please explain to me how to get an answer, and how you came to this conclusion with detailed steps...
(4c + 5)(5c - 6)
(2x - 5y)^2
(4c + 5)(5c - 6)
(2x - 5y)^2
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Have you leared FOIL yet? If not then its this pattern. Take a curved line and draw it from 4c to 5c. That's what you multiply first. Then 4c * -6. Then 5 * 5c. Then lastly 5 *-6. Then do the same pattern for the other equation.
Ok so take 4c(5c) to get 20c^2
Then take 4c(-6) to get -24c
Then 5(5c) to get 25c
And then 5(-6) to get -30
So your equation for the 1st one is 20c^2-24c+25c-30 which simplifies to
20c^2+c-30. This cannot be simplified any further so this is it.
Then for the next one same thing
Take 2x(2x) to get 4x^2
then do 2x(-5y) to get -10xy
Next do -5y(2x) to get -10xy
and then finally -5y(-5y) to get 25y^2
Therefore you get 4x^2-10xy-10xy+25y^2 which simplifies to
4x^2-20xy+25y^2
Hope this helps
Ok so take 4c(5c) to get 20c^2
Then take 4c(-6) to get -24c
Then 5(5c) to get 25c
And then 5(-6) to get -30
So your equation for the 1st one is 20c^2-24c+25c-30 which simplifies to
20c^2+c-30. This cannot be simplified any further so this is it.
Then for the next one same thing
Take 2x(2x) to get 4x^2
then do 2x(-5y) to get -10xy
Next do -5y(2x) to get -10xy
and then finally -5y(-5y) to get 25y^2
Therefore you get 4x^2-10xy-10xy+25y^2 which simplifies to
4x^2-20xy+25y^2
Hope this helps
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For these you use a method called FOIL.
(4c +5)(5c -6)
([4c*5c] +[4c*-6] +[5*5c]+ [5*-6])
first outer inner last
so the first number of each set, then the "outer ones" ( 4c and -6)
( [20c^2-24c+25c-30])
(20c^2 +c -30) simplified
for your second problem, its better to expand it out first.
(2x-5y)(2x-5y)
( [ 2x*2x] + [2x*-5y] +[-5y*2x]+ [-5y*-5y] )
( [4x^2] + [-10xy] + [-10xy] + [25y^2] )
( 4x^2 +25y^2) the 10xy term cancels out b/c one is positive and the other is negative.
hope it helped!
(4c +5)(5c -6)
([4c*5c] +[4c*-6] +[5*5c]+ [5*-6])
first outer inner last
so the first number of each set, then the "outer ones" ( 4c and -6)
( [20c^2-24c+25c-30])
(20c^2 +c -30) simplified
for your second problem, its better to expand it out first.
(2x-5y)(2x-5y)
( [ 2x*2x] + [2x*-5y] +[-5y*2x]+ [-5y*-5y] )
( [4x^2] + [-10xy] + [-10xy] + [25y^2] )
( 4x^2 +25y^2) the 10xy term cancels out b/c one is positive and the other is negative.
hope it helped!
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(4c + 5)(5c - 6) --> 20(c^2) + c - 30
(2x - 5y)^2 --> (2x - 5y)(2x - 5y) --> 4(x^2) - 20xy + 25(y^2)
*note* the parentheses in my answers aren't necessary, they just help to show the the ^2 only goes to the variable
when multiplying polynomials, just sure each term gets multiplied by all the terms outside of its parentheses, so 4c(5c - 6) + 5(5c - 6) --> 4c(5c) + 4c(- 6) + 5(5c) + 5(- 6), and the same for the second problem
hope this helps
(2x - 5y)^2 --> (2x - 5y)(2x - 5y) --> 4(x^2) - 20xy + 25(y^2)
*note* the parentheses in my answers aren't necessary, they just help to show the the ^2 only goes to the variable
when multiplying polynomials, just sure each term gets multiplied by all the terms outside of its parentheses, so 4c(5c - 6) + 5(5c - 6) --> 4c(5c) + 4c(- 6) + 5(5c) + 5(- 6), and the same for the second problem
hope this helps
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(4c+5)(5c-6) = 20c^2 - 24c + 25c - 30 = 20c^2 + c - 30 (FOIL, First Outer Inner Last, then simplify.)
(2x-5y)^2 = (2x-5y)(2x-5y) = 4x^2 -10xy - 10xy + 25y^2 = 4x^2 + 25y^2 -20xy (That's as far as I can get it. FOIL again, then simplify again.)
(2x-5y)^2 = (2x-5y)(2x-5y) = 4x^2 -10xy - 10xy + 25y^2 = 4x^2 + 25y^2 -20xy (That's as far as I can get it. FOIL again, then simplify again.)
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1) 20c + 15c - 24c - 30 = 11c -30
2) 4x^2 - 25y^2
2) 4x^2 - 25y^2
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these don't really get to much simpler than there current form
you can like factor it out, but thats about it
you can like factor it out, but thats about it