When the product (3x+2y+1)(x+4y+5) is expanded, what is the sum of the coefficients of the terms which contain a nonzero power of y?
I got
3x^2+14xy+16x+8y^2+14y+5
would the answer be 36?
2 answers
If you multiplied it correctly, the sum is 36
The sum of the coefficients of all the terms will be equal to the value obtained when the final polynomial in terms of x and y is evaluated with x=1 and y=1, since products or powers of x and y will collapse to 1 and the result will be the sum of all the coefficients. Since the un-expanded version is equivalent, we can plug in: (3(1)+2(1)+1)((1)+4(1)+5)=(6)(10)=60. However, we are only interested in those terms with powers of y in them. There are four possible types of terms: those with powers of x and y, those with powers of y only, those with powers of x only, and those with neither variable (the constant term). The sum of the coefficients of all 4
types of terms must add to 60. We are interested in the sum of the coefficients of the first two types of term. We use complimentary counting, and find the sum of the coefficients of the terms containing only x and the constant term. We find this sum by plugging in 1 for x and 0 for y, since this will make any term containing y (including the ones containing x and y) drop out and leave only terms with x and constants. This gives (3(1)+2(0)+1)((1)+4(0)+5)=(4)(6)=24. Thus the value we seek is the sum of the coefficients of the rest of the terms, or 60-24=36
types of terms must add to 60. We are interested in the sum of the coefficients of the first two types of term. We use complimentary counting, and find the sum of the coefficients of the terms containing only x and the constant term. We find this sum by plugging in 1 for x and 0 for y, since this will make any term containing y (including the ones containing x and y) drop out and leave only terms with x and constants. This gives (3(1)+2(0)+1)((1)+4(0)+5)=(4)(6)=24. Thus the value we seek is the sum of the coefficients of the rest of the terms, or 60-24=36
3 answers