Question

Multiply (2x^2-3x)(3x^2+2x-1)

Answer 1

Answer:

Step-by-step explanation:            -9X3+12x2-x-2

Answer 2

ANSWER:  

Multiplication of $\left(2 x^{2}-3 x\right) \text { and }\left(3 x^{2}+2 x-1\right) \text { is } 6 x^{4}-5 x^{3}-8 x^{2}+3 x$

SOLUTION:

We need to multiply $\left(2 x^{2}-3 x\right) \text { and }\left(3 x^{2}+2 x-1\right)$

$=\left(2 x^{2}-3 x\right)\left(3 x^{2}+2 x-1\right)$

$=2 x^{2}\left(3 x^{2}+2 x-1\right)-3 x\left(3 x^{2}+2 x-1\right)$

$=\left(2 x^{2} \times 3 x^{2}\right)+\left(2 x^{2} \times 2 x\right)+\left(2 x^{2} \times(-1)\right)-\left(3 x \times 3 x^{2}\right)-(3 x \times 2 x)-(3 x \times(-1))$

$=6 x^{4}+4 x^{3}-2 x^{2}-9 x^{3}-6 x^{2}+3 x$

$=6 x^{4}+(4-9) x^{3}-(2+6) x^{2}+3 x$

$=6 x^{4}-5 x^{3}-8 x^{2}+3 x$

Hence, multiplication of $\left(2 x^{2}-3 x\right) \text { and }\left(3 x^{2}+2 x-1\right) \text { is } 6 x^{4}-5 x^{3}-8 x^{2}+3 x$