Pi R Squared!?

Question

thanks for your help in the polynomial question, but I got stuck again after getting the remainder. I don't know how to get from (15a + 7b + 7)x + (-14a - 6b -3) to 2 equations given that the remainder is also equal to x + 1.
Sorry to have to put this out here, but I don't know any other way to to get to you. Could you help please?
Thanks again :)

Answer 1

I hope I'm reading this right. The method is sometimes called "equating coefficents".



Set (15a + 7b + 7)x + (-14a - 6b -3) = x + 1



Then the coefficent in front of the x on the LEFT must equal the coefficent in front of the x on the RIGHT



Same with the constant on the left, and the constant on the right.



It must follow that:



15a + 7b + 7 = 1

-14a - 6b -3 = 1



And there're your two eqn's in two unknowns.

Answer 2

Hi,



Math goddess explained correctly that because the problem told us the x term would have a coefficient of 1 that the coefficient of the x term in the remainder of the division had to equal 1. Therefore, 15a + 7b + 7, in front of x, has to equal the 1, in front of x in the given remainder.



Likewise, the constant in the remainder of the division has to equal the given constant for the remainder, 1. So -14a - 6b - 3 = 2.



I hope that helps to clarify it for you. :-)

Answer 3

Pie R round; cornbread R squared!





;)