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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 : 138 of 155

Marks: +1, -0

What is the largest root of Equation 1?

Solution:

Concept:

Use the polynomial division identity to find

P

, then solve Equation 1 for its roots.

Explanation:

Given:

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

, divisor

g(x) = 3x + 1

, quotient

= 2x + 9

, remainder

= 19

By the division identity:

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

So,

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

Expand:

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

Thus,

6x^2 + 29x + P = 6x^2 + 29x + 9 + 19 = 6x^2 + 29x + 28

Comparing coefficients,

P = 28

, so Equation 1 becomes

6x^2 + 29x + 28 = 0

Factorise:

(3x + 4)(2x + 7) = 0

Therefore, the roots are

x = -\frac{4}{3}

and

x = -\frac{7}{2}

Since

-\frac{4}{3} > -\frac{7}{2}

, the largest root is

-\frac{4}{3}

Answer:

-\frac{4}{3}

(Option B).

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