Factor polynomial $$ 4x^{2}+54x+140 $$

$$ 4x^{2}+54x+140 = 2( 2x+7 )( x+10 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 4x^{2}+54x+140 = 2 ( 2x^{2}+27x+70 ) $$

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:

$$ a = 2 , b = 27 ~ \text{ and } ~ c = 70 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 70} $.

$$ a \cdot c = 140 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 140 $ and add to $ b = 27 $.

Step 5: All pairs of numbers with a product of $ 140 $ are:

PRODUCT = 140
1     140	-1     -140
2     70	-2     -70
4     35	-4     -35
5     28	-5     -28
7     20	-7     -20
10     14	-10     -14

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 27 }$

PRODUCT = 140 and SUM = 27
1     140	-1     -140
2     70	-2     -70
4     35	-4     -35
5     28	-5     -28
7     20	-7     -20
10     14	-10     -14

Step 7: Replace middle term $ 27 x $ with $ 20x+7x $:

$$ 2x^{2}+27x+70 = 2x^{2}+20x+7x+70 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ 7 $ out of the last two terms.

$$ 2x^{2}+20x+7x+70 = 2x\left(x+10\right) + 7\left(x+10\right) = \left(2x+7\right) \left(x+10\right) $$

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