Factor polynomial $$ 14x^{2}-10x+8 $$

$$ 14x^{2}-10x+8 = 2( 7x^{2}-5x+4 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 14x^{2}-10x+8 = 2 ( 7x^{2}-5x+4 ) $$

Step 2 :

Step 2: 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:

$$ a = 7 , b = -5 ~ \text{ and } ~ c = 4 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 4} $.

$$ a \cdot c = 28 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 28 $ and add to $ b = -5 $.

Step 5: All pairs of numbers with a product of $ 28 $ are:

PRODUCT = 28
1     28	-1     -28
2     14	-2     -14
4     7	-4     -7

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -5 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -5 }$, we conclude the polynomial cannot be factored.

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