Additional Mathematics 0606 / Series / 2.142.14: Binomial product: find aaa, bbb and ccc0606/23/O/N/19 — Question 3 · 8 marks · BinomialMark as done · Save for later · Show all solutionsThe first four terms in the expansion of (1+ax)5(2+bx)(1 + ax)^{5}(2 + bx)(1+ax)5(2+bx) are 2+32x+210x2+cx32 + 32x + 210x^{2} + cx^{3}2+32x+210x2+cx3, where aaa, bbb and ccc are integers. Show that 3a2−16a+21=03a^{2} - 16a + 21 = 03a2−16a+21=0 and hence find the values of aaa, bbb and ccc.[8]▸ Answer▸ Official mark scheme← 2.5: Binomial expansion with parameter ppp1.2: Binomial constants and term independent of xxx →