# How do you factor the trinomial 5-6x+x^2?

Jun 15, 2018

$\left(x - 5\right) \left(x - 1\right)$

#### Explanation:

As per the question, we have

$5 - 6 x + {x}^{2}$

By arranging the terms, we get

${x}^{2} - 6 x + 5$

Now, here we have an equation of the form $a {x}^{2} + b x + c$ and to further factorise this equation we have to split the middle term $b x$ in such a way that it is able to get summed up by the factors of $c$.

As $c = 5$

$\therefore$ Factors of $c$ : $1 \mathmr{and} 5$

Now, $b x = - 6 x$

So, we can write $- 6 x$ as $- 1 x - 5 x$.

Note : $\left(- 1\right) \times \left(- 5\right) = 5 = c$

So, we have the equation as,

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

Rearranging the terms,

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

Taking the common terms out,

$x \left(x - 1\right) - 5 \left(x - 1\right)$

Simplifying it further,

$\left(x - 5\right) \left(x - 1\right)$