Answer to Assignment 3B
Factor:
1. 5x2y6z4 – 15x4y5z3 + 20x3y5z2
= 5x2y5z2 (yz2 – 3x2z + 4x)
2. 12xy + 20yz – 40wy
= 4y (3x + 5z -10w)
3. 16×2 – 49
= (4x -7) (4x+7)
4. 4y3 – 64u2y
= 4y (y2 – 16u2)
= 4y (y – 4u) (y + 4u)
5. y2 + 8y + 7
= (y + 7) (y+1)
6. 15z2 – 26z – 21
= (5z +3) (3z – 7)
7. 21uv3 – 46uv2 + 24uv
= uv (21v2 – 46v + 24)
8. 32y2 – 12y – 2
= 2 ( 16y2 – 6y -1)
= 2 (8y + 1) (2y -1)
9. 36×2 + 36x + 9
= 9 (4×2 + 4x + 1)
= 9 (2x + 1) (2x +1)
10. 3×2 + 54x + 195
= 3 (x2 + 18x + 65)
= 3 (x + 5) (x + 13)
Answers to Assignment 4B
Solve:
1. (2x – 3)(x – 4) = 0
(2x – 3) = 0 (x – 4) = 0
2x = 3 x = 4
x = 3/2
2. x2 + 3x – 28 = 0
(x +7) (x – 4) = 0
(x +7) = 0 (x – 4) = 0
x = -7 x = 4
3. x2 + 16 = -10x
x2 + 10x +16 = 0
(x + 2) (x +8) = 0
(x + 2) = 0 (x +8) = 0
x = – 2 x = – 8
4. 5×2 = 6x – 16x2
5×2 + 16×2 – 6x = 0
21×2 – 6x = 0
3x (7x – 2) = 0 (7x – 2) = 0
x = 0 x = 2/7
5. 6×2 + 24x = 2×2 – 36
6×2 – 2×2 +24x +36 = 0
4×2 + 24x +36 = 0
(2x + 6) (2x + 6) = 0
(2x + 6) = 0 (2x + 6) = 0
2x = -6 2x = -6
x = – 3 x = – 3
6. 3×2 – 17x + 20 = 0
(3x – 5) (x – 4) = 0
(3x – 5) = 0 (x – 4) = 0
x = 5/3 x = 4
7. x2 – 36 = 0
x2 = 36
x = √36
x = + 6
8. The area of a triangle is 13.5 square meters. Find the base and height of the rectangle if its height is 6 meters greater than its base. Use an equation and the formula area of a triangle = 0.5(base)(height).
Area of Triangle = 0.5 (base) height)
13.5 m2 = 0.5 (base) (height)
Base = x
Height = x + 6
13.5 m2 = 0.5 (x) (x+6)
13.5 m2 = 0.5×2 + 3x
0.5×2 + 3x – 13.5 = 0
(x2 +6x -27) = 0
(x + 9) (x – 3) = 0
(x + 9) = 0 (x – 3) = 0
x = -9 x = 3
But, there is restriction whereas: there is no negative side, and we only need to take the 3 m as the value of x or the value of the base.
Base = 3m and the Height = 9 m
To check if this is correct:
Area = 0.5 (3m) (9m)
Area = 0.5(27m2)
Area = 13.5 m2
This is the given area of the triangle.
Credit:ivythesis.typepad.com
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