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

Dec 13, 2017

$= \left(2 x - 1\right) \left(x - 2\right)$

#### Explanation:

Ok, we are just going to factor this one out!

$= 2 {x}^{2} - 5 x + 2$

Break it into groups

$= \left(2 {x}^{2} - x\right) + \left(- 4 x + 2\right)$

At this point we can factor out x and -2 respectively

$= x \left(2 x - 1\right) - 2 \left(2 x - 1\right)$

Then factor out the common term

$= \left(2 x - 1\right) \left(x - 2\right)$

Hope this helped!
~Chandler Dowd

Dec 13, 2017

$\left(2 x - 1\right) \left(x - 2\right)$

#### Explanation:

$\text{the factors of + 4 which sum to - 5 are - 4 and - 1}$

$\text{splitting the middle term}$

$2 {x}^{2} - 4 x - x + 2 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{2 x} \left(x - 2\right) \textcolor{red}{- 1} \left(x - 2\right)$

$\text{take out "color(blue)"common factor } \left(x - 2\right)$

$= \left(x - 2\right) \left(\textcolor{red}{2 x - 1}\right)$

$\Rightarrow 2 {x}^{2} - 5 x + 2 = \left(x - 2\right) \left(2 x - 1\right)$

Dec 13, 2017

$\left(2 x - 1\right) \left(x - 2\right)$

#### Explanation:

$\textcolor{m a \ge n t a}{2} {x}^{2} - \textcolor{b l u e}{5} x \textcolor{red}{+} \textcolor{\lim e}{2}$

All the information is in the trinomial.

Find the factors of $\textcolor{m a \ge n t a}{2} \mathmr{and} \textcolor{\lim e}{2}$ which $\textcolor{red}{\text{ADD}}$ to make $\textcolor{b l u e}{5}$

The $\textcolor{red}{+}$ sign gives two clues:

• ADD the factors to get the middle number
• the sign in the brackets will be the SAME.

Note that $2 \times 2 = 4 \text{ } \mathmr{and} 1 \times 1 = 1$

Now $4 + 1 = 5$, so $2 \mathmr{and} 2$ are the factors.
As $2$ is a prime number there are only two ways of combining the factors.

SO know we know:

$\left(2 x \text{ "1)(x" } 2\right)$

We also know the signs will be the SAME.

The sign of the middle term $\textcolor{b l u e}{- 5}$ shows they will be negative.

$\left(2 x - 1\right) \left(x - 2\right)$

Check by multiplying:

$= 2 {x}^{2} - 4 x - 1 x + 2$

$= 2 {x}^{2} - 5 x + 2$