Factoring Trinomials w/ Decimals: w^2+.14w-.048
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Well, there's nothing inherently special about factoring a quadratic where some (or all) of its coefficients are decimals. You say you were able to solve a similar problem with 140 and -48. What process did you follow to solve that problem? What happens if you follow that same process but use 0.14 and -0.048? Please share with us all of your work, even if you know it's wrong, so we can see exactly where you're getting bogged down. Thank you.
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I couldn't figure it out using 140 and 48
My normal method, which I see now is just not sufficient is to break down the product of the trinominal into it's factors and identify which combination can be combined to get the other number. So with this one I found the factors of -48, 1(-48), 2(-24),3(-16),4(-12),6(-8) but how could any of these numbers be added to equal 140? There must be something I'm missing.
I couldn't figure it out using 140 and 48
My normal method, which I see now is just not sufficient is to break down the product of the trinominal into it's factors and identify which combination can be combined to get the other number. So with this one I found the factors of -48, 1(-48), 2(-24),3(-16),4(-12),6(-8) but how could any of these numbers be added to equal 140? There must be something I'm missing.
D
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Suppose you re-write the given expression as:
w^2+.14w-.048 = 1/1000 * (1000*w^2 + 140 *w - 48) = 1/250 * (250 * w^2 + 35 * w - 12)
... can you proceed now?
Do you know the existence of quadratic equation?
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Unfortunately, since you haven't shown your work, we can't know what this means, nor where it led. So we'll have to start over.
To clear the decimals, one would need to multiply through by 1000. Of course, this isn't an equation, so there is no "equals" that we can multiply "through". Instead, we'll need to multiply and divide by 1,000, so we're multiplying by 1. But it's a useful form of 1.
. . . . .\(\displaystyle w^2\, +\, 0.14w\, -\, 0.048\, =\, \dfrac{1000}{1000}\, (w^2\, +\, 0.14w\, -\, 0.048)\, =\, \dfrac{1}{1000}\, (1000w^2\, +\, 140w\, -\, 48)\)
Then divide out a 4 to get:
. . . . .\(\displaystyle \dfrac{1}{250}\, (250x^2\, +\, 35w\, -\, 12)\)
Then factor in the usual way. (here) You'll need factors of -3000 which are 35 units apart.
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Steven G
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I get the feeling that you are saying that w2+.14w-.048 = w2+140w - 48 or possibly (1/1000)(w2+140w - 48).
Without you showing any work it is hard to help.
You should get \(\displaystyle w^2+.14w-.048 =\dfrac{1}{1000}(1000w^2+ 140w - 48) = \dfrac{1}{250}(250w^2+ 35w - 7)\)
Now try to factor (250w2+ 35w - 7). Note that does not mean to forget the 1/250 !!
The constant term becomes -12, not -7. Note the two posts above yours with the -12.
HallsofIvy
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Not the simplest thing to do but it always works when you can't see any "relation" between the coefficients:
To factor \(\displaystyle ax^2+ bx+ c\), solve the quadratic equation \(\displaystyle ax^2+ bx+ c= 0\) using the quadratic formula:
\(\displaystyle x_1= \frac{-b+ \sqrt{b^2- 4ac}}{2a}\) and \(\displaystyle x_2= \frac{-b- \sqrt{b^2- 4ac}}{2a}\)
Then \(\displaystyle ax^2+ bx+ c= a(x- x_1)(x- x_2)\)
To factor \(\displaystyle ax^2+ bx+ c\), solve the quadratic equation \(\displaystyle ax^2+ bx+ c= 0\) using the quadratic formula:
\(\displaystyle x_1= \frac{-b+ \sqrt{b^2- 4ac}}{2a}\) and \(\displaystyle x_2= \frac{-b- \sqrt{b^2- 4ac}}{2a}\)
Then \(\displaystyle ax^2+ bx+ c= a(x- x_1)(x- x_2)\)
Steven G
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Thank you.