Help please?

Question

1) remove the parentheses and simplify

5m-(6m-6n)

2) find the values of the polynomial
when x=2
4x^2-2x+1

3) solve
(x-1)^2=5

4) rewrite the middle term as the sum of two terms and then factor by grouping
x^2+4x-21

5) multiply
2m(5m-3)(4m+7)

6)the product of two consecutive postive even integers is 224. Find the intergers.

Answer 1

1) remove the parentheses and simplify

5m-(6m-6n) = 5m -6m +6n = 6n-m

2) find the values of the polynomial
when x=2
4x^2-2x+1 = 4*2^2 -2*2 + 1 = 16 -4 +1 = 13

3) solve
(x-1)^2=5
x^2-2x+1 = 5
x^2-2x-4 = 0
x= [2 +/- sqrt(4 -4(1)(-4)]/2 = 1 +/- sqrt(5)

4) rewrite the middle term as the sum of two terms and then factor by grouping
x^2+4x-21
x^2+7x -3x -21
=x(x+7)- 3(x+7)
=(x+7)(x-3)

5) multiply
2m(5m-3)(4m+7)=2m(20m^2+23m-21)=40m^2 +47m^2-42m

6)the product of two consecutive postive even integers is 224. Find the intergers
2x = 1st integer and 2x+2 = 2nd integer
2x(2x+2)= 224
4x^2+4x-224 = 0
x^2+x-56
(x+8)(x-7) =0
x-7 = 0 --> x=7, so 2x = 14 = 1st digit
14+2 = 16 = 2nd digit

Answer 2

1) 5m - (6m-6n) = 5m-6m+6n = -m+6n

2) 4(2)^2 - 2(2) + 1 = 16 - 4 + 1 = 13

3) (x-1)^2 = 5
Take sqr of both sides: x-1 = +-sqr[5]
x = 1+-sqr[5]

4) x^2+4x-21 = x^2 + (2x + 2x) - 21 = (x^2+2x+1) + (2x-22)
= (x+1)^2 + 2(x-11)

5) (5m-3)(4m+7) = 20m^2 -12m + 35m - 21 = 20m^2 -23m -21
2m(5m-3)(4m+7) = 40m^3 - 46m^2 - 42m

6) Let 2n be the first positive integer so that 2n+2 is the second
2n(2n+2) = 224
4n(n+1) = 224 by factoring out the 2's
n(n+1) = 224/4 = 56
n^2 + n - 56 = 0
(n-7)(n+8) = 0 so that n = 7 rejecting the negative n of -8
The first is 2n = 2(7) = 14
The second is 2n+2 = 16