pls answer step by step
and don't spam.
pls use only the grouping method of factorisation
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Answer:
2a + b + 3c-d + (2a + b)³ + (2a + b)² (3c - d)
= (2a + b + 3c - d) + ((2a + b)³ + (2a +b) ² (3c - d)]
= 1(2a + b + 3c - d) + (2a + b)² (2a + b + 3c
d)
= (2a + b + 3c-d) [1 + (2a + b)²].
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