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28 September, 21:25

What is the product?

(4y-3) (2y^2+3y-5)

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Answers (2)
  1. 28 September, 23:02
    0
    8y^3 + 6y^2 - 29y + 15

    Step-by-step explanation:

    8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

    then

    8y^3 + 6y^2 - 29y + 15 is the product

    Best regards
  2. 28 September, 23:31
    0
    =8y^3+6y^2-29y+15

    Step-by-step explanation:

    (4y-3) (2y^2+3y-5)

    = (4y+-3) (2y^2+3y+-5)

    = (4y) (2y^2) + (4y) (3y) + (4y) (-5) + (-3) (2y^2) + (-3) (3y) + (-3) (-5)

    =8y^3+12y^2-20^y-6y^2-9y+15

    =8y^3+6y^2-29y+15
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