplied by k of x, which is x squared plus 2. And to multiply these together, we can just use the foil method to multiply everything out. So I have 2x cubed multiplied by x squared. Notice we have the x's in common here, and what we're going to end up getting is this 2 out front multiplied by x to the 5th power, because when multiplying these 2 x's, you just have to add the exponent since we have this common base here. Now what I'm going to do is take 2x cubed and multiply it by this next term, which is 2. So we're going to end up with the 2's multiplied, which is 4, and then we have the x cubed. Now from here, I'm going to take this negative 4 and multiply it by x squared, which will give me minus 4x squared, and then I'm going to take this negative 4 and multiply it by this 2, which 4 times 2 is 8, and since we have this minus sign, we're going to have negative 8. So this right here is our expression, and it doesn't look like there's really anything more I can do to combine anything, So this is pretty close to simplified, although if I wanna simplify this even farther, I could divide everything by 2 and factor a 2 out because notice that we have a 2 in common with each of these. So what we're going to end up with is 2 multiplied by x to the 5th plus, and then 4 divided by 2 is 2, and then we have x cubed, minus, and then dividing this by 2 we get 2x squared, and then 8 divided by 2 is 4. So So this right here would be our fully simplified answer for h multiplied by k of x. That's the solution.

Given the functions h ( x ) = 2 x 3 − 4 h\left(x\right)=2x^3-4 h ( x ) = 2 x 3 − 4 and k ( x ) = x 2 + 2 k\left(x\right)=x^2+2 k ( x ) = x 2 + 2 , find and fully simplify h ⋅ k ( x ) h\cdot k\left(x\right) h ⋅ k ( x )

h ⋅ k ( x ) = 2 ( x 5 + 2 x 3 − 2 x 2 − 4 ) h\cdot k\left(x\right)=2\left(x^5+2x^3-2x^2-4\right) h ⋅ k ( x ) = 2 ( x 5 + 2 x 3 − 2 x 2 − 4 )

h ⋅ k ( x ) = 2 x 5 − 8 h\cdot k\left(x\right)=2x^5-8 h ⋅ k ( x ) = 2 x 5 − 8

h ⋅ k ( x ) = 2 x 5 + 4 x 3 − 8 h\cdot k\left(x\right)=2x^5+4x^3-8 h ⋅ k ( x ) = 2 x 5 + 4 x 3 − 8

h ⋅ k ( x ) = x 2 + 4 x + 4 h\cdot k\left(x\right)=x^2+4x+4 h ⋅ k ( x ) = x 2 + 4 x + 4

3. Functions

Function Operations

College Algebra

3. Functions

Function Operations

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Multiple Choice

Given the functions $h\left(x\right)=2x^3-4$ and $k\left(x\right)=x^2+2$ , find and fully simplify $h\cdot k\left(x\right)$

$h\cdot k\left(x\right)=2\left(x^5+2x^3-2x^2-4\right)$

$h\cdot k\left(x\right)=2x^5-8$

$h\cdot k\left(x\right)=2x^5+4x^3-8$

$h\cdot k\left(x\right)=x^2+4x+4$

Verified step by step guidance

First, understand that h(x) and k(x) are two functions, and we need to find their product, denoted as h⋅k(x). This means we will multiply the expressions for h(x) and k(x).

Write down the expressions for the functions: h(x) = 2x^3 - 4 and k(x) = x^2 + 2.

To find h⋅k(x), multiply the two expressions: (2x^3 - 4) * (x^2 + 2).

Use the distributive property to expand the product: Multiply each term in the first polynomial by each term in the second polynomial.

Simplify the resulting expression by combining like terms to get the final simplified form of h⋅k(x).

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