Factor polynomial $$ 5x^{2}-51x+10 $$

$$ 5x^{2}-51x+10 = ( x-10 )( 5x-1 ) $$

Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 5 , b = -51 ~ \text{ and } ~ c = 10 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 10} $.

$$ a \cdot c = 50 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 50 $ and add to $ b = -51 $.

Step 4: All pairs of numbers with a product of $ 50 $ are:

PRODUCT = 50
1     50	-1     -50
2     25	-2     -25
5     10	-5     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -51 }$

PRODUCT = 50 and SUM = -51
1     50	-1     -50
2     25	-2     -25
5     10	-5     -10

Step 6: Replace middle term $ -51 x $ with $ -x-50x $:

$$ 5x^{2}-51x+10 = 5x^{2}-x-50x+10 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -10 $ out of the last two terms.

$$ 5x^{2}-x-50x+10 = x\left(5x-1\right) -10\left(5x-1\right) = \left(x-10\right) \left(5x-1\right) $$

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