In Exercises 41–50, subtract the polynomials. Assume that all var...

Question

In Exercises 41–50, subtract the polynomials. Assume that all variable exponents represent whole numbers.(3x⁴y² + 5x³y − 3y) − (2x⁴y² − 3x³y − 4y + 6x)

Accepted Answer

Hey, everyone in this problem, we're asked to perform subtraction. Give an expression. The expression we're given is five X to the exponent four Y squared plus seven X cubed, Y minus Y minus the second part of the expression nine X to the exponent four Y squared minus 23 X cubed, Y minus six, Y plus 17 X. We're given four answer choices. Option A four X to the exponent four Y squared plus 30 X cube, Y plus five, Y plus 17 X. Option B four X to the exponent four Y squared plus 30 X cubed, Y plus five, Y minus 17 X. Option C negative four X to the exponent four Y squared plus 30 X cube, Y plus five, Y plus 17 X. And option D negative four, X to the exponent four Y squared plus 30 X cubed, Y plus five, Y minus 17 X. We're gonna start by rewriting the expression. We were given five X to the exponent four Y squared plus seven X cubed Y minus Y. And all of this is subtracted by nine X to the exponent four Y squared minus X cube, Y minus six Y plus X. Now we have minus this big expression on the right hand side of our problem. So what we wanna do is we want to distribute that minus, treat it as a negative and bring it inside the brackets and expand it to each term inside of our bracket. So if we do this, we get five X to the exponent four or Y squared plus 7 x cubed, Y minus Y minus nine X to the exponent four Y squared Plus 23 x cubed, Y plus 6, Y -17 X OK. So our 23 X cubed Y term and our six Y term that were originally subtracting are now becoming plus because we've multiplied by a negative, OK. Negative times a negative gives us a positive. So now we've expanded this expression as much as we can, I wanna simplify and combine like terms. We have a five X to the exponent four Y squared term and we have minus nine X to the exponent for Y squared. This is gonna give us -4 X to the exponent four Y squared. Moving to our next term, seven X cube to the exponent Y. Do we have any other like terms? Well, yes, we have 23 X Q Y so seven X cube, Y plus 23 X cube, Y gives us 30 X Q Y moving to our next term. We have minus Y and we also have plus 6 y this gives us plus 5 1. OK. Our next term is minus nine X to the exponent for Y squared, which we've already dealt with. In our first term, we have. Our next term is our X Q Y term, which we've also already dealt with. And then our six Y term, again, we've already dealt with that. So we're left with just our last term negative 17 X. So we have minus 17 X in our expression and that's it. We've subtracted the expressions we were given, we've simplified as much as we can and we found a result and that result is negative four X to the exponent four Y squared plus 30 X cubed Y plus five Y minus 17 X which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with College Algebra?

In Exercises 41–50, subtract the polynomials. Assume that all variable exponents represent whole numbers.(3x⁴y² + 5x³y − 3y) − (2x⁴y² − 3x³y − 4y + 6x)

Watch next

Multiplying Polynomials practice set