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Tom [10]
2 years ago
14

Given a leading coefficient of 8, polynomial roots of 1 & 2, and the known point on the graph (4,5). Write an equation that

will find the
missing 3rd root (r). You do not have to solve the problem. See the example "Find the Missing Root - Interesting Twist" if you need help.


_=_(_-1)(_-2)(_-_)
Mathematics
1 answer:
m_a_m_a [10]2 years ago
8 0

Given:

The leading coefficient of a polynomial is 8.

Polynomial roots are 1 and 2.

The graph passes through the point (4,5).

To find:

The 3rd root and the equation of the polynomial.

Solution:

The factor form of a polynomial is:

y=a(x-c_1)(x-c_2)...(x-c_n)

Where, a is a constant and c_1,c_2,...,c_n are the roots of the polynomial.

Polynomial roots are 1 and 2. So, (x-1) and (x-2) are the factors of the polynomial.

Let the third root of the polynomial by c, then (x-c) is a factor of the polynomial.

The leading coefficient of a polynomial is 8. So, a=8 and the equation of the polynomial is:

y=8(x-1)(x-2)(x-c)

The graph passes through the point (4,5). Putting x=4,y=5, we get

5=8(4-1)(4-2)(4-c)

5=8(3)(2)(4-c)

5=48(4-c)

Divide both sides by 48.

\dfrac{5}{48}=4-c

c=4-\dfrac{5}{48}

c=\dfrac{192-5}{48}

c=\dfrac{187}{48}

Therefore, the 3rd root on the polynomial is \dfrac{187}{48}.

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3 years ago
Gary wants to buy a video game with a selling price of $48, on sale for 50% off. The sales tax in his state is 4.5%. How much wi
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Answer:

$25.20

Step-by-step explanation:

First, take the 50% off the sales price:

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7 0
3 years ago
The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum
wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information:P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000, where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2

Equating it to zero we get,

x^2 + 36x - 6400 = 0

We use the quadratic formula to find the values of x:

x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}, where a, b and c are coefficients of x^2, x^1 , x^0 respectively.

Putting these value we get x = -100, 64

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\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x

At x = 64,  \displaystyle\frac{d^2(P(x))}{dx^2} < 0

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

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3 years ago
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