How would I factor these equations?
1. 6x² + 3x = 0
2. -2x² + 5x + 3 = 0
I know how to factor equations without the number before x². But when the number is there I don't know how to factor it.
for the first, 3x is a common term. just factor it out
for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler.
so $\displaystyle -2x^2 + 5x + 3 = 0$
$\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$
$\displaystyle \Rightarrow (2x ~~a)(x ~~b) = 0$
now in the place of $\displaystyle a$ and $\displaystyle b$, you want numbers that multiply to give you -3, but $\displaystyle a$ and $\displaystyle 2b$ will add to give -5
when you have a common term, you can factor it out
example:
factorize $\displaystyle 3xy + 5xy^2$
we want to factor out the lowest powers of all things common. i have $\displaystyle x$ common to both terms, and i have $\displaystyle y$ common to both terms. there is a $\displaystyle y^2$ in the second term, but i can't factor that out, because i only have $\displaystyle y$ in the first term. so $\displaystyle x$ and $\displaystyle y$ are common to both terms, both terms have them, so i will factor out $\displaystyle xy$, to get:
$\displaystyle xy(3 + 5y)$
so i took out the $\displaystyle xy$ from the 3, and i have none left in the brackets. as for the $\displaystyle 5xy^2$, i took out the $\displaystyle x$ and one of the $\displaystyle y$'s, so i am left with one $\displaystyle y$ on the inside, hence the $\displaystyle 5y$
now you have,
$\displaystyle 6x^2 + 3x = 0$
i told you that $\displaystyle 3x$ is common to both, so factor it out
No, the numbers $\displaystyle a$ and $\displaystyle b$ multiply to give you -6 (not -3) because 2*(-3) = -6, and their sum will be -5
See here
$\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$
$\displaystyle \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)$
$\displaystyle \Rightarrow 2x(x-3) +1(x - 3) = 0$
$\displaystyle \Rightarrow (2x + 1)(x-3) = 0$
$\displaystyle \Rightarrow x=\frac{-1}{2}\;and\;x=3$