Question

1)(3y-5)(2y-12) 2)(2x^2-3)(x^2+4) 3)b(6b+7)(3b-4)

Answer 1

$1) (3y-5)(2y-12)= 6y^2-36y-10y+60=6y^2-46y+60$

$2) (2x^2-3)(x^2+4)= 4x^4+8x^2-6x^2-12=4x^4+2x^2-12$

$3) b(6b+7)(3b-4)=(6b^2+7b)(3b^2-4b)=b^2(18b^2-3b-28)$