IPMAT Indore 2022 · Quantitative Ability · Question 8

The principle. For any polynomial P\left( x \right), the sum of all its coefficients is simply P\left( 1 \right) — substituting x=1 turns every power of x into 1, leaving just the coefficients added together.

Apply it.

P\left( 1 \right)={\left( 5\cdot 1-9 \right)}^{4}={\left( -4 \right)}^{4}=256

Why this works, written out. Expanding,

{\left( 5x-9 \right)}^{4}=\sum_{r=0}^{4}{\left( \genfrac{}{}{0pt}{}{4}{r}\right)}{\left( 5x \right)}^{4-r}{\left( -9 \right)}^{r}=\sum_{r=0}^{4}{\underset{\text{the coefficients}}{\underset{⏟}{\left( \genfrac{}{}{0pt}{}{4}{r}\right){5}^{4-r}{\left( -9 \right)}^{r}}}}{x}^{4-r}

and setting x=1 adds exactly those coefficients.

Answer: 256

Shortcut. Substitute x=1. That is the entire question.

{\left( 5-9 \right)}^{4}={\left( -4 \right)}^{4}=256

Two companions worth memorising alongside it:

Sum of all coefficients =P\left( 1 \right)

Sum of coefficients of even powers minus odd powers =P\left( -1 \right)

So the even-power coefficients sum to P\left( 1 \right)+P\left( -1 \right)/2. Note also that the answer is positive because the exponent 4 is even — a sign slip here gives -256, which is the trap.