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3 years ago

Find the remainder when the polynomial $x^5 x^4 x^3 x^2 x$ is divided by $x^3-4x$.

Mathematics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

Remainder would be $5x^2+21$

Step-by-step explanation:

Given,

Dividend = $x^5+x^4+x^3+x^2+x$

Divisor = $x^3-4x$

Using long division ( shown below ),

We get,

$\frac{x^5+x^4+x^3+x^2+x}{x^3-4x}=x^2+x+5+\frac{5x^2+21}{x^3-4x}$

Therefore,

Remainder would be $5x^2+21$

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EASY Work!! Please look at the picture and yes it’s easy for other people but not me for people asking.

Keith_Richards [23]

Answer:

7()=11

8()=42

I could barely see the answer for number 9 sorry :(

Step-by-step explanation:

the 7() is 2(3)+5

2(3)=6

6+5=11

Question (8)

c^2+3a x b

c(6)^2+ 3(3) x 2/3

36+9 x 2/3

36+6=42

7 0

3 years ago

Read 2 more answers

2. Consider the circle x² + y2 = 1, given in figure. Let OP makes an angle 30° with the x axis.

PolarNik [594]

The equation of the tangent line to the circle passing through the point P is; y = (1/√3)x ± 2/√3

<h3>How to find the equation of the tangent?</h3>

I) We are given the equation of the circle as;

x² + y² = 1

Since angle of inclination is 30°, then slope is;

m = tan 30 = 1/√3

Then equation of the tangent will be;

y = (1/√3)x + c

Put (√3)x + c into the given circle equation to get;

x² + ((1/√3)x + c)² = 1

x² + ¹/₃x² + (2/√3)cx + c² = 1

⁴/₃x² + (2/√3)x + (c² - 1) = 0

Since we need to find value of c for equation to become tangent, then the above quadratic equation must have real and equal roots.

Thus;

((2/√3)c)² - 4(⁴/₃)(c² - 1) = 0

⁴/₃c² - ¹⁶/₃(c² - 1) = 0

⁴/₃c² - ¹⁶/₃c² + ¹⁶/₃ = 0

4c² = ¹⁶/₃

c² = ⁴/₃

c = √⁴/₃

c = ±²/√3

Thus, equation of tangent is;

y = (1/√3)x ± 2/√3

II) Radius from the given equation is 1. Thus, we will use trigonometric ratio to find the x and y intercept;

x-intercept is at y = 0;

0 = (1/√3)x ± 2/√3

-(1/√3)x = ±2/√3

Intercept is positive. Thus;

x = (2/√3)/(1/√3)

x = 1

y - intercept is positive at x = 0;

y = (1/√3)0 ± 2/√3

y = 2/√3

Read more about Equation of tangent at; brainly.com/question/17040970

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6 0

2 years ago

Soft drinks are being purchased in case of 24. If there are 1008 soft drinks to packed,how many cases are needed?

galben [10]

Answer:

42 Cases.

Step-by-step explanation:

1008 divided by 24 = 42

7 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

Idek what to do, help

stepladder [879]

Distribute the 5 to the (2+y). Parenthisis always come first. after that combine the 5y and the -y. you should end up with 2-(+4y) +2. add the -2 and +2. you should be left with 4y

8 0

3 years ago