Question

Binomial Multiplication, idk what is done!

Answer 1

Use the formula.

Answer 2

i marked the step which is messing up with me!

Answer 3

??

Answer 4

expand the terms

Answer 5

like when we multiply them, the terms are getting less.

Answer 6

(a+b)^2 = a^1 + 2ab + b^2 right?

Answer 7

except for the typo

Answer 8

\begin{eqnarray*}
(1-10x+40x^2+...)(1+27x+324x^2+...&=&(1)(1+27x+324x^2+...)-10x(1+27x+324x^2+...)+40x^2(1+27x+324x^2+...) \\
&=&(1+27x+324x^2+...)-10x(1+27x+...)+40x^2(1+...)
\end{eqnarray*}
since powers larger than x^2 are ignored.

Answer 9

that's why they're becoming less... powers greater than x^2 are ignored.

Answer 10

\begin{array}c
&&&&&&1\\
&&&&&1&&1\\
&&&&1&&2&&1\\
&&&1&&3&&3&&1\\
&&1&&4&&6&&4&&1\\
&1&&5&&10&&10&&5&&1\\
1&&6&&15&&20&&15&&6&&1
\end{array}
pascals triangle

Answer 11

why?? why are they ignored?

Answer 12

Because the question asks you to write down the first three terms, so x^0, x^1 and x^2 only are needed.

Answer 13

brb.

Answer 14

why that is not happening?q

Answer 15

.... i never really tried multiply binomial stuff together tho ..

Answer 16

why that is not happening???
i mean that multiplication?

Answer 17

look:
(1-10x+40x^2+...)(1+27x+324x^2+...)
=(1)(1-27x+324x^2+...)
-10x(1-27x+324x^2+...)
+40x^2(1+27x+324x^2+...)
+...
Now first multiplication is needed completely, since all powers will still be x^2 or less.
But second multiplication is needed only up till 27x, since 10x times 324x^2 is a cubic, so is not needed.
In the third multiplication, only 40x^2 times 1 is needed since all the others are cubics or higher.

Clearer now? :)

Answer 18

u here?

Answer 19

yes

Answer 20

Oh Alright:
(1−10x+40x2+...)(1+27x+324x2+...==(1)(1+27x+324x2+...)−10x(1+27x+324x2+...)+40x2(1+27x+324x2+...)(1+27x+324x2+...)−10x(1+27x+...)+40x2(1+...)

that step of yours is right, it's with dots.

Like in my book there are no dots, that thing confused me.

Answer 21

Cool, thanks. :]