Posted by Chirs on Thursday, February 19, 2009 at 10:49pm.
Were doing factoring and i don't get these (^ means squared)
1) m^3m2
2) 6c^5c4
3) ab+mn+an+mb
4) The length of a rectangular garden is 5ft more than the width. The garden has a 3ft wide cement walk sorrounding it. The total area of the garden and the walk is 546ft^. Find the dimensions of the garden
5) Find 2 consecutive integers such that the sum of their squares is 52
6) The perimeter of a rectangle is 22ft and the area is 24ft^. Find the dimensions of the rectangle

Algebra  jane, Thursday, February 19, 2009 at 11:42pm
4)
GARDEN
width = w
length = w +5
GARDEN w/CEMENT WALK
width = w + 6
length = w + 5 + 6 = w + 11
a = 546 ft sq.
a = lw
a = (w+6)(w+11)
546 = (w+6)(w+11)
546 = w^2 + 17w + 66
0 = w^2 + 17w  480
w = (17 +/ sqrt(289  4(1)(480))/2
w = (17 +/ sqrt(289 + 1920))/2
w = (17 +/ sqrt(2209))/2
w = (17 +/ 47)/2
has to be + because it must be positive
w = 30/2 = 15
l = w + 5 = 20

Algebra  jane, Thursday, February 19, 2009 at 11:46pm
6)
P = 22
A = 24
2l + 2w = 22
lw = 24
l = 24/w
2(24/w) + 2w = 22
48/w + 2w = 22
48 + 2w^2 = 22w
w^2  11w + 24 = 0
(w  8)(w  3) = 0
w = 8 or 3
l = 3 or 8
dimensions = 3 x 8

Algebra  jane, Thursday, February 19, 2009 at 11:47pm
3) ab+mn+an+mb
ab + an + mn + mb
a(b + n) + m(n+b)
(a + m)(b + n)

Algebra  drwls, Friday, February 20, 2009 at 12:05am
5) No solution. Obviously 4 and 5 give a sum of squares too low (41) and 5 and 6 give a sum of squares that is too high (61).
If you had said two consecutive EVEN integers, the answer would be 4 and 6.
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