Factor polynomial $$ -x^{3}+x^{2}+90x $$

$$ -x^{3}+x^{2}+90x = -x( x+9 )( x-10 ) $$

Step 1 :

After factoring out $ -x $ we have:

$$ -x^{3}+x^{2}+90x = -x ( x^{2}-x-90 ) $$

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:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = -90 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ -90 } $.

Step 3: 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 4: Find out which pair sums up to $\color{blue}{ b = -1 }$

PRODUCT = -90 and SUM = -1
-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 5: Put 9 and -10 into placeholders to get factored form.

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

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