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3 years ago

Use polynomial division to completely factor y=x^3+3x^2-13x-15 by x+5

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

(x - 3)(x + 1)(x + 5)

Step-by-step explanation:

I'd use synthetic division instead. If we were to find the roots of the given polynomial, we could from them write the factors as well.

The divisor x + 5 corresponds to root x = -5. Setting up synthetic div.,

-5 ) 1 3 -13 -15

-5 10 +15

-----------------------------

1 -2 -3 0

Since the remainder is 0, we know that -5 is a root and (x + 5) is a factor. Moreover, we know that the coefficients of the quotient are 1, -2 and -3.

1x² - 2x - 3 can be factored: the factors are (x - 3) and (x + 1).

So the end result for this problem is (x - 3)(x + 1)(x + 5).

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How do i do this?! Please Help me

Dafna1 [17]

Answer:

The answer to your question is 50 inches

Step-by-step explanation:

Data

Area = 200 in²

diagonal 1 = 8 in

diagonal 2 = ?

Formula

Area = (diagonal 1 x diagonal 2) / 2

Solve for diagonal 2

diagonal 2 = 2 x Area 2 / diagonal 1

-Substitution

diagonal 2 = 2 x 200 / 8

-Simplification

diagonal 2 = 400 / 8

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diagonal 2 = 50 in

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2 years ago

A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t

Andreas93 [3]

Answer:

a) Side Parallel to the river: 200 ft

b) Each of the other sides: 400 ft

Step-by-step explanation:

Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be $F=L+2W$ .

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

$LW=80,000...(1)$

We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be $C=20L+5(2W)\Rightarrow 20L+10W$ .

From equation (1), we will get:

$L=\frac{80,000}{W}$

Upon substituting this value in cost equation, we will get:

$C=20(\frac{80,000}{W})+10W$

$C=\frac{1600,000}{W}+10W$

$C=1600,000W^{-1}+10W$

To minimize the cost, we need to find critical points of the the derivative of cost function as:

$C'=-1600,000W^{-2}+10$

$-1600,000W^{-2}+10=0$

$-1600,000W^{-2}=-10$

$-\frac{1600,000}{W^2}=-10$

$-10W^2=-1,600,000$

$W^2=160,000$

$\sqrt{W^2}=\pm\sqrt{160,000}$

$W=\pm 400$

Since width cannot be negative, therefore, the width of the fencing would be 400 feet.

Now, we will find the 2nd derivative as:

$C''=-2(-1600,000)W^{-3}$

$C''=3200,000W^{-3}$

$C''=\frac{3200,000}{W^3}$

Now, we will substitute $W=400$ in 2nd derivative as:

$C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05$

Since 2nd derivative is positive at $W=400$ , therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting $W=400$ in $L=\frac{80,000}{W}$ , we will get:

$L=\frac{80,000}{400}\\\\L=200$

Therefore, the side parallel to the river will be 200 feet.

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3 years ago

A flagpole is 10 feet tall. It’s shadow is 12 feet long how far is it from the top of the flagpole to the end of its shadow?

Genrish500 [490]

Answer:

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Step-by-step explanation:

I just usually do this trick in situations like these where if x equals 1, and y equals 2, then z is gonna be 3. 2 goes after 1, and 3 goes after 2. This trick can work in other counting-up situations like this, as long as the number go up, and are right next to each other.

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3 years ago

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