Online Course Discussion Forum

Math Challenge I-C Handout 2 Questions

 
 
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Re: Math Challenge I-C Handout 2 Questions
by John Lensmire - Wednesday, 27 September 2017, 11:41 PM
 

Taking it one step at a time, we first note that $-3x^3+18x^2-36x+24$ has a common factor of $3$. Hence we can divide by $-3$ on both sides to get $x^3 - 6x^2 + 12x - 8 = 0$. (We divide by $-3$ so we have a positive $x^3$.)

At this point it is a little trickier to see the pattern. We notice that $x^3 = (x)^3$ and $-8 = (-2)^3$, so we want to start examining our patterns involving cubes. From here we try $a = x$ and $b=-2$ in $(a+b)^3$. This gives $$x^3 + 3(x^2)(-2) + 3(x)(-2)^2 + (-2)^3 = x^3 - 6x^2 + 12x - 8$$ so we are left with $(x-2)^3 = 0$. This is where we get $x = 2$.

Do you mean the example questions covered in the video? Yes, you can ask those questions here. The forums are meant for any questions relevant to the courses.