Factor polynomial $$ 5x^{2}-27x+28 $$

$$ 5x^{2}-27x+28 = ( x-4 )( 5x-7 ) $$

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:

$$ a = 5 , b = -27 ~ \text{ and } ~ c = 28 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 28} $.

$$ a \cdot c = 140 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 140 $ and add to $ b = -27 $.

Step 4: 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 5: 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 6: Replace middle term $ -27 x $ with $ -7x-20x $:

$$ 5x^{2}-27x+28 = 5x^{2}-7x-20x+28 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 5x^{2}-7x-20x+28 = x\left(5x-7\right) -4\left(5x-7\right) = \left(x-4\right) \left(5x-7\right) $$

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