IPMAT Indore 2025 (SA) - If the polynomial ax^2 + bx + 5 leaves a remainder 3 when divided by x - 1, and a remainder 2 when divided by x + 1, then 2b - 4a equals | PYQs + Solutions

IPMAT Indore 2025

Number System

Remainder

Easy

If the polynomial $ax^2 + bx + 5$ leaves a remainder $3$ when divided by $x - 1$ , and a remainder $2$ when divided by $x + 1$ , then $2b - 4a$ equals

Entered answer:

✅ Correct Answer: 11

When a polynomial

P(x)

is divided by

(x-k)

, the remainder equals

P(k)

P(x) = ax^2 + bx + 5

Dividing by

(x-1)

, the remainder is

P(1)=3

P(1) = a(1)^2 + b(1) + 5 = 3

\newline

\Rightarrow a + b + 5 = 3

\newline

\Rightarrow a + b = -2 \quad \rightarrow

(1)

Dividing by

(x+1) = (x-(-1))

, the remainder is

P(-1)=2

P(-1) = a(-1)^2 + b(-1) + 5 = 2

\newline

\Rightarrow a - b + 5 = 2

\newline

\Rightarrow a - b = -3 \quad \rightarrow

(2)

Adding equations (1) and (2):

2a = -5

a = -\frac{5}{2}

Substituting into equation (1):

-\frac{5}{2} + b = -2

b = \frac{1}{2}

Calculating

2b - 4a

=2(\frac{1}{2}) - 4(-\frac{5}{2})

\newline

= 1 + 10

\newline

= 11