Factor polynomial $$ 10x^{2}-29x+10 $$

$$ 10x^{2}-29x+10 = ( 2x-5 )( 5x-2 ) $$

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:

$$ a = 10 , b = -29 ~ \text{ and } ~ c = 10 $$

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

$$ a \cdot c = 100 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 100 $ and add to $ b = -29 $.

Step 4: All pairs of numbers with a product of $ 100 $ are:

PRODUCT = 100
1     100	-1     -100
2     50	-2     -50
4     25	-4     -25
5     20	-5     -20
10     10	-10     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -29 }$

PRODUCT = 100 and SUM = -29
1     100	-1     -100
2     50	-2     -50
4     25	-4     -25
5     20	-5     -20
10     10	-10     -10

Step 6: Replace middle term $ -29 x $ with $ -4x-25x $:

$$ 10x^{2}-29x+10 = 10x^{2}-4x-25x+10 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 10x^{2}-4x-25x+10 = 2x\left(5x-2\right) -5\left(5x-2\right) = \left(2x-5\right) \left(5x-2\right) $$

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