Question

which is a trinomial with a leading coefficient of 2
a) 2x^4 +3x -sqrt(x)
b) 2x^4 + sqrt(3)x +1

Answer 1

Trinomial - polynomial with 3 terms

Answer 2

Polynomials don't contains roots of x (only whole number powers).
(rather they are defined for all x, including negative x values)

Answer 3

However, a polymial CAN have an irrational coefficient in one of the terms.

Answer 4

Ohhh so it would be B then

Answer 5

Yes

Answer 6

Also --
a polynomial can have all terms with irrational coefficients ... I said 1 irrational coefficient because that is the case in your example...

Answer 7

(but not to restrict the polynomial to only 1 irrational coefficient)

Answer 8

Okay thanks! can you help me with another problem?

Answer 9

Yes \( : ) \)

Answer 10

Factor completely [4x(x+1)^1/2+ 3(x+1)^-1/2] / (x+1)^1/2

Answer 11

\(\large\color{#000000 }{ \displaystyle \frac{4x(x+1)^{1/2}+3(x+1)^{-1/2}}{(x+1)^{1/2}} }\)

this is how you wrote it, and I just want to verify that this is indeed the equations.

Answer 12

equation **

Answer 13

yes that's is

Answer 14

What I would do, is to multiply times \((x+1)^{1/2}\) on top and bottom.

Answer 15

Not that;

\((x+1)^{1/2} \times (x+1)^{1/2} = (x+1)^{1/2+1/2}=(x+1)^1=x+1\)

and

\((x+1)^{1/2} \times (x+1)^{-1/2} = (x+1)^{1/2+-1/2}=(x+1)^0=1\)

Answer 16

okay so the denominator is equal to x+1

Answer 17

so (after multiplying times \((x+1)^{1/2}\) on top and bottom), you get the following equation:

\(\large\color{#000000 }{ \displaystyle \frac{\color{red}{\left\{\color{black}{4x(x+1)^{1/2}+3(x+1)^{-1/2}}\right\}\times (x+1)^{1/2}}}{(x+1)^{1/2}\color{red}{\times (x+1)^{1/2}}} }\)

you got the denominator (correctly)

\(\large\color{#000000 }{ \displaystyle \frac{\color{red}{\left\{\color{black}{4x(x+1)^{1/2}+3(x+1)^{-1/2}}\right\}\times (x+1)^{1/2}}}{x+1} }\)

Answer 18

\(\large\color{#000000 }{ \displaystyle \frac{\color{red}{\left\{\color{black}{4x(x+1)^{1/2}+3(x+1)^{-1/2}}\right\}\times (x+1)^{1/2}}}{x+1} }\)

expanding;

\(\large\color{#000000 }{ \displaystyle \frac{\color{red}{\left\{\color{black}{4x(x+1)^{1/2}}\right\}\times (x+1)^{1/2}}+\color{red}{\left\{\color{black}{3(x+1)^{-1/2}}\right\}\times (x+1)^{1/2}}}{x+1} }\)

Answer 19

REMINDER

\((x+1)^{1/2} \times (x+1)^{1/2} = (x+1)^{1/2+1/2}=(x+1)^1=x+1\)

and

\((x+1)^{1/2} \times (x+1)^{-1/2} = (x+1)^{1/2+-1/2}=(x+1)^0=1\)

Answer 20

okay so it = 4x(x+1)+3?

Answer 21

Yes the numerator is 4x(x+1)+3 , correct!

Answer 22

\(\large\color{#000000 }{ \displaystyle \frac{4x(x+1)+3}{x+1} }\)

Answer 23

Yay! thank you so much @SolomonZelman :)

Answer 24

but, the instructions say "factor", and I am not sure whteher I fullfilled the instructions (althouhg we certainly simplified it).

Answer 25

yes one of the answer choices is (4x^2 +4x+3)/x+1

Answer 26

Oh, I guess they don't know what the word factor means, but even better to us - we can move on away from this problem.

Answer 27

Do you have any questions regarding anything what we have done here (or anything (math-related) you have done elsewhere)?

Answer 28

Hmm well im also doing some log problems and they always confuse me

Answer 29

Sure :)

Answer 30

You can ask me now if you want...

Answer 31

Okay thanks I will tag you in a new post with another question. You are extremely helpful!

Answer 32

Some good properties (whether related to what you are doing or not, but they are crucial for the rest of your mathematical life)...

Product inside the log,
\(\large \color{#000000 }{ [1]\quad \displaystyle \log_{\color{red}{\rm A}}(\color{green}{\rm B}\times \color{blue}{\rm C})=\log_{\color{red}{\rm A}}(\color{green}{\rm B})+\log_{\color{red}{\rm A}}( \color{blue}{\rm C}) }\)

Quotient inside the log,
\(\large \color{#000000 }{ [2] \quad \displaystyle \log_{\color{red}{\rm A}}(\color{green}{\rm B}\div \color{blue}{\rm C})=\log_{\color{red}{\rm A}}(\color{green}{\rm B})-\log_{\color{red}{\rm A}}( \color{blue}{\rm C}) }\)

Answer 33

Exponent inside the log,
\(\large \color{#000000 }{ [3]\quad \displaystyle \log_{\color{red}{\rm A}}(\color{green}{\rm B}^\color{purple}{\bf C})=\color{purple}{\bf C}\cdot \log_{\color{red}{\rm A}}(\color{green}{\rm B}) }\)

Answer 34

Changing the base of log,
\(\large \color{#000000 }{ [4]\quad \displaystyle \log_{\color{red}{\rm A}}(\color{green}{\rm B})=\frac{\log_\color{magenta}{\bf W}(\color{green}{\rm B})}{\log_\color{magenta}{\bf W}(\color{red}{\rm A})} }\)

You can substitute any (positive) base \(\color{magenta}{\bf W}\) using this formula.

Answer 35

(As I am posting some properties of logarithms you can post any question if you want)

Answer 36

For as much as I am aware of, this is it.

Answer 37

One more, one of my famous properties;

\(\large \color{#000000 }{ [5]\quad \displaystyle {\color{red}{\rm A}}^{\Large \log_{\color{red}{\rm A}}(\color{green}{\rm x})}=\color{green}{\rm x} }\)


I always use this as;

\(\Large \color{#000000 }{ \displaystyle e^{\ln(\color{green}{\rm x})}= {\color{red}{\rm e}}^{\Large \log_{\color{red}{\rm e}}(\color{green}{\rm x})}=\color{green}{\rm x} }\)

Answer 38

if you got questions, don't hesitate; you know --:)