1. ## Factoring equation...

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.

2. Originally Posted by lp_144
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

3. Originally Posted by Jhevon
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
So it would be $\displaystyle (2x+1)(x-3)$?

4. Originally Posted by lp_144
So it would be $\displaystyle (2x+1)(x-3)$?
yes, very good

now finish the problem. what about the first problem?

5. Originally Posted by Jhevon
yes, very good

now finish the problem. what about the first problem?

6. Originally Posted by lp_144
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

7. Originally Posted by Jhevon
now you have,

$\displaystyle 6x^2 + 3x = 0$

i told you that $\displaystyle 3x$ is common to both, so factor it out
When I factored $\displaystyle 3x$ out, I got $\displaystyle 3x(3x)$.
What do I do from here?

8. Originally Posted by lp_144
When I factored $\displaystyle 3x$ out, I got $\displaystyle 3x(3x)$.
What do I do from here?
$\displaystyle 6x^2 + 3x = 0$

$\displaystyle \Rightarrow 3x(2x + 1) = 0$

step 1: look carefully at what i did and make sure you understand it.

step 2: finish the problem. how would we solve for $\displaystyle x$?

9. Originally Posted by Jhevon
$\displaystyle 6x^2 + 3x = 0$

$\displaystyle \Rightarrow 3x(2x + 1) = 0$

step 1: look carefully at what i did and make sure you understand it.

step 2: finish the problem. how would we solve for $\displaystyle x$?
I think I got it.

$\displaystyle 3x(2x+1)=0$

$\displaystyle 3x=0$
$\displaystyle x=0$

$\displaystyle 2x+1=0$
$\displaystyle 2x=-1$
$\displaystyle x= -\frac {1}{2}$

10. Originally Posted by Jhevon
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
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

$\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$

$\displaystyle \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)$

$\displaystyle \Rightarrow 2x(x-3) +1(x - 3) = 0$

$\displaystyle \Rightarrow (2x + 1)(x-3) = 0$

$\displaystyle \Rightarrow x=\frac{-1}{2}\;and\;x=3$

11. Originally Posted by Shyam
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

$\displaystyle \Rightarrow 2x^2 - 5x - 3 = 0$

$\displaystyle \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)$

$\displaystyle \Rightarrow 2x(x-3) +1(x - 3) = 0$

$\displaystyle \Rightarrow (2x + 1)(x-3) = 0$

$\displaystyle \Rightarrow x=\frac{-1}{2}\;and\;x=3$
note that your a and b multiply to give -6 as i said. i was not doing the AC-method here.