quick simple math question help plz!!

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Alex

Diamond Member
Oct 26, 1999
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i have the equation x^3 - 8x^2 + 17x -4 = 0 ("x cubed - 8x squared + 17x.....")

my textbook says we want to reduce this to a quadratic so we can use the formula...
so they say an integer solution has to be a factor of -4, and after substituting +-1, +-2 and +-4, +4 seems to work
ok, so x - 4 must be a factor of the original cubic equation.

so the next step is divide throughout by x-4 to give:

x^2 - 4x + 1

and then we can use (x-4)(x^2 - 4x + 1) = 0 and solve it like that...

the problem is: i dont get how to divide that cubic formula by x-4 and the book just skips all the steps in between... can someone walk me thru this real quick please? :)
 
L

lancestorm

Platinum Member
Oct 7, 2003
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This'll be hard to see but i'll do my best...let's split it up

let's take the first part:

(x - 4) into x^3 .... or just ask what can be multiplied against (x - 4) to get x^3 (and some added junk to deal with later in the equation)?

The answer of course is: x^2. Let's multiply it out. x^2 * (x - 4) = x^3 - 4x^2.

Take the original nominator (x^3 - 8x^2 + 17x - 4) and subtract out the (x^3 - 4x^2). Well to subtract out the (x^3 - 4x^2) we have to multiply the whole thing by -1 (so that the x^3's cancel out each other). Do this and we are left with a nominator of (0 - 4x^2 + 17x - 4), or simply (-4x^2 + 17x - 4).

Now second part:

(x-4) into (-4x^2) ....follow steps above and continue....you will get -4x for this part, and then +1 for the last part....thus we finally get x^2 -4x + 1.
 
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