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6 February, 21:28

What is th product? y2+3y+7 8y2+y+1

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  1. 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
  2. 6 February, 21:34
    0
    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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