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

- lisa123

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

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- SolomonZelman

Trinomial - polynomial with 3 terms

- SolomonZelman

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

- SolomonZelman

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

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## More answers

- lisa123

Ohhh so it would be B then

- SolomonZelman

Yes

- SolomonZelman

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

- SolomonZelman

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

- lisa123

Okay thanks! can you help me with another problem?

- SolomonZelman

Yes \( : ) \)

- lisa123

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

- SolomonZelman

\(\large\color{#000000 }{ \displaystyle \frac{4x(x+1)^{1/2}+3(x+1)^{-1/2}}{(x+1)^{1/2}} }\)
this is how you wrote it, and I just want to verify that this is indeed the equations.

- SolomonZelman

equation **

- lisa123

yes that's is

- SolomonZelman

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

- SolomonZelman

Not that;
\((x+1)^{1/2} \times (x+1)^{1/2} = (x+1)^{1/2+1/2}=(x+1)^1=x+1\)
and
\((x+1)^{1/2} \times (x+1)^{-1/2} = (x+1)^{1/2+-1/2}=(x+1)^0=1\)

- lisa123

okay so the denominator is equal to x+1

- SolomonZelman

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

- SolomonZelman

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

- SolomonZelman

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

- lisa123

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

- SolomonZelman

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

- SolomonZelman

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

- lisa123

Yay! thank you so much @SolomonZelman :)

- SolomonZelman

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

- lisa123

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

- SolomonZelman

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.

- SolomonZelman

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

- lisa123

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

- SolomonZelman

Sure :)

- SolomonZelman

You can ask me now if you want...

- lisa123

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

- SolomonZelman

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

- SolomonZelman

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

- SolomonZelman

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.

- SolomonZelman

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

- SolomonZelman

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

- SolomonZelman

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

- SolomonZelman

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

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