Question 83 (xxi)
Multiply the following:
(3x2+4x−8),(2x2−4x+3)
We have, (3x2+4x−8) and 2x2−4x+3
∴(3x2+4x−8)(2x2−4x+3)=3x2(2x2−4x+3)+4x(2x2−4x+3)−8(2x2−4x+3)=6x4−12x3+9x2+8x3−16x2+12x−16x2+32x−24
=6x4−12x3+8x3+9x2−16x2−16x2+12x+32x−24 [grouping like terms]
=6x4−4x3−23x2+44x−24