Zennestinala
Answered

Let f(x) = 16x5 – 48x4 – 8x3 and g(x) = 8x2. Find f of x over g of x. a) 2x2 + 6x + 1 b)2x2 – 6x – 1 c)2x3 + 6x2 + x d)2x3 – 6x2 – x

Answer :

JcAlmighty
The answer is d)2x3 – 6x2 – x.

It is given:
[tex]f(x)=16 x^{5}-48 x^{4} -8 x^{3} [/tex]
[tex]g(x)=8 x^{2} [/tex]

We should find:
[tex]f( \frac{f(x)}{g(x)} )[/tex]

So, we just need to replace it in the equation:
[tex]f( \frac{f(x)}{g(x)} )= \frac{16 x^{5}-48 x^{4} -8 x^{3}}{8 x^{2}} [/tex]

Let's now separate each member of equation:
[tex] f( \frac{f(x)}{g(x)} )= \frac{16 x^{5}}{8 x^{2}} - \frac{48 x^{4}}{8 x^{2}} - \frac{8 x^{3}}{8 x^{2}} = \frac{16}{8} x^{5-2} -\frac{48}{8} x^{4-2} -\frac{8}{8} x^{3-2} [/tex]

[tex]f( \frac{f(x)}{g(x)} )=2 x^{3} -6 x^{2} -x[/tex]
InesWalston

Answer:

[tex]\boxed{\boxed{\dfrac{f(x)}{g(x)}=2x^3-6x^2-x}}[/tex]

Step-by-step explanation:

Here given that,

[tex]f(x) = 16x^5-48x^4-8x^3\\\\g(x) = 8x^2[/tex]

So [tex]\dfrac{f(x)}{g(x)}[/tex] will be,

[tex]\dfrac{f(x)}{g(x)}=\dfrac{16x^5-48x^4-8x^3}{8x^2}[/tex]

Applying single division,

[tex]\dfrac{16x^5-48x^4-8x^3}{8x^2}=\dfrac{16x^5}{8x^2}-\dfrac{48x^4}{8x^2}-\dfrac{8x^3}{8x^2}[/tex]

Simplifying further,

[tex]\dfrac{16x^5}{8x^2}-\dfrac{48x^4}{8x^2}-\dfrac{8x^3}{8x^2}=2x^3-6x^2-x[/tex]

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