Find the term in the expansion of (x² + y²)^5 containing x^4 as a...

Question

Find the term in the expansion of (x² + y²)^5 containing x^4 as a factor.

Accepted Answer

Hello everybody and I today that are going to begin this question that states identify which term in the expansion of open parentheses. X Q U plus Y to the fourth exponent closed parentheses to the exponent seven has X to the exponent nine. Now the answers provided are a seven X to the ninth power Y to the fourth power B X to the ninth power, Y to the 12th power C 35 X to the ninth power Y to the 16th power N D 21 X to the ninth power Y to the 20th power. No, for this question, we're going to use different ways of expansion. So the, we're trying to expand open parentheses, X cubed plus Y to the fourth power closed parentheses to the seventh power. So a rule we have of expansion here. Let's just recall that open parentheses are plus one, that term, the R plus one term of expansion of something in the form of open parentheses. A plus B closed parentheses to the exponent N is the binomial coefficient. So open parentheses, N on top and R on the bottom, the vinyl coefficient of that multiplying A to the exponent N minus R multiplying B to the exponent R. So that is just a rule that we have for our expansion and something that we should keep in mind for this question. So with that in mind, let's see how that applies to our formula. So in our case, A is equal to X cubed B is equal to Y to the fourth power and N is equal to seven. And all we're doing here is comparing our recall equation equation with the equation that we have at hand. So now we know the A, the B and the N, we can plug that into a recall equation. So we'll have the binomial coefficient of seven on top and R on the bottom closed parentheses, multiplying X cubed to the exponent seven minus R multiplying Y to the fourth power to the exponent R. So now we have to equate the term where we have X cubed to exponent seven minus R we're trying to solve for the term that has X to the ninth power. So we have to equate our X to the cubed X cubed to the exponent seven minus R that has to be equal to X to the ninth power since that's the term that we want. So we want the X to be raised to the ninth power. And that's what we equate these two equations. So then we'll have that X to the exponent three multiplying seven minus R will be equal to X to the ninth power. So now that we have two axes equal to each other raised to a power, that means that those powers have to be equal to each other. So we have that three multiplying and open parentheses, seven minus R is equal to nine. So all we did was remove the X base and now we just have the exponents equal to each other. And to solve this, we just have, we move the three over by dividing. So we have that seven minus R is equal to three, meaning that R is equal to four. Now remember that we're solving for the R plus one term, so R plus one equals fifth time. So the fifth term in this expansion is the term that's going to have the X C exponent nine. So now that we have our, we can go back to our recall expression that we had and plug everything in. So we'll have the binomial coefficient of seven on top and four on the bottom since R was on the bottom. And we close that parentheses, multiplying an open parentheses of X cubed, closed parentheses to the exponent seven minus four, multiplying Y to the fourth power to the exponent four. And that will be equal to the binomial coefficient of open parentheses seven on top and four on the bottom, close parentheses of X. So the ninth power multiplying Y to the 16th power. So all we did there was expand the exponent. So if we have seven minus four, when we look at, when we looked at the X seven minus four is three, so we have X cubed to the third power. And when you have a power to another power, you just multiply them. So three times three or three multiplied by three is nine. So we have X to the ninth power and as for the Y, we had Y to the fourth to the fourth. So we multiply the four and the four and we have six Y to the 16th. So now we can expand the binomial coefficient and we'll have that seven factorial divided by four factorial multiplying an open parentheses, seven minus four, closed parentheses. Factorial, all of that would be on the denominator and that fraction is multiplying X to the ninth, multiplying Y to the 16. So we can expand some of that factorial and we'll have on the numerator, we'll have seven multiplying five, multiplying six, multiplying four, multiplying three factorial. And I'll stop there. I, we, we, we should continue to what? Buying a two and what's buying a one. But I'll stop at three factorial because as I show you now we're going to be able to do some canceling. So if we complete the parentheses, we have seven minus four in the denominator will have four factorial multiplying three factorial. And now we can treat the three factorial like any other number or any other variable since we have it in the numerator and the denominator, we can cancel out the three factorial and we cannot forget that we're still multiplying that entire fraction by X to the ninth power and Y to the 16th power. So now we can continue to do some expansion. So once after we cancel the three factorial, we'll have under the numerator seven, multiplying five, multiplying six multiplying four divided by four, multiplying three, multiplying two multiplying one. And that entire fraction multiplying X to the ninth and Y to the 16th. So now let's do some canceling. We have a four on the numerator and a four in the denominator. So that four can cancel out in the denominator. We have a three multiplying a two which is six. And in the numerator we have a six. So we can cancel the six and the, and the numerator and the three and the two and the denominator and will be left with seven multiplying five in the numerator divided by one in the denominator. And that fraction multiplying X to the ninth and Y to the 16th. And we'll be left with seven, multiplying five is 35 divided by one is 35. So the answer will be 35 X to the ninth, Y to the 16th. So this is answer choice C and that means that the term and the expansion that will have the X to the ninth exponent is 35 X to the ninth y to the 16th. So I hope that was helpful and until next time.

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Find the term in the expansion of (x² + y²)^5 containing x^4 as a factor.