Q) While factorizing a given polynomial, a student finds that (2x + 1) is a factor of 2 x3 + 7 x 2 + 2 x – 3
(a) Is the student’s solution correct stating that (2x + 1) is a factor of the given polynomial?
(b) Give a valid reason for your answer.
(c) Factorize the given polynomial completely.

ICSE Specimen Question Paper (SQP)2025

Ans:

(a) Check if (2x + 1) is the factor of polynomial or not:

If (2 x + 1) is the factor of the polynomial, then (2 x + 1) = 0

∴ x = \frac{- 1}{2}

Now if (2 x + 1) is the factor of the given polynomial, then x = \frac{- 1}{2} should satisfy the polynomial equation.

Given polynomial is f(x) = 2 x3 + 7 x 2 + 2 x – 3

f (\frac{- 1}{2}) = 2 (\frac{- 1}{2})^3 + 7 (\frac{- 1}{2})^2 + 2 \frac{- 1}{2} - 3

= 2 (\frac{- 1}{8}) + 7 (\frac{- 1}{4}) + 2 \frac{- 1}{2} - 3

(\frac{- 1}{4}) + (\frac{- 7}{4}) - 1 - 3

= – 2 – 4 = – 6

Clearly, (2 x + 1) is not the factor of the given polynomial.

(b) Reason:

for (2 x +1) to be the factor of the given polynomial, it should satisfy f(x) = 0

When we keep x = \frac{- 1}{2}, we found f(\frac{- 1}{2}) = – 6

Since f(\frac{- 1}{2}) \neq 0, therefore (2 x + 1) is not a factor of the given polynomial.

(c) Factors of the polynomial:

Polynomial is: 2 x 3 + 7 x 2 + 2 x – 3 = 0

∴ 2 x 3 – x 2 + 8 x 2 + 2 x – 3 = 0

∴ x 2 (2 x – 1) + 8 x 2  – 4 x + 6 x – 3 = 0

∴ x 2 (2 x – 1) + 4 x (2 x – 1) + 6 x – 3 = 0

∴ x 2 (2 x – 1) + 4 x (2 x – 1) + 3 (2 x – 1) = 0

∴ (2x – 1) (x 2 + 4 x + 3) = 0

∴ (2 x – 1) (x + 3) (x + 1) = 0

Therefore, factors of the given polynomial are (2 x – 1), (x + 3) and (x + 1).

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