how to easily multiply
(x-2)(x-3)(x-5)?

Question

anonymous · Answer

Distributive law

amistre64 · Answer

do you know how to multiply numbers?  like 35(56)?

aravindg · Answer

easy way plz

amistre64 · Answer

math isnt about "easy" its about discipline

aravindg · Answer

like not opening brackets mechanically

amistre64 · Answer

if you wanna memorize a bunch of formulas to try to sort thru, thats up to you i spose

anonymous · Answer

lol
there is no royal road to algebra

anonymous · Answer

ok really it was "there is no royal road to geometry" but the point  is the same

anonymous · Answer

It doesn't matter how you do it, you will be using the following property of the real number field: distributivity of multiplication over addition

amistre64 · Answer

(x-2)(x-3)(x-5) ; x=10
(8)(7)(5)

56(5) = 280;  200 + 50 - 30;  2x^2 +5x -30 .. got no idea

anonymous · Answer

lost him/her i am sure

amistre64 · Answer

indeed, but the notion is intriguing lol

amistre64 · Answer

x^2-5x +6
           x-5
---------
x^3-5x^2+6x
      -5x^2+25x-30
------------------
x^3-10x^2+31x-30 ... i wasnt even close :)

amistre64 · Answer

the "easy" way for me is to stack them up like numbers (they are numbers) and multiply like usual

anonymous · Answer

Applying the property recursively until expansion is complete:
$$\begin{eqnarray*}
(x-2)(x-3)(x-5) &=&
(x(x-2)-3(x-2))(x-5) \ &=& 
(x^2 - 2x - 3x + 6)(x-5) \ &=&
x^2(x-5) - 2x(x-5) - 3x(x-5) + 6(x-5) \ &=&
x^3-10x^2+31x-30
\end{eqnarray*}$$