# How do you factor completely 50a^2b^5 – 35a^4b^3 + 5a^3b^4?

Jun 16, 2016

$50 {a}^{2} {b}^{5} - 35 {a}^{4} {b}^{3} + 5 {a}^{3} {b}^{4}$

$= 5 {a}^{2} {b}^{3} \left(10 {b}^{2} - 7 {a}^{2} + a b\right)$

$= 50 {a}^{2} {b}^{3} \left(b + \frac{1 - \sqrt{281}}{20} a\right) \left(b + \frac{1 + \sqrt{281}}{20} a\right)$

#### Explanation:

Note that all three terms are divisible by $5$, ${a}^{2}$ and ${b}^{3}$, so therefore by $5 {a}^{2} {b}^{3}$.

Separate that out as a factor:

$50 {a}^{2} {b}^{5} - 35 {a}^{4} {b}^{3} + 5 {a}^{3} {b}^{4} = 5 {a}^{2} {b}^{3} \left(10 {b}^{2} - 7 {a}^{2} + a b\right)$

Note that the remaining term is an homogeneous quadratic which we can factor by completing the square:

$10 {b}^{2} + a b - 7 {a}^{2}$

$= 10 \left({b}^{2} + 2 \cdot \frac{1}{20} a b + \frac{1}{20} ^ 2 {a}^{2} - \frac{1}{20} ^ 2 {a}^{2} - \frac{7}{10} {a}^{2}\right)$

$= 10 \left({\left(b + \frac{1}{20} a\right)}^{2} - \frac{1}{400} {a}^{2} - \frac{280}{400} {a}^{2}\right)$

$= 10 \left({\left(b + \frac{1}{20} a\right)}^{2} - \frac{281}{20} ^ 2 {a}^{2}\right)$

$= 10 \left({\left(b + \frac{1}{20} a\right)}^{2} - {\left(\frac{\sqrt{281}}{20} a\right)}^{2}\right)$

$= 10 \left(\left(b + \frac{1}{20} a\right) - \left(\frac{\sqrt{281}}{20} a\right)\right) \left(\left(b + \frac{1}{20} a\right) + \left(\frac{\sqrt{281}}{20} a\right)\right)$

$= 10 \left(b + \frac{1 - \sqrt{281}}{20} a\right) \left(b + \frac{1 + \sqrt{281}}{20} a\right)$

Putting it all together:

$50 {a}^{2} {b}^{5} - 35 {a}^{4} {b}^{3} + 5 {a}^{3} {b}^{4}$

$= 50 {a}^{2} {b}^{3} \left(b + \frac{1 - \sqrt{281}}{20} a\right) \left(b + \frac{1 + \sqrt{281}}{20} a\right)$