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2 years ago

13

How can you apply the remainder theorom

1 answer:

Goshia [24]2 years ago

6 0

In remainder theorem, first we set the given factor or the divisor to 0 and solve for the variable. Then we put the value of the vriable that we got on setting the divisor to 0, in the given polynomial and check the result . If the result is 0, then the given divisor is a factor, else it is not. For e.g. Suppose we have to see whether x-1 is a factor of x square -2x+1 or not. So we set x-1to 0 that is x-1=0 which gives x=1 . Now we put 1 for x in x square -2x+1 that is 1-2+1 =0, so x-1 is a factor of the given polynomial .

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Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for d<em>y</em>/d<em>x</em> :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If <em>y</em> ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

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3 years ago

Paha777 [63]

Step-by-step explanation:

30 minutes x an hour and half past prolly around 1 hr and 30 sec

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3 years ago

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2 years ago

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lys-0071 [83]

It would be B)no.

Hope This Helps!

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2 years ago

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3 years ago

Read 2 more answers