Factor polynomial $$ 40y^{2}+y-6 $$

$$ 40y^{2}+y-6 = ( 8y-3 )( 5y+2 ) $$

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:

$$ a = 40 , b = 1 ~ \text{ and } ~ c = -6 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 40 }$ by the constant term $\color{blue}{c = -6} $.

$$ a \cdot c = -240 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -240 $ and add to $ b = 1 $.

Step 4: All pairs of numbers with a product of $ -240 $ are:

PRODUCT = -240
-1     240	1     -240
-2     120	2     -120
-3     80	3     -80
-4     60	4     -60
-5     48	5     -48
-6     40	6     -40
-8     30	8     -30
-10     24	10     -24
-12     20	12     -20
-15     16	15     -16

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

PRODUCT = -240 and SUM = 1
-1     240	1     -240
-2     120	2     -120
-3     80	3     -80
-4     60	4     -60
-5     48	5     -48
-6     40	6     -40
-8     30	8     -30
-10     24	10     -24
-12     20	12     -20
-15     16	15     -16

Step 6: Replace middle term $ 1 x $ with $ 16x-15x $:

$$ 40x^{2}+x-6 = 40x^{2}+16x-15x-6 $$

Step 7: Apply factoring by grouping. Factor $ 8x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 40x^{2}+16x-15x-6 = 8x\left(5x+2\right) -3\left(5x+2\right) = \left(8x-3\right) \left(5x+2\right) $$

Polynomial Factoring Calculator