How do you factor $3 b^{2} - 13 b + 4$ $3b^2-13b+4$ ?

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How do you factor $3 b^{2} - 13 b + 4$ $3b^2-13b+4$ ?

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Answer 1 · 2015-05-24T23:15:25

If $3 b^{2} - 13 b + 4 = 0$ $3b^2-13b+4 = 0$ has a rational root $\frac{p}{q}$ $p/q$ in lowest terms, then $p$ $p$ is a divisor of the constant term $4$ $4$ and $q$ $q$ is a divisor of the coefficient $3$ $3$ of the highest order term $3 b^{2}$ $3b^2$ .

That means any rational root must be one of:

$\pm 1$ $+-1$ , $\pm \frac{1}{3}$ $+-1/3$ , $\pm 2$ $+-2$ , $\pm \frac{2}{3}$ $+-2/3$ , $\pm 4$ $+-4$ or $\pm \frac{4}{3}$ $+-4/3$

Notice that $4$ $4$ is a root (try substituting $b = 4$ $b=4$ and find the result is $0$ $0$ ), so $(b - 4)$ $(b-4)$ is a factor of $3 b^{2} - 13 b + 4$ $3b^2-13b+4$ .

That leaves $(3 b - 1)$ $(3b-1)$ as the other factor - check $\frac{1}{3}$ $1/3$ is a root too - yes.

So $3 b^{2} - 13 b + 4 = (b - 4) (3 b - 1)$ $3b^2-13b+4 = (b-4)(3b-1)$

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How do you factor $3 b^{2} - 13 b + 4$ $3b^2-13b+4$ ?

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