Factor polynomial $$ 28x^{2}+x-15 $$

$$ 28x^{2}+x-15 = ( 7x-5 )( 4x+3 ) $$

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:

$$ a = 28 , b = 1 ~ \text{ and } ~ c = -15 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 28 }$ by the constant term $\color{blue}{c = -15} $.

$$ a \cdot c = -420 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -420 $ and add to $ b = 1 $.

Step 4: All pairs of numbers with a product of $ -420 $ are:

PRODUCT = -420
-1     420	1     -420
-2     210	2     -210
-3     140	3     -140
-4     105	4     -105
-5     84	5     -84
-6     70	6     -70
-7     60	7     -60
-10     42	10     -42
-12     35	12     -35
-14     30	14     -30
-15     28	15     -28
-20     21	20     -21

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 1 }$

PRODUCT = -420 and SUM = 1
-1     420	1     -420
-2     210	2     -210
-3     140	3     -140
-4     105	4     -105
-5     84	5     -84
-6     70	6     -70
-7     60	7     -60
-10     42	10     -42
-12     35	12     -35
-14     30	14     -30
-15     28	15     -28
-20     21	20     -21

Step 6: Replace middle term $ 1 x $ with $ 21x-20x $:

$$ 28x^{2}+x-15 = 28x^{2}+21x-20x-15 $$

Step 7: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 28x^{2}+21x-20x-15 = 7x\left(4x+3\right) -5\left(4x+3\right) = \left(7x-5\right) \left(4x+3\right) $$

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