Question

Use the binomial theorem to write the binomial expansion

(x-2)^4

Answer 1

@mathstudent55

Answer 2

1
11
121
1331
14641
pascals triangle, (x-2)^4 is a fourth order polynomial so I'll use the fifth row.
\[=x^{4}+4(-2)x ^{3}+6(-2)^{2}x ^{2}+4(-2)^{3}x+1(-2)^{4}\]\[=x^4-8x^3+24x^2-32x+16\]

Answer 3

wait so is that simplified?

Answer 4

the bottom expression is fully simplified.

Answer 5

oh k

Answer 6

how about this one: (2x+5)^5

Answer 7

let me expand my pascals triangle then see if you can figure it out.
1
11
121
1331
14641
1 5 10 10 5 1
use the bottom row.

Answer 8

i dont get how u set it up like the tree thing can u please explain to me

Answer 9

alright, pascals triangle, also know as binomial theorem is a method for multiplying binomial , that is equations of the form (a+b)^x, the triangle is found by simple addition. starting with 1, the next row is then the sum of the two numbers above it in this case 1+0=1, then 1+0=1. the same goes for the third row 0+1=1, 1+1=2, 1+0=1 so the third row is 1 2 1. the fourth works the same way 0+1=1, 1+2=3, 2+1=3, 1+0=1 so the fourth row is 1 3 3 1. all rows follow this pattern. To find which row you need simply look at your exponent in this case 5 then add 1, so we get 6; thus you would use the 6th row 1 5 10 10 5 1 to do this problem. Furthermore, the coefficients, found with the triangle effect the terms (a and b, or (2x) and 5). simply take each term with the coefficient and the exponent varying. I'm going to explain this using (x+1)^3 for my own ease, the function expands to x^(3-0)*1^(0+0) {notice how the zero interacts in this step, the next term will use 1 instead of zero} +x^(3-1)*1^(0+1) + x^(3-2)*1(0+2) + x^(3-3)*1^(0+3) I stop here because I am using a third order polynomial. Then reduce to x^3 + x^2 + x +1; this maybe a bad example because 1^n=1 for all integers.

Answer 10

ohhk so can u help me set up 2x+5^5

Answer 11

1 5 10 10 5 1; ((2x)+5)^5; that is all the information we need it becomes\[1*5^0(2x)^5 + 5*5^1*(2x)^4+10*5^2*(2x)^3+10*5^3(2x)^2+5*5^4(2x)^1+1*5^5(2x)^0\]

Answer 12

that final term is cut off on my screen it says, 1*5^5*(2x)^0