anonymous
  • anonymous
(5x-8) (2x-5)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
5x(2x-5)-8(2x-5) <-- try to solve this
anonymous
  • anonymous
um
anonymous
  • anonymous
use multiplication

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
okay i got 10x-25 -16x+40
anonymous
  • anonymous
10x^2-25x-16x+25 which would be 10x^2-41x+25
anonymous
  • anonymous
great now solve further..
johnweldon1993
  • johnweldon1993
Correct but remember 10x is 10 x ²
anonymous
  • anonymous
um
anonymous
  • anonymous
idk what to do now
anonymous
  • anonymous
yea when u multiply x's it becomes square
anonymous
  • anonymous
x.x=x^2
johnweldon1993
  • johnweldon1993
When you multiply 2 numbers with x's the x's multiply too...which becomes x times x which is x squared...or x²
anonymous
  • anonymous
10x2
anonymous
  • anonymous
and then what
anonymous
  • anonymous
what about the 16x
anonymous
  • anonymous
-4x2
anonymous
  • anonymous
now u hv 10x^2-41x+25.. after adding x's
anonymous
  • anonymous
i mean -6x2
johnweldon1993
  • johnweldon1993
16x stays the same...do you have any other x's? ***you cannot combine x² and x.
anonymous
  • anonymous
oh
johnweldon1993
  • johnweldon1993
wait perhaps I've looked wrong let me do this out
anonymous
  • anonymous
so its 10x2 -41x+25
johnweldon1993
  • johnweldon1993
(5x-8) (2x-5) 10x² - 25x - 16x + 40 so yes 10x² -41x + 40
anonymous
  • anonymous
i have like 13 problems like these
anonymous
  • anonymous
did u get some idea how to do it?
anonymous
  • anonymous
so thats your final answer
anonymous
  • anonymous
Use the FOIL method i used above for this e.g
anonymous
  • anonymous
(7w+5)(11w-3)
anonymous
  • anonymous
let me do it over: 5x(2x-5)-8(2x-5) 10x^2-25x-16x+40 10x^2-41x+40
anonymous
  • anonymous
so 77w2+55w-33w-15
anonymous
  • anonymous
yea use the same trick with this: 7w(11w-3)+5(11w-3) <- they just replaced x with w's
anonymous
  • anonymous
is that right
anonymous
  • anonymous
u got first one right 7w x 3 = 21w
anonymous
  • anonymous
so it will be 77w^2-21w+55w-15 77w^2+34w-15
anonymous
  • anonymous
yup the next one is *b-2)(b2-b+1)
anonymous
  • anonymous
try it..
anonymous
  • anonymous
b(b^2-b-1)-2(b^2-b+1)
anonymous
  • anonymous
um
anonymous
  • anonymous
its not tht hard, give it a try
anonymous
  • anonymous
i got 2b2-2b-1b-2b2-2b+2
anonymous
  • anonymous
ur second part is right..
anonymous
  • anonymous
my first part is wrong
anonymous
  • anonymous
when u multiply bxb^2 it will be b^3
anonymous
  • anonymous
so its b3
anonymous
  • anonymous
right!
anonymous
  • anonymous
same goes with the next one.. u just add power to the variable when u multiply
anonymous
  • anonymous
so what will it look i like
anonymous
  • anonymous
b(b^2-b-1)-2(b^2-b+1) \[b ^{3}-b ^{2}-b-2b ^{2}+2b-2\]
anonymous
  • anonymous
can you solve that

Looking for something else?

Not the answer you are looking for? Search for more explanations.