Algebra

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Were doing factoring and i don't get these (^ means squared)
1) m^-3m-2

2) 6c^-5c-4

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 - ,

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 - ,

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 - ,

3) ab+mn+an+mb
ab + an + mn + mb
a(b + n) + m(n+b)
(a + m)(b + n)

• Algebra - ,

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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