Answer:
Part A : 2y( x³ + 9x - 5x² - 45 ), Part B : 2y( x - 5 )( x² + 9 )
Step-by-step explanation:
Part A : Let's break every term down here to their " prime factors ", and see what is common among them,
2x³y + 18xy − 10x²y − 90y -
2x³y = 2 [tex]*[/tex] x³ [tex]*[/tex] y,
18xy = 2 [tex]*[/tex] 3 [tex]*[/tex] 3 [tex]*[/tex] x [tex]*[/tex] y,
− 10x²y = 2 [tex]*[/tex] - 5 [tex]*[/tex] x² [tex]*[/tex] y, - so as you can see for this example I purposely broke down - 10 into 2 and - 5. I could have placed the negative on the 2, but as that value was must likely common among all the terms, I decided to place it on the 5. The same goes for " − 90y. " I placed the negative there on the 5 once more.
− 90y = 2 [tex]*[/tex] - 5 [tex]*[/tex] 3 [tex]*[/tex] 3 [tex]*[/tex] y
The terms common among each term are 2 and y. Therefore, the GCF ( greatest common factor ) is 2x. Let's now factor the expression using this value.
2y( x³ + 9x - 5x² - 45 )
Part B : Let's simply factor this entire expression. Of course starting with the " factored " expression : 2y( x³ + 9x - 5x² - 45 ),
[tex]2y\left(x^3+9x-5x^2-45\right)[/tex] - Factor out " [tex](x^3+9x-5x^2-45\right))[/tex] " by grouping,
[tex]\left(x^3-5x^2\right)+\left(9x-45\right)[/tex] - Factor 9 from 9x - 45 and x² from x³ - 5x²,
[tex]9\left(x-5\right)+x^2\left(x-5\right)[/tex] - Factor out common term x - 5,
[tex]\left(x-5\right)\left(x^2+9\right)[/tex] - And our solution is thus 2y( x - 5 )( x² + 9 )