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4 years ago

Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1?

Mathematics

2 answers:

Stella [2.4K]4 years ago

7 0

Answer:

f(x) = 3x² - 15x + 12

Step-by-step explanation:

Given f(x) has roots x = a and x = b, then

(x - a) and (x - b) are the factors

f(x) is then the product of the factors

f(x) = a(x - a)(x - b) ← where a is a multiplier

Given roots are x = 4 and x = 1, then

(x - 4) and (x - 1) are the factors

With a = 3, then

f(x) = 3(x - 4)(x - 1) ← expand factors using FOIL

= 3(x² - 5x + 4) ← distribute

= 3x² - 15x + 12

Sophie [7]4 years ago

5 0

Answer:

f(x) = 3x^2 - 15x + 12

Step-by-step explanation:

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Answer
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Move the terms
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3 years ago

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3 years ago

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Find the volume of the parallelepiped with adjacent edges t = 3j – 4j – 6k, u = 3i – 7j – 7k and v = 5i + j + 3k.

Assoli18 [71]

Answer:

94 cubic units

Step-by-step explanation:

The volume of this parallelepiped (the product of area of its base and height) is equal to the absolute value of the scalar triple product $|\vec{t}\cdot (\vec{u}\times \vec{v})|$

Since

$\vec{t}=3\vec{i}-4\vec{j}-6\vec{k}=(3,-4,-6)\\ \\\vec{u}=3\vec{i}-7\vec{j}-7\vec{k}=(3,-7,-7)\\ \\\vec{v}=5\vec{i}+\vec{j}+3\vec{k}=(5,1,3)$

we have

$|\vec{t}\cdot (\vec{u}\times \vec{v})|=\left|\left|\begin{array}{ccc}3&-4&-6\\3&-7&-7\\5&1&3\end{array}\right|\right|=|3\cdot (-7)\cdot 3+(-4)\cdot (-7)\cdot 5+3\cdot 1\cdot (-6)-5\cdot (-7)\cdot (-6)-3\cdot (-7)\cdot 1-3\cdot (-4)\cdot 3|=|-63+140-18-210+21+36|=|-94|=94\ un^3.$

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3 years ago