How do you factor 24x^3- 40x^2+ 21x - 35?

Apr 14, 2015

Notice that the coefficients have the following property of proportionality:
$\frac{24}{40} = \frac{21}{35} = \frac{3}{5}$

This should prompt us to transform the expression by factoring out something from the first two members and from the last two members counting on the remainder of this factorization to be the same:
$24 {x}^{3} - 40 {x}^{2} + 21 x - 35 = 8 {x}^{2} \left(3 x - 5\right) + 7 \left(3 x - 5\right)$

Indeed, the remainder is the same in both cases - $3 x - 5$.

Now we can factor out $3 x - 5$ getting
$\left(3 x - 5\right) \left(8 {x}^{2} + 7\right)$.