How many millimetres are there in 5 and a half litres
1 answer:
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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 x³
y,
18xy = 2 3
3
x
y,
− 10x²y = 2 - 5
x²
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 - 5
3
3
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 ),
- Factor out "
" by grouping,
- Factor 9 from 9x - 45 and x² from x³ - 5x²,
- Factor out common term x - 5,
- And our solution is thus 2y( x - 5 )( x² + 9 )