Factor polynomial $$ -x^{2}+20x-91 $$

$$ -x^{2}+20x-91 = -( x-7 )( x-13 ) $$

Step 1 :

After factoring out $ -1 $ we have:

$$ -x^{2}+20x-91 = - ~ ( x^{2}-20x+91 ) $$

Step 2 :

Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -20 } ~ \text{ and } ~ \color{red}{ c = 91 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -20 } $ and multiply to $ \color{red}{ 91 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 91 }$.

PRODUCT = 91
1     91	-1     -91
7     13	-7     -13

Step 4: Find out which pair sums up to $\color{blue}{ b = -20 }$

PRODUCT = 91 and SUM = -20
1     91	-1     -91
7     13	-7     -13

Step 5: Put -7 and -13 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-20x+91 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-20x+91 & = (x -7)(x -13) \end{aligned} $$

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