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creativ13 [48]
3 years ago
10

(4y − 3)(2y2 + 3y − 5)

Mathematics
1 answer:
IRINA_888 [86]3 years ago
6 0

Answer:

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

Step-by-step explanation:

(4y-3)(2y^2+3y-5)\\4y(2y^2 + 3y - 5) - 3(2y^2+3y-5)\\8y^3+12y^2-20y-3(2y^2+3y-5)\\8y^3+12y^2-20y-6y^2-9y+15\\8y^3+6y^2-29y-9y+15\\= 8y^3+6y^2-29y+15

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Step-by-step explanation:

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The roots of a quadratic equation depends on the discriminant \Delta.

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---------------------

A quadratic equation has the following format:

y = ax^2 + bx + c

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ax^2 + bx + c = 0

They are given by:

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The discriminant is \Delta.

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  • If it is negative, \sqrt{\Delta} is a complex number, and thus, the roots will be complex and will not touch the x-axis.

A similar problem is given at brainly.com/question/19776811

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