Question
Simplify:
$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$

Answer

To simplify, we will proceed as follows:
$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$
$=2 x^2\left(3 x^2-5 x+4\right)+3 x\left(3 x^2-5 x+4\right)-5\left(3 x^2-5 x+4\right) \text { (Distributive law) }$
$=6 x^4-10 x^3+8 x^2+9 x^3-15 x^2+12 x-15 x^2+25 x-20$
$=6 x^4-10 x^3+9 x^3+8 x^2-15 x^2-15 x^2+12 x+25 x-20 \text { (Rearranging) }$
$=6 x^4-x 3-22 x^2+36 x-20$
Thus, the answer is $6 x^4-x 3-22 x^2+36 x-20$.

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