IPMAT Indore 2023 · Quantitative Ability · Question 8
IPMAT Indore 2023 previous year question on Polynomials and Higher Degree Equations, with the answer and a step-by-step solution.
Answer & solution
Answer: 5
Step 1 — apply the Remainder Theorem. Dividing p\left( x \right) by \left( x-1 \right) leaves remainder p\left( 1 \right). So
p\left( 1 \right)=2
Step 2 — evaluate p\left( 1 \right). At x=1 every power of x equals 1, so p\left( 1 \right) is simply the sum of the coefficients:
4-1+3-5+c+2-1+1-4+6-2
Add the known numbers in order: 4-1=3; +3=6; -5=1; +2=3; -1=2; +1=3; -4=-1; +6=5; -2=3.
p\left( 1 \right)=3+c
Step 3 — solve.
3+c=2 \Rightarrow c=-1
Step 4 — answer what was asked.
c+6=-1+6=5
Answer: 5
Shortcut. “Divided by x-1” means “add up the coefficients”. That is the entire question — the ten-term polynomial is there to waste your time, not to be divided.
Sum the knowns to get 3, so 3+c=2 and c=-1.
Then read the question again. It asks for c+6, not c — answer 5. Questions that ask for a shifted or scaled version of the unknown are common in the SA section precisely because candidates stop one step early; circle what is being asked before you start.
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