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# HELP

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how to divide 45x^5+6x^4+3x^3+8x+12 by 3x-12 using synthetic division

Jan 8, 2016

#1
+10

45x^5+6x^4+3x^3+8x+12 by 3x-12

This is certainly an unusual one!!!!

First, we want  to factor the denominator  as 3[x - 4]

Then, we want to divide all the coefficients by 3   [don't forget to include the "missing"  power of x^2....it's coefficient will be 0

[45x^5 + 6x^4 + 3x^3 + 0x^2 + 8x + 12]  / [3 (x - 4)] =

This results in :

[15x^5 + 2x^4 + x^3 + 0x^2 + (8/3)x + 4] / (x - 4)

Now, proceed normally

4  [   15        2        1          0          8/3              4          ]

60      248     996      3984          47840/3

--------------------------------------------------------------

15     62      249    996       11960/3      47852/3

And the resultant polynomial  is :

15x^4 + 62x^3 + 249x^2 + 996x + 11960/3  +   remainder of 47852 / 3   Jan 8, 2016
edited by CPhill  Jan 8, 2016
edited by CPhill  Jan 8, 2016

#1
+10

45x^5+6x^4+3x^3+8x+12 by 3x-12

This is certainly an unusual one!!!!

First, we want  to factor the denominator  as 3[x - 4]

Then, we want to divide all the coefficients by 3   [don't forget to include the "missing"  power of x^2....it's coefficient will be 0

[45x^5 + 6x^4 + 3x^3 + 0x^2 + 8x + 12]  / [3 (x - 4)] =

This results in :

[15x^5 + 2x^4 + x^3 + 0x^2 + (8/3)x + 4] / (x - 4)

Now, proceed normally

4  [   15        2        1          0          8/3              4          ]

60      248     996      3984          47840/3

--------------------------------------------------------------

15     62      249    996       11960/3      47852/3

And the resultant polynomial  is :

15x^4 + 62x^3 + 249x^2 + 996x + 11960/3  +   remainder of 47852 / 3   CPhill Jan 8, 2016
edited by CPhill  Jan 8, 2016
edited by CPhill  Jan 8, 2016
#2
0

Hi Chris,

I have used synthaetic division before but if i remember rightly it can only be used when the divisor is a polynomial of degree one.

I can't much see the point of commiting such a limiting proceedure to memory.

I do understand that in this case you were expressely asked to use synthetic division.

Why is this one unusual?

Jan 8, 2016
#3
0

It's unusual - to me - because I don't remember one where we were asked to divide by something in the form of ax + b where "a" was something other than "1"

So....I had to actually factor out the "a" and divide the polynomial in the numerator by it before I could perform the synthetic division.....   Jan 8, 2016
edited by CPhill  Jan 8, 2016