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Oreo
Mathematics
6 February, 21:28
What is th product? y2+3y+7 8y2+y+1
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Daugherty
6 February, 21:30
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Product is : (y²+3y+7) * (8y²+y+1) = 8y^4 + 25y^3 + 60y² + 10y + 7
Step-by-step explanation:
We have given two equations:
y²+3y+7 and 8y²+y+1
Product of above two equations can be given as:
(y²+3y+7) * (8y²+y+1)
or (y².8y² + y². y + y².1) + (3y. 8y² + 3y. y + 3y. 1) + (7.8y²+7. y+7.1)
or 8y^4 + y^3 + y² + 24y^3 + 3y² + 3y + 56y² + 7y + 7
or 8y^4 + 25y^3 + 60y² + 10y + 7, this is the answer
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Belen Cordova
6 February, 21:34
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8y⁴+25y³+60y²+10y+7
Step-by-step explanation:
Given two functions
y²+3y+7 and 8y²+y+1
To take the product of both functions means multiplying both functions together.
(y²+3y+7) (8y²+y+1)
Opening up the bracket we have:
= 8y⁴+y³+y²+24y³+3y²+3y+56y²+7y+7
Collecting the like terms
= 8y⁴+y³+24y³+y²+3y²+56y²+3y+7y+7
Adding up the functions with the same degree and simplifying the expression will give;
= 8y⁴+25y³+60y²+10y+7
This gives the product of both functions
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