# How do you factor 2x^2 - 11x + 5 = 0?

Jun 9, 2016

$\left(2 x - 1\right) \left(x - 5\right) = 0$, which gives the solution as $x = \frac{1}{2}$ or $x = 5$.

#### Explanation:

To factorize $a {x}^{2} + b x + c$, one should split the middle term $b$ in two parts, whose product is $a c$

Hence in $2 {x}^{2} - 11 x + 5$, we should split $- 11$ in two parts, so that their product is $2xx5=10#. It is apparent these numbers are $- 10$and $- 1$. Hence $2 {x}^{2} - 11 x + 5 = 0$can be written as $2 {x}^{2} - 10 x - x + 5 = 0$or $2 x \left(x - 5\right) - 1 \left(x - 5\right) = 0$or $\left(2 x - 1\right) \left(x - 5\right) = 0$, which gives the solution as $x = \frac{1}{2}$or $x = 5$. Jun 9, 2016 $\left(x - 5\right) \left(2 x - 1\right) = 0$#### Explanation: What we have to do is to find two factors of 10 =$2 \cdot 5$(co-eff. of term ${x}^{2}$multiplied by constant term 5) that add up to 11 (coeff. of term x). Clearly, these are 10 & 1. Now we split 11x as (10+1)x and proceed as under :- $2 {x}^{2} - 11 x + 5 = 0$$2 {x}^{2} - \left(10 + 1\right) x + 5 = 0$$2 {x}^{2} - 10 x - x + 5 = 0$$2 x \left(x - 5\right) - 1 \left(x - 5\right) = 0$$\left(x - 5\right) \left(2 x - 1\right) = 0$$x = 5 , x = \frac{1}{2}\$.

A note : I think that the problem is wrongly questioned. It should have been asked as to solve the eqn. It is OK that to solve it, we have to factorise the given quadratic polynomial