Factor polynomial $$ 10k^{2}-51k+56 $$

$$ 10k^{2}-51k+56 = ( 2k-7 )( 5k-8 ) $$

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:

$$ a = 10 , b = -51 ~ \text{ and } ~ c = 56 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 10 }$ by the constant term $\color{blue}{c = 56} $.

$$ a \cdot c = 560 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 560 $ and add to $ b = -51 $.

Step 4: All pairs of numbers with a product of $ 560 $ are:

PRODUCT = 560
1     560	-1     -560
2     280	-2     -280
4     140	-4     -140
5     112	-5     -112
7     80	-7     -80
8     70	-8     -70
10     56	-10     -56
14     40	-14     -40
16     35	-16     -35
20     28	-20     -28

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -51 }$

PRODUCT = 560 and SUM = -51
1     560	-1     -560
2     280	-2     -280
4     140	-4     -140
5     112	-5     -112
7     80	-7     -80
8     70	-8     -70
10     56	-10     -56
14     40	-14     -40
16     35	-16     -35
20     28	-20     -28

Step 6: Replace middle term $ -51 x $ with $ -16x-35x $:

$$ 10x^{2}-51x+56 = 10x^{2}-16x-35x+56 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 10x^{2}-16x-35x+56 = 2x\left(5x-8\right) -7\left(5x-8\right) = \left(2x-7\right) \left(5x-8\right) $$

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