Question

$(2x-1{)}^{20}-(ax+b{)}^{2}0=(x^2+px+q{)}^{10}$ find value of $\frac{80q+6p}{2}$.

Answer 1

Let $F=(2x-1{)}^{20}-(ax+b{)}^{20},G=(x^2+px+q{)}^{10}$

$F=G$

Equating $x^{20}$ coefficient in $F,G$

$2^{20}-a^{20}=1$

Equating $x^{19}$ coefficient in $F,G$

${}^{20}{C}_1\left(2{)}^{19}\right(-1\left){-}^{20}{C}_1\right(a{)}^{19}\left(b\right)=\left(10\right)\left(p\right)$ $(\because$ for $x^{19}$, we have to take nine $x^2$ terms and one $px$ which has probability of occuring as ten $)$

$⟹p=-2(a^{19}b+2)$

equating constant terms $1-b^{20}=q^{10}$

equating $x$ term $10p{q}^{19}=20(2+a{b}^{19})$

$p{q}^{19}=2a{b}^{19}+4$

solving all these equations we get

$a=(1048575{)}^{\frac{1}{20}},b=0,p=-4,q=-1$

$\frac{80q+6p}{2}=-40-12=-52$