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Problem Solving - don't understand missing step

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by Ninedot » Sat Mar 02, 2013 4:38 pm
Hi Everyone,
Just curious if anyone could explain to me how the second step leads to the third step:
Step 1)[3x^2(x-2)-x+2]/[x-2] where x does not equal 2
Step 2)= [3x^2(x-2)+(-1)(x-2)]/[x-2]
Step 3)= [(x-2)(3x^2 +(-1))]/[x-2]

Specifically, I don't understand how a)3x^2 turned into (3x^2 + (-1)) - wouldn't BEDMAS suggest that 3x^2 cannot be seperated from (x-2) because addition comes after multiplication? And b) where did the other (x-2) go? Shouldn't it be (x-2)^2?

Thanks in advance for any help you guys can provide!
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by Anurag@Gurome » Sat Mar 02, 2013 7:10 pm
Ninedot wrote:Step 1)[3x^2(x-2)-x+2]/[x-2] where x does not equal 2
Step 2)= [3x^2(x-2)+(-1)(x-2)]/[x-2]
Step 3)= [(x-2)(3x^2 +(-1))]/[x-2]
Here 3x^2(x-2) means (3x²)*(x - 2), i.e. this is the product of 3x² and (x - 2)
And, (-x + 2) = -(x - 2) = (-1)*(x - 2)

Hence, the numerator = [(3x²)*(x - 2) - (-1)*(x - 2)]
Now, we are taking (x - 2) out of bracket as it is a common factor of both the term,
Hence, the numerator = (x - 2)*[(3x²) - (-1)] = (x - 2)*[(3x²) + 1]

Hope that helps.
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by Ninedot » Sun Mar 03, 2013 6:30 pm
Anurag@Gurome wrote:
Ninedot wrote:Step 1)[3x^2(x-2)-x+2]/[x-2] where x does not equal 2
Step 2)= [3x^2(x-2)+(-1)(x-2)]/[x-2]
Step 3)= [(x-2)(3x^2 +(-1))]/[x-2]
Here 3x^2(x-2) means (3x²)*(x - 2), i.e. this is the product of 3x² and (x - 2)
And, (-x + 2) = -(x - 2) = (-1)*(x - 2)

Hence, the numerator = [(3x²)*(x - 2) - (-1)*(x - 2)]
Now, we are taking (x - 2) out of bracket as it is a common factor of both the term,
Hence, the numerator = (x - 2)*[(3x²) - (-1)] = (x - 2)*[(3x²) + 1]

Hope that helps.
Thank you Anurag@Gurome for the response. I now see how to get from step 2 to step 3. I didn't realize that (x-2) could be factored out of the whole equation.
Thanks again!
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