Question

How do you factor `2x^2-0.6x-0.08=0`?

Answer 1

`2x ^2 - 0.6 x - 0.08 = 2( x - .4) (x + .1) = 1/25 (10 x + 1)(5x - 2)`

Explanation:

Hi Portsmouth, I'm here in Lee, NH! Two year old question; you've probably graduated and moved away by now.

First a minor point, we usually factor expressions and solve equations. The factoring requires keeping track of the constant multiple, which we'll do here.

There are a couple of ways to do it. Since we already have decimals, let's just take out the two:

`2x ^2 - 0.6 x - 0.08 = 2( x ^2 - .3x - .04 )`

Now it's a regular factoring, two numbers that add up to -.3 and multiply to -.04, obviously -.4 and .1, so

`2x ^2 - 0.6 x - 0.08 = 2( x - .4) (x + .1)`

That's a perfectly good factorization, and we can read off the zeros.

Let's try another way where we first clear the decimals by dividing and multiplying by 100.

`2x ^2 - 0.6 x - 0.08 = 1/100 ( 200 x ^2 - 60 x - 8 )`

There a common factor of 4 we can pull out,

`= 1/25 (50 x^2 - 15 x - 2)`

Now we have a factoring problem that's normal. The factoring puzzle is to find a pair of factors of `pq=50` and `rs=-2` whose sum of products adds `pr+qs=-15.` It doesn't take much thought to come up with `-2(10)+1(5)=-15` so a factorization of

`= 1/25 (10 x + 1)(5x - 2)`