Factor polynomial $$ 40d^{4}+92d^{3}+48d^{2} $$

$$ 40d^{4}+92d^{3}+48d^{2} = 4d^{2}( 5d+4 )( 2d+3 ) $$

Step 1 :

After factoring out $ 4d^{2} $ we have:

$$ 40d^{4}+92d^{3}+48d^{2} = 4d^{2} ( 10d^{2}+23d+12 ) $$

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:

$$ a = 10 , b = 23 ~ \text{ and } ~ c = 12 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 10 }$ by the constant term $\color{blue}{c = 12} $.

$$ a \cdot c = 120 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 120 $ and add to $ b = 23 $.

Step 5: All pairs of numbers with a product of $ 120 $ are:

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

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 23 }$

PRODUCT = 120 and SUM = 23
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 7: Replace middle term $ 23 x $ with $ 15x+8x $:

$$ 10x^{2}+23x+12 = 10x^{2}+15x+8x+12 $$

Step 8: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ 4 $ out of the last two terms.

$$ 10x^{2}+15x+8x+12 = 5x\left(2x+3\right) + 4\left(2x+3\right) = \left(5x+4\right) \left(2x+3\right) $$

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