[FONT=KaTeX_Main]I have a question about this equation: [/FONT][FONT=KaTeX_Main]
[/FONT]2y2 + 3y -27
[FONT=KaTeX_Main]
The answer says that we should factor it into (2y + 9)(x - 3). Why is this? I can't seem to get the answer. I understand the basics of factoring because if we did it like what equals to 3 but multiplies
to -27? Then I would write 9 * -3. But doesn't this only apply to normal x2? Why do you have to multiply. Aren't you supposed to multiply the coefficient by the 27 then find what equals to 3. That is 9 * -6 = -54.
so that becomes 2y2 + 9y -6y - 27. No? What am I doing wrong. I don't understand this. [/FONT]
First off, I'll guess that
these are typos and are meant to read 'y'. That aside, your method is a bit flawed because it only works for polynomials where the coefficient on the squared term is 1. That is to say, if the equation were of the form y
2 + by + c = 0, you could that factor into \(\displaystyle (y + \alpha)(y + \beta)\), such that \(\displaystyle \alpha \cdot \beta = c\) and \(\displaystyle \alpha + \beta = b\)
When the coefficient on the y term is not 1, things are a bit more difficult. The general form ay
2 + by + c = 0 factors into \(\displaystyle (\gamma \cdot y + \alpha)(\delta \cdot y + \beta)\), this time obeying the constraints that \(\displaystyle \gamma \cdot \delta = a\), \(\displaystyle \gamma \cdot \beta + \delta \cdot \alpha = b\), and \(\displaystyle \alpha \cdot \beta = c\).
As an example of the above, with "real" numbers, to hopefully help it sink it a bit, consider the polynomial 4y
2 + 4y - 3. That factors into (2y - 1)(2y + 3). Here, we have \(\displaystyle \gamma = 2\), \(\displaystyle \alpha = -1\), \(\displaystyle \delta = 2\), and \(\displaystyle \beta = 3\). The three required relationships hold with these values:
- \(\displaystyle \gamma \cdot \delta = 2 \cdot 2 = 4 = a\)
- \(\displaystyle \gamma \cdot \beta + \delta \cdot \alpha = 2 \cdot 3 + 2 \cdot (-1) = 4 = b\)
- \(\displaystyle \alpha \cdot \beta = (-1) \cdot 3 = -3 = c\)
However, to be honest, I don't even bother factoring "manually" for polynomials where the coefficient of the squared term isn't 1. I just jump straight to using the quadratic formula, which works for any polynomial with any real-valued coefficients, even if they're not integers. You may have seen the formula floating about. It says that any polynomial \(\displaystyle ax^2 + bx + c\) can be factored into \(\displaystyle (x - \alpha)(x - \beta)\), where \(\displaystyle \alpha \text{ and } \beta\) are given by:
\(\displaystyle \alpha = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}\)
\(\displaystyle \beta = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}\)
In your problem you have a = 2, b = 3, and c = -27. What happens if you plug in these values?
