How do you factor the expression #35x^3-61x^2+8x#?

1 Answer
Mar 11, 2017

Answer:

The answer is #=x(7x-1)(5x-8)#

Explanation:

We start by factoring #x#

#35x^3-61x^2+8x=x(35x^2-61x+8)#

We need the roots of

#35x^2-61x+8#

We compare this to

#ax^2+bx+c#

We calculate the discriminant

#Delta=b^2-4ac=(-61)^2-4*35*8=2601#

As #Delta>0#, there are 2 real roots

#x=(-b+-sqrtDelta)/(2a)#

#sqrtDelta=sqrt2601=51#

#x_1=(61+51)/(70)=1.6=16/10=8/5#

#x_2=(61-51)/70=10/70=1/7#

Therefore,

#35x^2-61x+8=35(x-8/5)(x-1/7)#

#=(7x-1)(5x-8)#

So,

#35x^3-61x^2+8x=x(35x^2-61x+8)=x(7x-1)(5x-8)#