Factor polynomial $$ 8r^{2}-67r+24 $$

$$ 8r^{2}-67r+24 = ( r-8 )( 8r-3 ) $$

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:

$$ a = 8 , b = -67 ~ \text{ and } ~ c = 24 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 24} $.

$$ a \cdot c = 192 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 192 $ and add to $ b = -67 $.

Step 4: All pairs of numbers with a product of $ 192 $ are:

PRODUCT = 192
1     192	-1     -192
2     96	-2     -96
3     64	-3     -64
4     48	-4     -48
6     32	-6     -32
8     24	-8     -24
12     16	-12     -16

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -67 }$

PRODUCT = 192 and SUM = -67
1     192	-1     -192
2     96	-2     -96
3     64	-3     -64
4     48	-4     -48
6     32	-6     -32
8     24	-8     -24
12     16	-12     -16

Step 6: Replace middle term $ -67 x $ with $ -3x-64x $:

$$ 8x^{2}-67x+24 = 8x^{2}-3x-64x+24 $$

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

$$ 8x^{2}-3x-64x+24 = x\left(8x-3\right) -8\left(8x-3\right) = \left(x-8\right) \left(8x-3\right) $$

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