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IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Question Numbers: 138-139

Directions: The following questions are based on the information given below.

Two polynomial equations are provided:

Equation 1: 6x^{2} + 29x + P = 0

Equation 2: 3x + 1 = 0

It is known that when the polynomial from Equation 1 is divided by the polynomial from Equation 2, the quotient is (2x + 9) and the remainder is 19.

Question : 139 of 155

Marks: +1, -0

What is the value of P?

Solution:

Concept:

Use the polynomial division identity: dividend equals divisor times quotient plus remainder.

Explanation:

The given polynomial is

f(x) = 6x^{2} + 29x + P

The divisor is

g(x) = 3x + 1

The quotient is

2x + 9

and the remainder is

19

By the division identity:

f(x) = g(x) \times \text{quotient} + \text{remainder}

Substitute the values:

6x^{2} + 29x + P = (3x + 1)(2x + 9) + 19

Expand the product:

(3x + 1)(2x + 9) = 6x^{2} + 27x + 2x + 9 = 6x^{2} + 29x + 9

So the equation becomes:

6x^{2} + 29x + P = 6x^{2} + 29x + 9 + 19

6x^{2} + 29x + P = 6x^{2} + 29x + 28

Comparing the constant terms gives:

P = 28

Answer:

P = 28

, which is option C.

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