Find the coefficient - calculus questions

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The discussion revolves around finding coefficients in polynomial expressions. The first question involves evaluating g(2+h) for the function g(x) = 2x^2 – 5x, leading to the simplified expression -2 + 3h + 2h^2. The second question focuses on determining the coefficient of x in the expansion of (2x – 7)^2, which is clarified to be -28 after expanding the polynomial. Participants emphasize that the coefficient refers specifically to the term directly in front of x, distinguishing it from coefficients of other terms like x^2. The final consensus confirms that the coefficient of x in the given expression is indeed -28.
vorcil
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1:
If g(x) = 2x^2 – 5x then g(2+h) =

my working

2(2+h)^2 -5(2+h)
( 2((2+h)*(2+h)) ) - 5(2+h)
2(4 + h^2 + 4h) - 10 -5h
8 + 2h^2 + 8h - 10 -5h

-2 + 3h + 2h^2

2:
Find the coefficient of x in (2x – 7)^2.
isn't it 2?
 
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#1 looks correct.
#2 you'll have to expand it first!
 
queenofbabes said:
#1 looks correct.
#2 you'll have to expand it first!

Find the coefficient of x in (2x – 7)2.

2x-7 * 2x-7
4x^2 - 14x - 14x + 49

4x^2 - 28x + 49

coefficient is 4-28?
so -24?
or 4, or 28

i'm pretty sure it's 4, because it's infront of the x^2, but the 28 infront of x could also be a coefficient?
 
Coefficient fof x is the number in front of x...so in 4x^2 + (- 28)x + 49 what is that?
 
rock.freak667 said:
Coefficient fof x is the number in front of x...so in 4x^2 + (- 28)x + 49 what is that?

there is a 4 in front of the x^2,
and -28 infront of the x.

if you meant the number infront of just the x, then it's -28
if you meant the number infront of the x, in x^2 then it's 4

which one do i choose? the power functioned x or the x without it?

i'm assuming it's the x without it because if there were more functions, like a cubic polynomial, with like ax^4 + bx^3 + cx^2 + dx + e, then it'd get confusing

so -28 is my awnser?
 
vorcil said:
so -28 is my awnser?

Yes.


But if you had

ax3+bx2+cx+d

The coefficient of x3 is a
The coefficient of x2 is b
The coefficient of x is c

and so forth.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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