Question

In the expansion of (1+ax)(1+bx)⁴ in ascending powers of x and x² are 0 and -40 respectively. Find the values of a and b​

Answer 1

`(1+ax)(1+bx)^4=(1+ax)\displaystyle\sum_(k=0)^4\binom4k(bx)^k`

When
`k=0`, the sum contributes
`\binom40(bx)^0=1`;

when
`k=1`, it contributes
`\binom41(bx)^1=4bx`;

when
`k=2`, it contributes
`\binom42(bx)^2=6b^2x^2`.

In the complete expansion, the
`x` term is then
`ax\cdot1+1\cdot4bx=(a+4b)x=0x`, and the
`x^2` term is
`ax\cdot4bx+1\cdot6b^2x^2=(4ab+6b^2)x^2=-40x`.

So we have

`\begin{cases}a+4b=0\\4ab+6b^2=-40\end{cases}\implies a=\pm8,b=\mp2`