Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 16

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\dfrac{1}{2}$

Substituting x = $-\dfrac{1}{2}$ in 2x³ + 7x² + 2x - 3, we get :

$\Rightarrow 2 \times \Big(-\dfrac{1}{2}\Big)^3 + 7 \times \Big(-\dfrac{1}{2}\Big)^2 + 2 \times \Big(-\dfrac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\dfrac{1}{8} + 7 \times \dfrac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\dfrac{1}{4} + \dfrac{7}{4} - 4 \\[1em] \Rightarrow \dfrac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \dfrac{6}{4} - 4 \\[1em] \Rightarrow \dfrac{6 - 16}{4} \\[1em] \Rightarrow \dfrac{-10}{4} \\[1em] \Rightarrow -\dfrac{5}{2}.$

Since, remainder is not equal to zero.

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

Substituting x = $\dfrac{1}{2}$ in 2x³ + 7x² + 2x - 3, we get :

$\Rightarrow 2 \times \Big(\dfrac{1}{2}\Big)^3 + 7 \times \Big(\dfrac{1}{2}\Big)^2 + 2 \times \dfrac{1}{2} - 3 \\[1em] \Rightarrow 2 \times \dfrac{1}{8} + 7 \times \dfrac{1}{4} + 1 - 3 \\[1em] \Rightarrow \dfrac{1}{4} + \dfrac{7}{4} - 2 \\[1em] \Rightarrow \dfrac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.$

Since, remainder is equal to zero.

∴ x - $\dfrac{1}{2}$ is factor of polynomial,

⇒ x - $\dfrac{1}{2}$ = 0

⇒ x = $\dfrac{1}{2}$

⇒ 2x = 1

⇒ 2x - 1 is factor of polynomial.

Dividing 2x³ + 7x² + 2x - 3 by 2x - 1, we get :

$\begin{array}{l} \phantom{2x - 1)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x - 3} \\ \phantom{2x - 1)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{{2x - 1}+2x^3 + 1}8x^2 + 2x \\ \phantom{{2x - 1)}+2x^31}\underline{\underset{-}{+}8x^2 \underset{+}{-} 4x} \\ \phantom{{2x - 1)}+2x^31+8x^212}6x - 3 \\ \phantom{{2x - 1)}+2x^31+8x^21}\underline{\underset{-}{+}6x \underset{+}{-} 3} \\ \phantom{{2x - 1)}+2x^31+8x^21+6}\times \end{array}$

2x³ + 7x² + 2x - 3 by 2x - 1 = (2x - 1)(x² + 4x + 3)

= (2x - 1)[x² + 3x + x + 3]

= (2x - 1)[x(x + 3) + 1(x + 3)]

= (2x - 1)(x + 1)(x + 3).

Hence, 2x³ + 7x² + 2x - 3 = (2x - 1)(x + 1)(x + 3).