How do I "solve by factoring" something with an = sign?

Like this: $\displaystyle 3x^2+5x=2$

?

Printable View

- Aug 31st 2007, 08:28 PMdeathtolife04How do I "solve by factoring" something with an = sign?
How do I "solve by factoring" something with an = sign?

Like this: $\displaystyle 3x^2+5x=2$

? - Aug 31st 2007, 08:37 PMKrizalid
$\displaystyle 3x^2+(6x-x)-2=0\iff(3x^2+6x)-(x+2)=0$, so $\displaystyle (x+2)(3x-1)=0$

- Aug 31st 2007, 10:51 PMtopsquark
$\displaystyle 3x^2 + 5x = 2$

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

This is something called the "ac method."

Multiply the coefficient of the $\displaystyle x^2$ term by the constant term:

$\displaystyle 3 \cdot (-2) = -6$

Now write out all the possible pairs of factors of -6:

1, -6

2, -3

3, -2

6, -1

Now add the pairs of factors:

1 + (-6) = -5

2 + (-3) = -1

3 + (-2) = 1

6 + (-1) = 5

*If*your quadratic equation factors over the integers (which is the long term for what they mean by "factoring") then the coefficient of the x term is somewhere in this list. In this case the coefficient of the x term is 5, so we have:

6 + (-1) = 5.

So here's the equation again:

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

We want to rewrite the linear (x) term as $\displaystyle 5x = 6x + (-1)x$:

$\displaystyle 3x^2 + (6x - x) - 2 = 0$

Now regroup:

$\displaystyle (3x^2 + 6x) + (-x - 2) = 0$

Factor each of the terms in parenthesis:

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

Now each term has a common factor of $\displaystyle x + 2$, so factor that:

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

Voila! It is factored. (You can, and should, multiply this back out to check that it matches the original expression.)

From here we will use the property that if the product ab = 0, then either a = 0 or b = 0.

So we know that

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

Thus either

$\displaystyle 3x - 1 = 0 \implies x = \frac{1}{3}$

or

$\displaystyle x + 2 = 0 \implies x = -2$

-Dan