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5x(2x-5)-8(2x-5) <-- try to solve this
um
use multiplication

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Other answers:

okay i got 10x-25 -16x+40
10x^2-25x-16x+25 which would be 10x^2-41x+25
great now solve further..
Correct but remember 10x is 10 x ²
um
idk what to do now
yea when u multiply x's it becomes square
x.x=x^2
When you multiply 2 numbers with x's the x's multiply too...which becomes x times x which is x squared...or x²
10x2
and then what
what about the 16x
-4x2
now u hv 10x^2-41x+25.. after adding x's
i mean -6x2
16x stays the same...do you have any other x's? ***you cannot combine x² and x.
oh
wait perhaps I've looked wrong let me do this out
so its 10x2 -41x+25
(5x-8) (2x-5) 10x² - 25x - 16x + 40 so yes 10x² -41x + 40
i have like 13 problems like these
did u get some idea how to do it?
so thats your final answer
Use the FOIL method i used above for this e.g
(7w+5)(11w-3)
let me do it over: 5x(2x-5)-8(2x-5) 10x^2-25x-16x+40 10x^2-41x+40
so 77w2+55w-33w-15
yea use the same trick with this: 7w(11w-3)+5(11w-3) <- they just replaced x with w's
is that right
u got first one right 7w x 3 = 21w
so it will be 77w^2-21w+55w-15 77w^2+34w-15
yup the next one is *b-2)(b2-b+1)
try it..
b(b^2-b-1)-2(b^2-b+1)
um
its not tht hard, give it a try
i got 2b2-2b-1b-2b2-2b+2
ur second part is right..
my first part is wrong
when u multiply bxb^2 it will be b^3
so its b3
right!
same goes with the next one.. u just add power to the variable when u multiply
so what will it look i like
b(b^2-b-1)-2(b^2-b+1) \[b ^{3}-b ^{2}-b-2b ^{2}+2b-2\]
can you solve that

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