Problem #12

poly1=(3x^2+2x+1)^2 (x+1)^3 (3x^2+5x+1)

poly2=(3x^2+2x+1)(x+1)^2 (3x+2)^2

Notice that poly1 and poly2 have as a common factor (3x^2+2x+1)(x+1)^2. However, if we use Mathematica to divide poly1 by poly2 the common factor will not cancel.

poly1 = Expand[(3 x^2 + 2 x + 1)^2 (x + 1)^3 (3 x^2 + 5 x + 1)]

poly2 = Expand[(3 x^2 + 2 x + 1) (x + 1)^2 (3 x + 2)^2]

1 + 12 x + 63 x^2 + 201 x^3 + 429 x^4 + 633 x^5 + 641 x^6 + 423 x^7 +

162 x^8 + 27 x^9

4 + 28 x + 89 x^2 + 164 x^3 + 180 x^4 + 108 x^5 + 27 x^6

poly1/poly2

(1 + 12 x + 63 x^2 + 201 x^3 + 429 x^4 + 633 x^5 + 641 x^6 +

423 x^7 + 162 x^8 + 27 x^9)/(4 + 28 x + 89 x^2 + 164 x^3 + 180 x^4 +

108 x^5 + 27 x^6)

To get the common factor to cancel, we can use the Cancel command:

Cancel[poly1/poly2]

(1 + 8 x + 23 x^2 + 37 x^3 + 30 x^4 + 9 x^5)/(2 + 3 x)^2

The cancel command is actually shorthand

for

PolynomialQuotient[m[x],PolynomialGCD[m[x],n[x]],x]/PolynomialQuotient[n[x], PolynomialGCD[m[x], n[x]], x]

Problem #12 asks you to check that the command above is equivalent to the Cancel command. In the command above we are dividing m[x] by the common factor between m[x] and n[x], then doing the same to n[x] and then dividing the two expressions. For numbers, if you have the fraction 15/24 you would be dividing 15 by the greatest common factor between 15 and 24 (which is 3) and then divide 24 by the greatest common factor between 15 and 24 (which as I just mentioned is 3) and then divide the two numbers to get the fraction in simplified form. That is 24/3/(15/3)=8/15. So to do the problem, in an input cell, type the command above that is highlighted in green where

m[x]=poly1 and n[x]=poly2 and then apply Simplify to your output. You should get the same result that we obtained above using the Cancel command.

The post Problem #12 poly1=(3x^2+2x+1)^2 (x+1)^3 (3x^2+5x+1) poly2=(3x^2+2x+1)(x+1)^2 (3x+2)^2 Notice that poly1 and poly2 appeared first on PapersSpot.