1
Question
Factorise:
(10) 24p3q2 + 28p2q3
(10) 24p3q2 + 28p2q3
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Solution
24p3q2 + 28p2q3
24p3q2 = 2 × 2 × 2 × 3 × p × p × p × q × q
28p2q3 = 2 × 2 × 7 × p × p × q × q × q
Here 2, 2, p, p, q and q are the common factors of the given terms.
Hence, 24p3q2 + 28p2q3 = 2 × 2 × p × p × q × q (2 × 3 × p + 7 × q)
= 4p2q2 (6p + 7q)
24p3q2 = 2 × 2 × 2 × 3 × p × p × p × q × q
28p2q3 = 2 × 2 × 7 × p × p × q × q × q
Here 2, 2, p, p, q and q are the common factors of the given terms.
Hence, 24p3q2 + 28p2q3 = 2 × 2 × p × p × q × q (2 × 3 × p + 7 × q)
= 4p2q2 (6p + 7q)
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