When factoring quadratics the most popular method we teach in high school and college is known as the AC method. The method is straightforward, given a quadratic ax^2 + bx + c, multiply a and c, find factors of AC that can add and make b, replace bx with those two factors written as additve statements, then group, factor, and combine).
For example:
9x^2 -30x + 25
9 * 25 = 225
15 * 15 = 225 and 15 + 15 = 30
9x^2 -15 x -15 x + 25
(9x^2 -15x) + (-15x + 25) **
3x(3x-5) -5(3x-5)
(3x-5)(3x-5)
Now, this approach works for many people but there are a large number of people who get stuck at the stage where I indicated the "**". Especially when it comes to factoring with the negative sign. So there is an alternative approach, I would like to present, that removes a lot of issues with factoring:
9x^2 -30x + 25
9 * 25 = 225
15 * 15 = 225 and -15 + -15 = -30
9x^2 / -15x and 9x^2 / -15x write each factor with x under ax^2
Reduce= 3x/-5
(Numerator + denominator)
(3x-5)(3x-5)
This approach seems much simpler and easier for students to handle and works every time when a,b, and c are mutually prime.
Your task, prove why the second method works just as often as the first.
When factoring quadratics the most popular method we teach in high school and college is known as the AC method. The method is straightforward, given a quadratic ax^2 + bx + c, multiply a and c, find factors of AC that can add and make b, replace bx with those two factors written as additve statements, then group, factor, and combine).
For example:
9x^2 -30x + 25
9 * 25 = 225
15 * 15 = 225 and 15 + 15 = 30
9x^2 -15 x -15 x + 25
(9x^2 -15x) + (-15x + 25) **
3x(3x-5) -5(3x-5)
(3x-5)(3x-5)
Now, this approach works for many people but there are a large number of people who get stuck at the stage where I indicated the "**". Especially when it comes to factoring with the negative sign. So there is an alternative approach, I would like to present, that removes a lot of issues with factoring:
9x^2 -30x + 25
9 * 25 = 225
15 * 15 = 225 and -15 + -15 = -30
9x^2 / -15x and 9x^2 / -15x write each factor with x under ax^2
Reduce= 3x/-5
(Numerator + denominator)
(3x-5)(3x-5)
This approach seems much simpler and easier for students to handle and works every time when a,b, and c are mutually prime.
Your task, prove why the second method works just as often as the first.
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