Resolve into factors: 1) 5a + 6bc + 15b + 2ca
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5a + 6bc + 15b + 2c
= > 5a + 15b + 2ac + 6bc
= > 5( a + 3b ) + 2c( a + 3b )
= > ( a + 3b ) ( 5 + 2c )
Hence,
5a + 6bc + 15b + 2ca = ( a + 3b ) ( 5 + 2c )
In Simple meaning we can say that ( a + 3b ) and ( 5 + 2c ) are the factors of 5a + 6bc + 15b + 2ca
= > 5a + 15b + 2ac + 6bc
= > 5( a + 3b ) + 2c( a + 3b )
= > ( a + 3b ) ( 5 + 2c )
Hence,
5a + 6bc + 15b + 2ca = ( a + 3b ) ( 5 + 2c )
In Simple meaning we can say that ( a + 3b ) and ( 5 + 2c ) are the factors of 5a + 6bc + 15b + 2ca
Answered by
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Factorise 5a + 6bc + 15b + 2ca
= 5a + 15b + 6bc + 2ac
= 5(a + 3b) + 2c(3b + a)
= 5(a + 3b) + 2c(a + 3b)
⇒ (5 + 2c) (a + 3b)
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