Factor polynomial $$ 3y^{2}+y-4 $$

$$ 3y^{2}+y-4 = ( y-1 )( 3y+4 ) $$

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:

$$ a = 3 , b = 1 ~ \text{ and } ~ c = -4 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -4} $.

$$ a \cdot c = -12 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -12 $ and add to $ b = 1 $.

Step 4: All pairs of numbers with a product of $ -12 $ are:

PRODUCT = -12
-1     12	1     -12
-2     6	2     -6
-3     4	3     -4

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

PRODUCT = -12 and SUM = 1
-1     12	1     -12
-2     6	2     -6
-3     4	3     -4

Step 6: Replace middle term $ 1 x $ with $ 4x-3x $:

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

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

$$ 3x^{2}+4x-3x-4 = x\left(3x+4\right) -1\left(3x+4\right) = \left(x-1\right) \left(3x+4\right) $$

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