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2 years ago

Use the rational zero theorem to create a list of all possible rational zeros of the function f(x)=14x7-4x2+2

Mathematics

1 answer:

frosja888 [35]2 years ago

6 0

Answer:

Factor this polynomial:

F(x)=x^3-x^2-4x+4

Try to find the rational roots. If p/q is a root (p and q having no factors in common), then p must divide 4 and q must divide 1 (the coefficient of x^3).

The rational roots can thuis be +/1, +/2 and +/4. If you insert these values you find that the roots are at

x = 1, x = 2 and x = -2. This means that

x^3-x^2-4x+4 = A(x - 1)(x - 2)(x + 2)

A = 1, as you can see from equation the coefficient of x^3 on both sides.

Typo:

The rational roots can be

+/-1, +/-2 and +/-4

Step-by-step explanation:

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On a given week, the classical literature section in a certain library had 40 different books, all of which were in stock on the

Montano1993 [528]

Answer:

The greatest number of the books that could have been borrowed during the week is 36

Step-by-step explanation:

Let's create an equation which represent this problem, where will rest the book borrow and add the book returned to the 40 initial amount of book on Monday, to get the 22 book in the shelf on Saturday morning

40 (book on the shelf on Monday) - x (book borrowed) + 0.5 * x (50 % of the book was returned before Saturday) = 22 (books in the shelf on Saturday morning)

40 - x + 0.5 * x = 22

40 - 0.5*x = 22

18 = 0.5 * x

18/0.5 = x

36 = x

The amount of borrowed books is 36

6 0

2 years ago

Read 2 more answers

Choose the correct simplification of the expression (−2x + 4y)(3x − 7y).

Nadya [2.5K]

(-2x + 4y)(3x - 7y) = -6x^2 - 28y^2 + 12yx

3 0

3 years ago

Read 2 more answers

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.