How do you factor #3x^3-7x^2+x#?

Question

2029 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-27T03:48:53

#x(3x^2-7x+1)#

Expression #= 3x^3-7x^2+x#

Clearly #x# is a factor

Expression #= x(3x^2-7x+1)#

Now consider the quadratic factor #(3x^2-7x+1)# of the form #ax^2+bx+c#

#b^2-4ac = 7^2 -4xx3xx1 = 49-12 = 37#

Since 37 is prime, the quadratic will have no rational factors.

Hence, Expression #= x(3x^2-7x+1)# cannot be factored further.