Factor polynomial $$ 28t^{2}-16t-12 $$

$$ 28t^{2}-16t-12 = 4( t-1 )( 7t+3 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 28t^{2}-16t-12 = 4 ( 7t^{2}-4t-3 ) $$

Step 2 :

Step 2: 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:

$$ a = 7 , b = -4 ~ \text{ and } ~ c = -3 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -3} $.

$$ a \cdot c = -21 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -21 $ and add to $ b = -4 $.

Step 5: All pairs of numbers with a product of $ -21 $ are:

PRODUCT = -21
-1     21	1     -21
-3     7	3     -7

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -4 }$

PRODUCT = -21 and SUM = -4
-1     21	1     -21
-3     7	3     -7

Step 7: Replace middle term $ -4 x $ with $ 3x-7x $:

$$ 7x^{2}-4x-3 = 7x^{2}+3x-7x-3 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 7x^{2}+3x-7x-3 = x\left(7x+3\right) -1\left(7x+3\right) = \left(x-1\right) \left(7x+3\right) $$

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