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Alika [10]
2 years ago
15

Write the expression as a polynomial: (5–2a+a^2)(4a^2–3a–1)

Mathematics
1 answer:
agasfer [191]2 years ago
6 0

Answer: 4a^4 - 11a^3 + 25a^2 - 13a - 5

Explanation:

Check out the attached image below for the box method diagram. It's basically a 3 by 3 box showing all the 9 terms, some of which combine together (only the like terms though).

Each inner box or inner cell is the result of multiplying the outer terms. For example, in row 2, column 2 we have the outer terms -3a and -2a multiply to get 6a^2. Figure 1 shows the full table filled out while figure 2 has that same table highlighted with different colors to show the like terms.

The like terms are:

-15a and 2a, which combine to -13a

20a^2, 6a^2 and -a^2 which combine to 25a^2

-8a^3 and -3a^3 which combine to -11a^3

Those results are added to the other terms but we do not combine them (as they are not like terms). That's how we end up with the final answer. This is a quartic polynomial as the degree (largest exponent) is 4. There are five terms in this polynomial when fully simplified.

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Use logarithmic differentiation to find the derivative of the function. y = x2cos x Part 1 of 4 Using properties of logarithms,
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ANSWER

{y}^{'}  = 2x \cos(x)  -   {x}^{2} \sin(x)

EXPLANATION

The given function is

y =  {x}^{2}  \cos(x)

We take natural log of both sides;

ln(y) =   ln({x}^{2}  \cos(x) )

Recall and use the product rule of logarithms.

ln(AB)  =  ln(A )  +  ln(B)

This implies that:

ln(y) =   ln({x}^{2}  ) +  ln( \cos(x) )

ln(y) =  2 ln({x} ) +  ln( \cos(x) )

We now differentiate implicitly to obtain;

\frac{ {y}^{'} }{y}  =  \frac{2}{x}   -  \frac{ \sin(x) }{ \cos(x) }

Multiply through by y,

{y}^{'} = y( \frac{2}{x}   - \frac{ \sin(x) }{ \cos(x) ) })

Substitute y=x²cosx to obtain;

{y}^{'} =  {x}^{2}  \cos(x) ( \frac{2}{x}   - \frac{ \sin(x) }{ \cos(x) ) } )

Expand:

{y}^{'}  = 2x \cos(x)  -   {x}^{2} \sin(x)

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