I missed a day of school but have the homework and need a little help figuring it out. How do I factor this:
3y^2 - 20y + 12
Please break this problem down step by step.
3y^2 - 20y + 12
Please break this problem down step by step.
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3y^2 - 20y + 12
3y^2 - 18y - 2y + 12
3y(y - 6) - 2(y - 6)
(y - 6)(3y - 2)
3y^2 - 18y - 2y + 12
3y(y - 6) - 2(y - 6)
(y - 6)(3y - 2)
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Formula: ax² + bx + c = 0
Factored equation: 1/a . (ax + x1).(ax + x2) = 0
When a =1, it can be simplified as: (x + x1).(x + x2) = 0
To factorize the quadratic equation, you need to find the factors by the following rules:
1. The multiplication of the factors equals a x c
2. The addition of factors equals b
For a simpler example:
x² - x - 2 = 0
using the formula ax² + bx + c = 0,
a = 1, b = -1, c = -2
a.c = -2
b = -1
Suppose the factors are x1 and x2, then
Find x1 and x2 such that x1 . x2 = a.c, and x1 + x2 = b
x1 . x2 = -2 and x1 + x2 = -1
Intuitively, x1 = -2 and x2 = 1 (can be applied interchangeably)
x1 . x2 = (-2) . 1 = -2
x1 + x2 = -2 + 1 = -1
The factored equation:
(x + x1) . (x + x2) = 0
(x - 2) . (x + 1) = 0, or
(x + 1) . (x - 2) = 0
For your question:
3y² - 20y + 12 = 0
using the formula ax² + bx + c = 0,
a = 3, b = -20, c = 12
a.c = 3.12 = 36
b = -20
Find the factors x1 & x2 such that x1 . x2 = 36 and x1 + x2 = -20
Intuitively, it would be x1 = -2 & x2 = -18 (can be applied interchangeably)
x1 . x2 = (-2) . (-18) = 36
x1 + x2 = -2 -18 = -20
The factored equation:
1/a . (ax + x1) . (ax + x2) = 0
1/3 . (3x - 2) . (3x - 18) = 0, we put 1/3 behind so that it can be eliminated with the other factor
(3x - 2) . 1/3 (3x - 18) = 0
(3x - 2) . (x - 6) = 0
In addition,
3x - 2 = 0
x = 2/3
or
x - 6 = 0
x = 6
x = 2/3 or x = 6
PS: Intuitively means it can be done by trial and error, by dividing the number.
PPS: Do make sure you don't miss the next class :P
My blog: http://programmerslab.blogspot.com ^^
Factored equation: 1/a . (ax + x1).(ax + x2) = 0
When a =1, it can be simplified as: (x + x1).(x + x2) = 0
To factorize the quadratic equation, you need to find the factors by the following rules:
1. The multiplication of the factors equals a x c
2. The addition of factors equals b
For a simpler example:
x² - x - 2 = 0
using the formula ax² + bx + c = 0,
a = 1, b = -1, c = -2
a.c = -2
b = -1
Suppose the factors are x1 and x2, then
Find x1 and x2 such that x1 . x2 = a.c, and x1 + x2 = b
x1 . x2 = -2 and x1 + x2 = -1
Intuitively, x1 = -2 and x2 = 1 (can be applied interchangeably)
x1 . x2 = (-2) . 1 = -2
x1 + x2 = -2 + 1 = -1
The factored equation:
(x + x1) . (x + x2) = 0
(x - 2) . (x + 1) = 0, or
(x + 1) . (x - 2) = 0
For your question:
3y² - 20y + 12 = 0
using the formula ax² + bx + c = 0,
a = 3, b = -20, c = 12
a.c = 3.12 = 36
b = -20
Find the factors x1 & x2 such that x1 . x2 = 36 and x1 + x2 = -20
Intuitively, it would be x1 = -2 & x2 = -18 (can be applied interchangeably)
x1 . x2 = (-2) . (-18) = 36
x1 + x2 = -2 -18 = -20
The factored equation:
1/a . (ax + x1) . (ax + x2) = 0
1/3 . (3x - 2) . (3x - 18) = 0, we put 1/3 behind so that it can be eliminated with the other factor
(3x - 2) . 1/3 (3x - 18) = 0
(3x - 2) . (x - 6) = 0
In addition,
3x - 2 = 0
x = 2/3
or
x - 6 = 0
x = 6
x = 2/3 or x = 6
PS: Intuitively means it can be done by trial and error, by dividing the number.
PPS: Do make sure you don't miss the next class :P
My blog: http://programmerslab.blogspot.com ^^
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(3y - 2)(y - 6)
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(3y - 2) (y - 6)
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