Factor polynomial $$ 5x^{2}-21x+4 $$

$$ 5x^{2}-21x+4 = ( x-4 )( 5x-1 ) $$

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:

$$ a = 5 , b = -21 ~ \text{ and } ~ c = 4 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 4} $.

$$ a \cdot c = 20 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 20 $ and add to $ b = -21 $.

Step 4: All pairs of numbers with a product of $ 20 $ are:

PRODUCT = 20
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -21 }$

PRODUCT = 20 and SUM = -21
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 6: Replace middle term $ -21 x $ with $ -x-20x $:

$$ 5x^{2}-21x+4 = 5x^{2}-x-20x+4 $$

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

$$ 5x^{2}-x-20x+4 = x\left(5x-1\right) -4\left(5x-1\right) = \left(x-4\right) \left(5x-1\right) $$

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