lisa123
  • lisa123
which is a trinomial with a leading coefficient of 2 a) 2x^4 +3x -sqrt(x) b) 2x^4 + sqrt(3)x +1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
Trinomial - polynomial with 3 terms
SolomonZelman
  • SolomonZelman
Polynomials don't contains roots of x (only whole number powers). (rather they are defined for all x, including negative x values)
SolomonZelman
  • SolomonZelman
However, a polymial CAN have an irrational coefficient in one of the terms.

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lisa123
  • lisa123
Ohhh so it would be B then
SolomonZelman
  • SolomonZelman
Yes
SolomonZelman
  • 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
  • SolomonZelman
(but not to restrict the polynomial to only 1 irrational coefficient)
lisa123
  • lisa123
Okay thanks! can you help me with another problem?
SolomonZelman
  • SolomonZelman
Yes \( : ) \)
lisa123
  • lisa123
Factor completely [4x(x+1)^1/2+ 3(x+1)^-1/2] / (x+1)^1/2
SolomonZelman
  • 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
  • SolomonZelman
equation **
lisa123
  • lisa123
yes that's is
SolomonZelman
  • SolomonZelman
What I would do, is to multiply times \((x+1)^{1/2}\) on top and bottom.
SolomonZelman
  • 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
  • lisa123
okay so the denominator is equal to x+1
SolomonZelman
  • 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
  • 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
  • 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
  • lisa123
okay so it = 4x(x+1)+3?
SolomonZelman
  • SolomonZelman
Yes the numerator is 4x(x+1)+3 , correct!
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{4x(x+1)+3}{x+1} }\)
lisa123
  • lisa123
Yay! thank you so much @SolomonZelman :)
SolomonZelman
  • SolomonZelman
but, the instructions say "factor", and I am not sure whteher I fullfilled the instructions (althouhg we certainly simplified it).
lisa123
  • lisa123
yes one of the answer choices is (4x^2 +4x+3)/x+1
SolomonZelman
  • 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
  • SolomonZelman
Do you have any questions regarding anything what we have done here (or anything (math-related) you have done elsewhere)?
lisa123
  • lisa123
Hmm well im also doing some log problems and they always confuse me
SolomonZelman
  • SolomonZelman
Sure :)
SolomonZelman
  • SolomonZelman
You can ask me now if you want...
lisa123
  • lisa123
Okay thanks I will tag you in a new post with another question. You are extremely helpful!
SolomonZelman
  • 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
  • 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
  • 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
  • SolomonZelman
(As I am posting some properties of logarithms you can post any question if you want)
SolomonZelman
  • SolomonZelman
For as much as I am aware of, this is it.
SolomonZelman
  • 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
  • SolomonZelman
if you got questions, don't hesitate; you know --:)

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