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

Question

Julia Adams · Accepted Answer

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

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) #

Isaiah Anthony · Answer

undefined To factor the quadratic equation $2x^2 - 0.6x - 0.08 = 0$, you can use the quadratic formula or factor by grouping. Alternatively, you can simplify the equation by multiplying all terms by 100 to eliminate the decimals. Then proceed with factoring as usual.