algebra intuitive understanding: (4a^4 + 11a^3 + 27a^2 + 17a + 5) / (a^2 + 2a + 5)

[amigo1]

this problem is an algebriac division, here it has been asked to divide by the use of formulae rather than using the long division method of algebra division:-
divide (4a^4 + 11a^3 + 27a^2 + 17a + 5) by (a^2 + 2a + 5)

answer given:- 4a^2 + 3a + 1
my solution:-

i dont know i do not get a whiff of it --- how to think of such a thing?
the problem remains that at least how do i know if this (of any other problem ) could be solved by use of formulae and does not warrant the use of long division techniques?
Can you recommend any reading / books ..etc

although i roughly started as (i dont know why i did this except for the fact that a^4 was somewhere in the dividend expression as given)
(a^2 + 2a + 5 )^2 =
(a^2)^2 + 4a^2 + 25 + 4a^3 + 10a + 5a^2
= a^4 + 4a^3 + 9a^2 + 10a + 25 ----- leads me nowhere

(you type maths like in a book -- can you lead me to an editor knowledge like that so that i may learn that/ latex )

My name was sujoy if you remember-- how are you teachers (Eliz s Stapel , PKA , Denis, Mathman et al)

 

[stapel]

this problem is an algebriac division, here it has been asked to divide by the use of formulae rather than using the long division method of algebra division:

What is the "formula" that you're supposed to be using? Is there a similar example in your book or class notes, that we could use to work backwards to whatever they're wanting from you?

Please be complete. Thank you!

 

[soroban]

Hello, Sujoy!

We have been asked to divide by the use of formulae
rather than using the long division method.

\(\displaystyle (4a^4 + 11a^3 + 27a^2 + 17a + 5) \,\div\,(a^2 + 2a + 5)\)

Answer: .\(\displaystyle 4a^2 + 3a + 1\)

I don't know what "formulae" we're expected to know.
But I can hammer it into a factorable form.

\(\displaystyle 4a^4 + \color{blue}{11a^3} + \color{red}{27a^2} +17a + 5\)

\(\displaystyle \qquad =\; 4a^4 + (\color{blue}{8a^3+3a^3}) +(\color{red}{20a^2+7a^2}) + 17a+ 5\)

\(\displaystyle \qquad =\;(4a^4 + 8a^3 + 20a^2) + 3a^3 +\color{blue}{7a^2} + \color{red}{17a} +5 \)

\(\displaystyle \qquad =\;4a^2(a^2+2a + 5) + 3a^3 + (\color{blue}{6a^2+a^2}) +(\color{red}{15a+2a}) + 5\)

\(\displaystyle \qquad =\;4a^2(a^2+2a+5) + (3a^3 + 6a^2 + 15a) + a^2 + 2a + 5\)

\(\displaystyle \qquad =\;4a^2\color{purple}{(a^2+2a+5)} + 3a\color{purple}{(a^2 + 2a + 5)} + 1\color{purple}{(a^2+2a+5)}\)

\(\displaystyle \qquad =\;(a^2+2a+5)(4a^2 + 3a + 1)\)

 

[Steven G]

(a^2 + 2a + 5 )^2 =
(a^2)^2 + 4a^2 + 25 + 4a^3 + 10a + 5a^2
= a^4 + 4a^3 + 9a^2 + 10a + 25 ----- leads me nowhere

You did not square correctly!

 

[amigo1]

What is the "formula" that you're supposed to be using? Is there a similar example in your book or class notes, that we could use to work backwards to whatever they're wanting from you?

the formulae to be used were the normal formulae we use in everyday maths as:- (a+b)^2, (a+b+c)^2, or (a+b+c)^3, (a+b)^2..... etc. There were no examples of this types as there were only few of them. The solution were also for very easy problems, --- nothing like this were discussed.

thank you teacher

 

[amigo1]

thank you teacher soroban --- it is nice after a very long time.

 

[amigo1]

thank you teacher Denis -- we speak after a very long time ---- how are you?

 

[Deleted member 4993]

this problem is an algebriac division, here it has been asked to divide by the use of formulae rather than using the long division method of algebra division:-
divide 4a^4 + 11a^3 + 27a^2 + 17a + 5 by a^2 + 2a + 5

answer given:- 4a^2 + 3a + 1
my solution:-

i dont know i do not get a whiff of it --- how to think of such a thing?
the problem remains that at least how do i know if this (of any other problem ) could be solved by use of formulae and does not warrant the use of long division techniques?
Can you recommend any reading / books ..etc

although i roughly started as (i dont know why i did this except for the fact that a^4 was somewhere in the dividend expression as given)
(a^2 + 2a + 5 )^2 =
(a^2)^2 + 4a^2 + 25 + 4a^3 + 10a + 5a^2
= a^4 + 4a^3 + 9a^2 + 10a + 25 ----- leads me nowhere

(you type maths like in a book -- can you lead me to an editor knowledge like that so that i may learn that/ latex )

My name was sujoy if you remember-- how are you teachers (Eliz s Stapel , PKA , Denis, Mathman et al)

Since the divisor is given to you _ I simply factor out the divisor from the dividend.


4a^4 + 11a^3 + 27a^2 + 17a + 5 = 4a^2*(a² + 2a + 5) - 8a³ - 20a² + 11a³ + 27a² + 17a + 5

= 4a^2*(a² + 2a + 5) + 3a³ + 7a² + 17a + 5 = 4a^2*(a² + 2a + 5) + 3*a*(a² + 2a + 5) - 6a² - 15a + 7a² + 17a + 5

= 4a^2*(a² + 2a + 5) + 3*a*(a² + 2a + 5) + a² + 2a + 5

= (4a² + 3a + 1)(a² + 2a + 5)

However, this process tantamount to doing the long-division, without doing the long-division.

 

[Deleted member 4993]

Why not use next method https://en.wikipedia.org/wiki/Polynomial_long_division ?

because OP said ....

..... rather than using the long division method of algebra division

 

[amigo1]

Since the divisor is given to you _ I simply factor out the divisor from the dividend.


4a^4 + 11a^3 + 27a^2 + 17a + 5 = 4a^2*(a² + 2a + 5) - 8a³ - 20a² + 11a³ + 27a² + 17a + 5

= 4a^2*(a² + 2a + 5)+ 3a³ + 7a² + 17a + 5 = 4a^2*(a² + 2a + 5) + 3*a*(a² + 2a + 5) - 6a² - 15a + 7a² + 17a + 5

= 4a^2*(a² + 2a + 5) + 3*a*(a² + 2a + 5) + a² + 2a + 5

= (4a² + 3a + 1)(a² + 2a + 5)

However, this process tantamount to doing the long-division, without doing the long-division.

Thank you teacher
When ever i am asking for help it is like a deluge of blessings that i get in this forum. Today i wonder that had there been such a senario when i was in my learning stages . However i am also looking at management entrance exams -- although not actively anyway ;and the merited answers are surely of much help