How do you factor #x^3 - 12x^2 + 35x#?

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Answer 1 · 2018-04-22T17:04:00

The answer is #x(x-5)(x-7)#.

First we take out the common factor x:
#x^3-12x^2+35x=x(x^2-12x+35)#

Then we use the quadratic formula:
#x=\frac {-b\pm (\sqrt {b^{2}-4ac\ })}{2a}#

#x=\frac {-(-12)\pm (\sqrt {(-12)^{2}-4*1*35\ })}{2*1}#

#x=\frac {12\pm (\sqrt {144-140\ })}{2}#

#x=\frac {12\pm \sqrt {4)}{2}#

#x=\frac {12\pm \2}{2}#

#x_1=5#

#x_2=7#

#ax^2+bx+c=a*(x-x_1)(x-x_2)#

Then we just write this instead of the quadratic equation we've just solved:

#x^3-12x^2+35x=x(x-5)(x-7)#

Answer 2 · 2018-04-22T17:08:21

#x(x -7)(x -5)#

Start by taking out the common factor of #x#

#x^3 -12x^2 +35x#

#=x(x^2-12x+35)#

Now factor the quadratic trinomial.

Find factors of #35# which add to make #12#

#7 xx 5# will do nicely: #7xx5=35 and 7+5=12#

#x(x" "7)(x" "5)#

For the signs, they both have to be the same to get #+35#, but they need to add to #-12#

#x(x -7)(x -5)#