Factor completely 10x2 + 2x − 8

1 answer:

4 0

First, notice every term has a common factor of 2, so factor out a 2.

$10x^2 + 2x - 8 =$

$= 2(5x^2 + x - 4)$

Now we need two numbers that multiply to 5 and two numbers that multiply to -4.
For the 5, it is a single possibility, 5 and 1, so we will have

$= 2(5x~~~~~~)(x~~~~~~)$

Pairs of numbers that multiply to -4 are:
-4, 1
4, -1
2, -2

We need to choose the correct pair and place it in the correct order.
To do that, we place the numbers as guesses and multiply out the factors to see if we get the correct middle term.

First guess: -4, 1

$= 2(5x - 4)(x + 1)$

The middle term would be: 5x - 4x = x, which is what we want, so we lucked out and we have the answer with just one guess.

Answer: $10x^2 + 2x - 8 = 2(5x - 4)(x + 1)$

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3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o

Fiesta28 [93]

Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true that f must be surjective? Is it true that g must be surjective? Justify your answers with either a counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not injective (since both 2, 3 7→ 4) but g ◦ f is injective. Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2 . Then g is not injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective. Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b. Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page. The part of this problem most groups had the most issue with was the second. Everyone should be comfortable with carefully proving a function is surjective by the time we get to the midterm.

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3 years ago

Write an inequality to describe the region represented on the number line.

vovangra [49]

Answer:

See explanation

Step-by-step explanation:

The question is incomplete, as the required number line is not included in the question.

A general explanation is as follows:

An open circle means > or <

A closed circle means >= or <=

Take, for instance, the attached number line: