Factor polynomial $$ b^{2}-19b+90 $$

$$ b^{2}-19b+90 = ( b-9 )( b-10 ) $$

Step 1: 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:

$$ \color{blue}{ b = -19 } ~ \text{ and } ~ \color{red}{ c = 90 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -19 } $ and multiply to $ \color{red}{ 90 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 90 }$.

PRODUCT = 90
1     90	-1     -90
2     45	-2     -45
3     30	-3     -30
5     18	-5     -18
6     15	-6     -15
9     10	-9     -10

Step 3: Find out which pair sums up to $\color{blue}{ b = -19 }$

PRODUCT = 90 and SUM = -19
1     90	-1     -90
2     45	-2     -45
3     30	-3     -30
5     18	-5     -18
6     15	-6     -15
9     10	-9     -10

Step 4: Put -9 and -10 into placeholders to get factored form.

$$ \begin{aligned} b^{2}-19b+90 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ b^{2}-19b+90 & = (x -9)(x -10) \end{aligned} $$

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