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

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Answer 1 · 2017-03-11T14:17:21

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

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,

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

How do you factor the expression 35x^3-61x^2+8x35x^3-61x^2+8x?