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

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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.

How do you factor 3x^3-7x^2+x3x^3-7x^2+x?