Factor polynomial $$ 10x^{2}+49x-5 $$

$$ 10x^{2}+49x-5 = ( 10x-1 )( x+5 ) $$

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:

$$ a = 10 , b = 49 ~ \text{ and } ~ c = -5 $$

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

$$ a \cdot c = -50 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -50 $ and add to $ b = 49 $.

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 $\color{blue}{ b = 49 }$

PRODUCT = -50 and SUM = 49
-1     50	1     -50
-2     25	2     -25
-5     10	5     -10

Step 6: Replace middle term $ 49 x $ with $ 50x-x $:

$$ 10x^{2}+49x-5 = 10x^{2}+50x-x-5 $$

Step 7: Apply factoring by grouping. Factor $ 10x $ out of the first two terms and $ -1 $ out of the last two terms.

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

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