Factor polynomial $$ 5x^{2}-50x+105 $$

$$ 5x^{2}-50x+105 = 5( x-3 )( x-7 ) $$

Step 1 :

After factoring out $ 5 $ we have:

$$ 5x^{2}-50x+105 = 5 ( x^{2}-10x+21 ) $$

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:

$$ \color{blue}{ b = -10 } ~ \text{ and } ~ \color{red}{ c = 21 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -10 } $ and multiply to $ \color{red}{ 21 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 21 }$.

PRODUCT = 21
1     21	-1     -21
3     7	-3     -7

Step 4: Find out which pair sums up to $\color{blue}{ b = -10 }$

PRODUCT = 21 and SUM = -10
1     21	-1     -21
3     7	-3     -7

Step 5: Put -3 and -7 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-10x+21 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-10x+21 & = (x -3)(x -7) \end{aligned} $$

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