Factor polynomial $$ 35r^{2}+23r-4 $$

$$ 35r^{2}+23r-4 = ( 7r-1 )( 5r+4 ) $$

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:

$$ a = 35 , b = 23 ~ \text{ and } ~ c = -4 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 35 }$ by the constant term $\color{blue}{c = -4} $.

$$ a \cdot c = -140 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -140 $ and add to $ b = 23 $.

Step 4: All pairs of numbers with a product of $ -140 $ are:

PRODUCT = -140
-1     140	1     -140
-2     70	2     -70
-4     35	4     -35
-5     28	5     -28
-7     20	7     -20
-10     14	10     -14

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 23 }$

PRODUCT = -140 and SUM = 23
-1     140	1     -140
-2     70	2     -70
-4     35	4     -35
-5     28	5     -28
-7     20	7     -20
-10     14	10     -14

Step 6: Replace middle term $ 23 x $ with $ 28x-5x $:

$$ 35x^{2}+23x-4 = 35x^{2}+28x-5x-4 $$

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

$$ 35x^{2}+28x-5x-4 = 7x\left(5x+4\right) -1\left(5x+4\right) = \left(7x-1\right) \left(5x+4\right) $$

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