Question

what is the product of $(p^{3})(2p^{2}-4p)(3p^{2}-1)?$○ $6p^{7}+4p^{4}$○ $6p^{7}-2p^{5}-12p^{5}+4p^{3}$○ $6p^{7}-12p^{6}-2p^{5}+4p^{4}$○ $6p^{12}-14p^{6}+4p^{3}$done

Explanation:

Step1: Multiply first two terms

$p^3(2p^2 - 4p) = 2p^5 - 4p^4$

Step2: Multiply result by third term

$(2p^5 - 4p^4)(3p^2 - 1)$

Step3: Distribute each term

$2p^5 \cdot 3p^2 - 2p^5 \cdot 1 - 4p^4 \cdot 3p^2 + 4p^4 \cdot 1$

Step4: Simplify each term

$6p^7 - 2p^5 - 12p^6 + 4p^4$

Step5: Rearrange by degree

$6p^7 - 12p^6 - 2p^5 + 4p^4$

Answer:

$6p^7 - 12p^6 - 2p^5 + 4p^4$ (the third option)