Factor polynomial $$ 2t^{2}-11t+15 $$

$$ 2t^{2}-11t+15 = ( t-3 )( 2t-5 ) $$

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:

$$ a = 2 , b = -11 ~ \text{ and } ~ c = 15 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 15} $.

$$ a \cdot c = 30 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 30 $ and add to $ b = -11 $.

Step 4: All pairs of numbers with a product of $ 30 $ are:

PRODUCT = 30
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -11 }$

PRODUCT = 30 and SUM = -11
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 6: Replace middle term $ -11 x $ with $ -5x-6x $:

$$ 2x^{2}-11x+15 = 2x^{2}-5x-6x+15 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 2x^{2}-5x-6x+15 = x\left(2x-5\right) -3\left(2x-5\right) = \left(x-3\right) \left(2x-5\right) $$

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