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Multiplication of a Binomial by a Monomial

Question 82 v...

Question

Question 82 (vi)

Subtract

7p(3q + 7p) from 8p(2p - 7q)

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Solution

We have, 7p (3q + 7p) and 8p(2p - 7q)
The required difference is given by 8p(2p -7q) -7p(3q + 7p)
$= 16 p^{2} - 56 p q - 21 p q - 49 p^{2} = (16 p^{2} - 49 p^{2}) + (- 56 p q - 21 p q)$
[grouping like terms]
$= - 33 p^{2} - 77 p q$

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Similar questions

Question 82 (vi)

Subtract

7p(3q + 7p) from 8p(2p - 7q)

7 p (3 q + 7 p)

8 p (2 p - 7 q)

.

Question 4 (iii)

Subtract $4 p^{2} q - 3 p q + 5 p q^{2} - 8 p + 7 q - 10 f r o m 18 - 3 p - 11 q + 5 p q - 2 p q^{2} + 5 p^{2} q .$

Question 82 (vii)

Subtract

$- 3 p^{2} + 3 p q + 3 p x$ from 3p(-p-a-r)

12 p^{2} - 7 p = - 1

6 q^{2} - 7 q + 2 = 0

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Multiplication of a Binomial by a Monomial

Standard VIII Mathematics

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