For this case we have:
By definition, the distributive property states that:
[tex](a + b) (c + d) = ac + ad + bc + bd[/tex]
In addition, we know that: [tex]ax * bx = (ab) x ^ 2[/tex], that is, the exponents are added together.
If we have:[tex](4y - 3)\ and\ (2y ^ 2 + 3y - 5)[/tex], your product is given by:
[tex](4y - 3) (2y ^ 2 + 3y - 5) = (4 * 2) y^{1 + 2} + (4 * 3) y^{1 + 1} - (4 * 5) y^{1 +0} - (3 * 2) y ^ {0 + 2} - (3 * 3) y ^ {0 + 1} - (3 * (-5))[/tex]
[tex](4y - 3) (2y ^ 2 + 3y - 5) = 8y ^ 3 + 12y ^ 2-20y ^ 1-6y ^ 2-9y ^ 1 - (- 15)[/tex]
[tex](4y - 3) (2y ^ 2 + 3y - 5) = 8y ^ 3 + (12-6) y^ 2 + (- 20-9)y - (-15)[/tex]
Taking into account that equal signs are added and the same sign is placed, and that [tex]- * - = +[/tex]we have:
[tex](4y - 3) (2y ^ 2 + 3y - 5) = 8y ^ 3 + 6y ^ 2-29y + 15[/tex]
So,[tex](4y - 3) (2y ^ 2 + 3y - 5) = 8y ^ 3 + 6y ^ 2-29y + 15[/tex]
Answer:
Option D
