factorize b2c3+8bc4+12c5
Q 21
We have,
b2c3 + 8bc4 + 12c5
= c3 (b2 + 8bc + 12c2)
= c3 (b2 + 6bc + 2bc + 12c2)
= c3 [b(b + 6c) + 2c(b + 6c)]
= c3 (b + 6c)(b + 2c)
factorize b2c3+8bc4+12c5
Q 21
We have,
b2c3 + 8bc4 + 12c5
= c3 (b2 + 8bc + 12c2)
= c3 (b2 + 6bc + 2bc + 12c2)
= c3 [b(b + 6c) + 2c(b + 6c)]
= c3 (b + 6c)(b + 2c)