{Use the binomial theorem to find the eight term of (3x-2y)^13}

Recall that y^2-16=(y+8)*(y-8). That is the common denominator of the three rational expressions .

a) It is a 3rd degree binomial because there are only 2 terms.

b) It is in descending order because the degree of each term goes down as you read the expression from left to right

Try to seek these: By using the suggested format:
Suggested su

1. Adding and Subtracting Binomials
   • 1

     Arrange each term in each binomial in order of degree from greatest to least. The degree of a binomial is the exponent attached to the term. For example, 4x^2 is a second degree term.
     

   • 2

     Multiply each term in the binomial that is being subtracted by -1 to turn it into an addition problem. For example, the problem (8x^2 + 8) - (x^2 - 2) becomes (8x^2 + 8) + (-x^2 + 2).
     

   • 3

     Combine like terms. In the example problem, the x^2 terms are combined and the constant terms are combined, yielding (8x^2 + 8) + (-x^2 + 2) = 7x^2 + 10.
     

   Multiplying Binomials
   • 4

     Understand the F.O.I.L. method. F.O.I.L. is an acronym standing for first, outside, inside and last. It means that you multiply the first number of the first binomial by the first number of the second, then the numbers on the outside (the first term of the first binomial by the second term of the second binomial) and so on. This ensures that both numbers in the first binomial are multiplied by both numbers in the second.
     

   • 5

     Use the F.O.I.L. method to multiply the two binomials together. For example, (3x + 4)(3x - 4) = 9x^2 +12x - 12x - 16. Notice that -12x is the product of the outside terms and -16 is the product of the last terms, 4 and -4.
     

   • 6

     Simplify. There will almost always be like terms to combine. In the example, 12x and -12x cancel out, yielding the answer 9x^2 - 16.
     

   Dividing Binomials
   • 7

     Use the distributive property to divide both terms in the binomial by the monomial divisor. For example, (18x^3 + 9x^2) / 3x = (18x^3 / 3x) + (9x^2 / 3x).
     

   • 8

     Understand how to divide by a term. If you are dividing a higher order term by a lower order term, you subtract the exponent. For example, y^3/y = y^2. The number part of each term is handled like any other division problem. For example, 20z / 4 = 5z.
     

   • 9

     Divide each term in the binomial by the divisor; (18x^3 / 3x) + (9x^2 / 3x) = 6x^2 + 3x.
     

x + 2y = -1

You have to type expand((38x-28y)^4,x). See captured images

Ok so first you need to find what x equals in terms of y. so use the first equation .2x+.7y=1.5 solve for x by the following steps
subract .7y from both sides: .2x= 1.5-.7y --->
divide by .2 on both sides: x= 7.5 - 3.5y. --->
Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->
Distribute the .3: 2.25 - 1.05y -.2y = 1--->
solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->
divide by -1.25 and you get y= 1
Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->
x=7.5-3.5(1) so x= 4 and y=1 yay your done!

Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:



Now to solve using the Ti 89.

Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.






The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.





This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.




This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y






This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2







This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3







This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4






This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5





This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6





Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6

(2x - 3y)(3x + 2y) = 2x*3x + 2x*2y - 3y*3x - 3y*2y
= 6x^2 + 4xy - 9xy - 6y^2
= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:
2x by 3x = 6x squared
2x by 2y = 4xy
-3y by 3x = -9xy
-3y by 2y = 6y squared

6x squared -5xy + 6y squared

Hi joanmae jmeh;

You can add these equations just like you would numbers

3x + 2y = 7
5x - 2y = 1
-----------------------
8x =8
x=1
Now plug x = 1 back into either equation and you can solve for y
Y = 2

When you add the 2 equations together the +2y and the -2y cancel
out

Hope this helps Loringh Please leave a rating for me Thks

just a simple multiplication solve first which is inside the brackets 9x(-7)=-63 3x5=15 (-63)-4x15 -67x15 -1005 .. i think am rite..