Factor polynomial $$ 8r^{2}+89r+90 $$

$$ 8r^{2}+89r+90 = ( 8r+9 )( r+10 ) $$

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:

$$ a = 8 , b = 89 ~ \text{ and } ~ c = 90 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 90} $.

$$ a \cdot c = 720 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 720 $ and add to $ b = 89 $.

Step 4: All pairs of numbers with a product of $ 720 $ are:

PRODUCT = 720
1     720	-1     -720
2     360	-2     -360
3     240	-3     -240
4     180	-4     -180
5     144	-5     -144
6     120	-6     -120
8     90	-8     -90
9     80	-9     -80
10     72	-10     -72
12     60	-12     -60
15     48	-15     -48
16     45	-16     -45
18     40	-18     -40
20     36	-20     -36
24     30	-24     -30

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 89 }$

PRODUCT = 720 and SUM = 89
1     720	-1     -720
2     360	-2     -360
3     240	-3     -240
4     180	-4     -180
5     144	-5     -144
6     120	-6     -120
8     90	-8     -90
9     80	-9     -80
10     72	-10     -72
12     60	-12     -60
15     48	-15     -48
16     45	-16     -45
18     40	-18     -40
20     36	-20     -36
24     30	-24     -30

Step 6: Replace middle term $ 89 x $ with $ 80x+9x $:

$$ 8x^{2}+89x+90 = 8x^{2}+80x+9x+90 $$

Step 7: Apply factoring by grouping. Factor $ 8x $ out of the first two terms and $ 9 $ out of the last two terms.

$$ 8x^{2}+80x+9x+90 = 8x\left(x+10\right) + 9\left(x+10\right) = \left(8x+9\right) \left(x+10\right) $$

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