Factor polynomial $$ 8y^{2}-9y+1 $$

$$ 8y^{2}-9y+1 = ( y-1 )( 8y-1 ) $$

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:

$$ a = 8 , b = -9 ~ \text{ and } ~ c = 1 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 1} $.

$$ a \cdot c = 8 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 8 $ and add to $ b = -9 $.

Step 4: All pairs of numbers with a product of $ 8 $ are:

PRODUCT = 8
1     8	-1     -8
2     4	-2     -4

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = 8 and SUM = -9
1     8	-1     -8
2     4	-2     -4

Step 6: Replace middle term $ -9 x $ with $ -x-8x $:

$$ 8x^{2}-9x+1 = 8x^{2}-x-8x+1 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 8x^{2}-x-8x+1 = x\left(8x-1\right) -1\left(8x-1\right) = \left(x-1\right) \left(8x-1\right) $$

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