In Exercises 49–52, use the Binomial Theorem to expand each expre...

Question

In Exercises 49–52, use the Binomial Theorem to expand each expression and write the result in simplified form. (x^3 +x^-2)^4

Accepted Answer

Hey everyone here we are asked to expand the expression the quantity of X to the fourth power plus X. To the negative third power all raised to the power of five. And we need to expand this expression using the binomial theorem. Now before we start expanding, we need to recall the binomial theorem which gives us a formula for expanding Bino meals. And it tells us for any positive in the N. That the quantity of A plus B to the N power is equal to N zero times A. To the power of N plus n. One times A. To the power of n minus one times B plus n. Two times A. To the power of n minus two times B squared plus n three times A. To the power of n minus three times B cubed. And then we continue to expand the expression until we reach the term where we have the coefficient of n N times B to the power of N. So with this formula in mind we can go ahead and start identifying the variables in relation to our given expression. So we can start expanding. So beginning with our A term here. Well, A is just the left most term in our given expression. So A will be X to the fourth power. And that leaves the other term to BB. So we have B. Is X to the power of negative three. And lastly our our end term is the power to which the binomial is raised to. So our end in this case is going to be five. So with these things identified, we can go ahead and start expanding. So we have the quantity of X. To the fourth power plus X. To the negative third power raised to the power of five expands to our first term being instead of n. For our coefficient we have five. So we have 50 as the coefficient times A. Which is X. To the fourth power raised to the end power which again is five. And then we can move on to our next term where we have the coefficient 51 times A. Which again is X. To the fourth power raised to the power of five minus one times B. Which is X. To the negative third power raised to the power of one. And then we can move on to our next term. We have the coefficient 52 times X. To the fourth power raised to the power of five minus two times X. To the negative third power raised to the power of two. For our next turn we have the coefficient times X. To the fourth power raised to five raised to the power of five minus three times X. To the negative third power raised to the power of three. For our next turn we have the coefficient 54 times X to the fourth power raised to the power of five minus four times X. To the negative third power raised to the power of four and now I'll continue below here, our final term takes the coefficient 55 times X. To the power of four raised to the power of five minus five times X. To the negative third power raised to the power of five. And with this we are done expanding And we just need to start simplifying So again our beginning expression we had X. The quantity of X to the power of four plus X. To the power of negative three. All raised to the power of five. And now keeping the coefficient the same as same. For now, our first term we have five zero and then our X to the fourth. Power raised to the fifth power simplifies to X to the power of 20. For our next term we have the coefficient 51 and our X term simplifies to X to the power of 13. And then for our next turn we have the coefficient 52. And our X terms here simplify two X. To the power of six. For our next term, with the coefficient 53 we have 53 times X. To the power of negative one plus the next term we have 54 times X to the power of negative eight. And lastly we have 55 times X to the power of negative 15. So with this, with our X term simplified, we can go ahead and start calculating our coefficients here. So if we recall the binomial coefficient N. R is equal to n factorial divided by the quantity of n minus r factorial times R factorial and now we need to note here that in our coefficients the upper term is the value for end and the lower term is the value for our So now we can start writing out our coefficients beginning with our first term. We have five factorial divided by the quantity of five minus zero factorial times zero factorial and this is the coefficient of X to the b Power of 20 plus. The next term our coefficient is five factorial divided by the quantity of 5 -1 factorial times one factorial and this is the coefficient of X to the power of 13. And now for our next term we have five factorial divided by the quantity of five minus two factorial times two factorial and this is the coefficient of X to the power of six. For our next term we have five factorial divided by the quantity of five minus three factor real times three factorial and this is the coefficient of X to the power of negative one. For our next term we have the coefficient five factorial divided by the quantity of five minus four factorial times four factorial which is the coefficient of X to the power of neck AIDS. And for our last term here we have the coefficient five factorial divided by the quantity of five minus five factorial times five factorial and this is the coefficient of our last term which is X to the power of negative 15. So now we just need to start simplifying. So first our simplification can just be the subtraction that happens in the denominator. So with our first coefficient we have five factorial divided by 5 0 is just five. So we have five factorial times zero factorial times X to the power of 20. Then for our next coefficient we have five factorial divided by five minus one, gives us four factorial and then we have times one factorial is all multiplied by X to the power of 13. Then for the next term we have five factorial divided by five minus two gives us three factorial times two factorial all multiplied by x to the power of six plus the next term we have five factorial divided by five minus three gives us two factorial times three factorial times X to the power of negative one plus the next turn we have five factorial divided by five minus four gives us one factorial times four factorial which is the coefficient of X to the power of negative aids. And for our last term here we have five factorial divided by five minus five gives us zero factorial times five factorial and this is the coefficient of X to the power of negative 15. So from here we need to keep simplifying. So beginning with our first term we can clearly see that the five factorial in the numerator cancels out with the five factorial in the denominator. And we also need to recall that zero factorial is equal to one. So with this our coefficient for X to the power of 20 is just one. So we have one times x to the power of 24 For our next term. Here we see that we have this four factorial in the denominator so we can expand our numerator until we also have four factorial in the numerator. So doing this, we have five times four factorial in the numerator. All divided by four factorial times one factorial. And again this is the coefficient of X to the power of 13. And so from here we can move on to the next term where we have this five, sorry, this three factorial in the denominator. So we can again expand our numerator. So doing this we have five times four times three factorial in the numerator. All divided by three factorial and then times two factorial but our two factorial can be expanded to just two times one and again this is the coefficient of X to the power of six. Now for our next term again we have the same situation as our previous coefficient where we have this three factorial in the denominator here so we can expand our numerator and we have five times four times three factorial, all divided by two factorial which expands to two times one, times three factorial and this is the coefficient of X to the power of negative one. For our next term here we have a four factorial in the denominator. So again we can expand our numerator and we get five times four factorial in the numerator, all divided by one factorial times four factorial. And this is the coefficient of X to the power of negative AIDS. And then for our last term here we can simplify it just as we did with our first term. So we see that the five factorial cancel out in the numerator and the denominator and then our zero factorial is just one. So we're left with the coefficient of one times X to the power of negative 15. So now we can make our final simplifications here. And so we see that our original expression where we have the quantity of X to the fourth power plus X to the negative third power all raised to the power of the eye. We can see the simplified expansion for our first term we just have X to the power of 20 for our next term. Again we see that these four factorial is canceled out. So we're left with five divided by one factorial which is the same as five divided by one. So our term here just turns out to be five Times X to the power of 13 then for our next term we see that the three factorial cancel out. So we're left with 20 divided by two giving us the term 10 times X. To the power of six for the next term we see the same thing and we have 10 times X to the power of negative one for the following term. Again we see the four factorial is canceled out and we're left with five divided by one factorial, which is just one. So we have the term five times X. To the power of negative AIDS. And for our last term we can just exclude the one as the coefficient since it's implied. So we have the last term as X to the power of negative 15. And with this we are done simplifying and expanding. So we are left with answer choice C. Thanks for watching. I hope you found this video helpful. And I'll see you again next time.

In Exercises 49–52, use the Binomial Theorem to expand each expression and write the result in simplified form. (x^3 +x^-2)^4

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