Factor polynomial $$ 5x^{2}-43x+24 $$

$$ 5x^{2}-43x+24 = ( x-8 )( 5x-3 ) $$

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:

$$ a = 5 , b = -43 ~ \text{ and } ~ c = 24 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 24} $.

$$ a \cdot c = 120 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 120 $ and add to $ b = -43 $.

Step 4: All pairs of numbers with a product of $ 120 $ are:

PRODUCT = 120
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -43 }$

PRODUCT = 120 and SUM = -43
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 6: Replace middle term $ -43 x $ with $ -3x-40x $:

$$ 5x^{2}-43x+24 = 5x^{2}-3x-40x+24 $$

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

$$ 5x^{2}-3x-40x+24 = x\left(5x-3\right) -8\left(5x-3\right) = \left(x-8\right) \left(5x-3\right) $$

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