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
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
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?
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.....