Factor polynomial $$ 7r^{2}-9r-10 $$

$$ 7r^{2}-9r-10 = ( r-2 )( 7r+5 ) $$

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:

$$ a = 7 , b = -9 ~ \text{ and } ~ c = -10 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -70 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -70 $ and add to $ b = -9 $.

Step 4: All pairs of numbers with a product of $ -70 $ are:

PRODUCT = -70
-1     70	1     -70
-2     35	2     -35
-5     14	5     -14
-7     10	7     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = -70 and SUM = -9
-1     70	1     -70
-2     35	2     -35
-5     14	5     -14
-7     10	7     -10

Step 6: Replace middle term $ -9 x $ with $ 5x-14x $:

$$ 7x^{2}-9x-10 = 7x^{2}+5x-14x-10 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 7x^{2}+5x-14x-10 = x\left(7x+5\right) -2\left(7x+5\right) = \left(x-2\right) \left(7x+5\right) $$

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