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Question

Subtract 4p2q3pq+5pq28p+7q10 from
183p11q+5pq2pq2+5p2q

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Solution

⇒4p2q−3pq+5pq2−8p+7q−10=x \Rightarrow 4p^{2}q-3pq+5pq^{2}-8p+7q-1Lert
18−3p−11q+5pq−2pq2+5p2q=y 18-3p-11q+5pq-2pq^{2}+5p^{2}q = y

Let x=4p2q3pq+5pq28p+7q10

and, y=183p11q+5pq2pq2+5p2q

yx=(183p11q+5pq2pq2+5p2q)(4p2q3pq+5pq28p+7q10)

=183p11q+5pq2pq2+5p2q4p2q+3pq5pq2+8p7q+10

=(5p2q4p2q)+(5pq22pq2)+(5pq+3pq)+(8p3p)+(11q7q)+(18+10)

=p2q7pq2+8pq+5p18q+28

Therefore, the required difference is p2q7pq2+8pq+5p18q+28


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