Factor polynomial $$ 7n^{2}-24n+9 $$

$$ 7n^{2}-24n+9 = ( n-3 )( 7n-3 ) $$

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:

$$ a = 7 , b = -24 ~ \text{ and } ~ c = 9 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 9} $.

$$ a \cdot c = 63 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 63 $ and add to $ b = -24 $.

Step 4: All pairs of numbers with a product of $ 63 $ are:

PRODUCT = 63
1     63	-1     -63
3     21	-3     -21
7     9	-7     -9

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -24 }$

PRODUCT = 63 and SUM = -24
1     63	-1     -63
3     21	-3     -21
7     9	-7     -9

Step 6: Replace middle term $ -24 x $ with $ -3x-21x $:

$$ 7x^{2}-24x+9 = 7x^{2}-3x-21x+9 $$

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

$$ 7x^{2}-3x-21x+9 = x\left(7x-3\right) -3\left(7x-3\right) = \left(x-3\right) \left(7x-3\right) $$

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