# How do you completely factor 25x^3+150x^2+131x+30 given 5x+3 is one of the factors?

Oct 4, 2016

$\left(5 x + 3\right) \left(5 x + 2\right) \left(x + 5\right)$.

#### Explanation:

Let, $p \left(x\right) = 25 {x}^{3} + 150 {x}^{2} + 131 x + 30.$

We are given that, $\left(5 x + 3\right)$ is a factor of $p \left(x\right)$.

To factorise $p \left(x\right)$ completely, Synthetic Division is usually

performed, but, we can proceed as under :-

$p \left(x\right) = 25 {x}^{3} + 150 {x}^{2} + 131 x + 30$

$= \underline{25 {x}^{3} + 15 {x}^{2}} + \underline{135 {x}^{2} + 81 x} + \underline{50 x + 30}$

$= 5 {x}^{2} \left(5 x + 3\right) + 27 x \left(5 x + 3\right) + 10 \left(5 x + 3\right)$

$= \left(5 x + 3\right) \left(5 {x}^{2} + 27 x + 10\right)$

$= \left(5 x + 3\right) \left\{\underline{5 {x}^{2} + 25 x} + \underline{2 x + 10}\right\}$
....................................................................$\left[25 \times 2 = 50 , 25 + 2 = 27\right]$

$= \left(5 x + 3\right) \left\{5 x \left(x + 5\right) + 2 \left(x + 5\right)\right\}$

$\therefore p \left(x\right) = \left(5 x + 3\right) \left(5 x + 2\right) \left(x + 5\right)$.