Step 1 :
After factoring out $ 2y^{2} $ we have:
$$ 6y^{4}+14y^{3}-10y^{2} = 2y^{2} ( 3y^{2}+7y-5 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -5} $.
$$ a \cdot c = -15 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -15 $ and add to $ b = 7 $.
Step 5: All pairs of numbers with a product of $ -15 $ are:
PRODUCT = -15 | |
-1 15 | 1 -15 |
-3 5 | 3 -5 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 7 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 7 }$, we conclude the polynomial cannot be factored.