Melody

Good work Juriemagic,

It is good to see you here again too  

Hi dp,

Why haven't you posted your LaTex properly?

Thanks Jenny, just another bug in the system (not being able to give likes I mean)

Thanks Ginger :)     And thanks ilorty too   

Why ?

You do not need to understand latex.

The latex coding that i have posted is for my benefit, and for the possible benefit of someone who specifically wants to learn LaTex, it is not a part of the answer.

The answer is written in normal maths symbols.

(it is rendered LaTex, but it is not LaTex coding)

Thanks very much guest.        

Thanks Loki and Heureka,

I went with a more straight forward (but not shorter) approach, 

I just divided by (x+1) three times to find the coefficients.

\(f(x)=ax^5+bx^4+cx^3\\ f(x)+2=ax^5+bx^4+cx^3+2\\ [f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\ \text{I did this by algebraic division and got a remainer of }-c+b-a+2\\ \text{The remainder must be 0 so}\\ -c+b-a+2=0\\ c=b-a+2 \)

So now I have

\(f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\ [f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\ \qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }6-b+2a\\ \text{The remainder must be 0 so}\\ 6-b+2a=0\\ b=6+2a\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ =(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\ =ax^3+6x^2-4x+2 \)

\((ax^3+6x^2-4x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }12-a\\ \text{The remainder must be 0 so}\\ a=12\\~\\ so\\ a=12\\ b=6+2a=30\\ c=b-a+2=30-12+2=20\\ so\\ f(x)=12x^5+30x^4+20x^3\)

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LaTex:

f(x)=ax^5+bx^4+cx^3\\
f(x)+2=ax^5+bx^4+cx^3+2\\
[f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\
\text{I did this by algebraic division and got a remainer of }-c+b-a+2\\
\text{The remainder must be 0 so}\\
-c+b-a+2=0\\
c=b-a+2

f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\
[f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\
\qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }6-b+2a\\
\text{The remainder must be 0 so}\\
6-b+2a=0\\
b=6+2a\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
=(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\
=ax^3+6x^2-4x+2

(ax^3+6x^2-4x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }12-a\\
\text{The remainder must be 0 so}\\
a=12\\~\\
so\\
a=12\\
b=6+2a=30\\
c=b-a+2=30-12+2=20\\
so\\
f(x)=12x^5+30x^4+20x^3

I do not know why it is blocked, there are a lot of bugs lately.

I have sent you a private message.

Thanks guest, here is a little more insight for others to see what you have done.

(a) Compute the sum 101^2 - 97^2 + 93^2 - 89^2 + .... + 5^2 - 1^2

I paired them first

(101^2 - 97^2) + (93^2 - 89^2) + .... + (5^2 - 1^2)

then turned them around

(5^2 - 1^2)+ ..................   +  (93^2 - 89^2) +  (101^2 - 97^2) 

\(T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\ T_2=(13^2-9^2)= [(11+2)^2 - (11-2)^2]\\ ...\\ T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ ...\\ T_{13}=(101^2-97^2)\\\quad= [(8*13-5+2)^2 - (8*13-5-2)^2]\\ \quad= [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\ \)

\(T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ T_n= [(8n-3)^2 - (8n-7)^2]\\ T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\ T_n= [-48n+9)] - [-112n+49)]\\ T_n= -48n+9 +112n-49\\ T_n= 64n-40\\ \text{This is an AP}\\ a=24, d=64,\;n=13\\ S_n=\frac{n}{2}(2a+(n-1)d]\\ S_{13}=\frac{13}{2}(2*24+12*64]\\ S_{13}=13(24+6*64]\\ S_{13}=13*408\\ S_{13}=5304\\\)

LaTex:

T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\
T_2=(13^2-9^2)=  [(11+2)^2 - (11-2)^2]\\
...\\
T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
...\\
T_{13}=(101^2-97^2)\\\quad=  [(8*13-5+2)^2 - (8*13-5-2)^2]\\
\quad=  [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\

T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
T_n= [(8n-3)^2 - (8n-7)^2]\\
T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\
T_n= [-48n+9)] - [-112n+49)]\\
T_n= -48n+9 +112n-49\\
T_n= 64n-40\\
\text{This is an AP}\\
a=24, d=64,\;n=13\\
S_n=\frac{n}{2}(2a+(n-1)d]\\
S_{13}=\frac{13}{2}(2*24+12*64]\\
S_{13}=13(24+6*64]\\
S_{13}=13*408\\
S_{13}=5304\\

Thanks Gwen,

It is such a pity that your pics have been blocked.

It is not your fault, I know this.