Bright Tutorials Bright Tutorials
ICSE Class 10 Mathematics Question 18 of 21

Solved 2024 Question Paper ICSE Class 10 Mathematics — Question 12

Back to all questions
12
Question

Question 8(i)

The polynomial 3x3 + 8x2 - 15x + k has (x - 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.

Answer

⇒ x - 1 = 0

⇒ x = 1.

Given, (x - 1) is a factor of 3x3 + 8x2 - 15x + k.

Thus, on substituting x = 1 in 3x3 + 8x2 - 15x + k, the remainder will be zero.

⇒ 3.(1)3 + 8.(1)2 - 15(1) + k = 0

⇒ 3.1 + 8.1 - 15 + k = 0

⇒ 3 + 8 - 15 + k = 0

⇒ 11 - 15 + k = 0

⇒ k - 4 = 0

⇒ k = 4.

Polynomial = 3x3 + 8x2 - 15x + 4

On dividing (3x3 + 8x2 - 15x + 4) by (x - 1), we get :

x−1)3x2+11x−4x−1)3x3+8x2−15x+4‾x−1))+−3x3−+3x2‾x−131x3−211x2−15xx−1)x3−2+−11x2−+11x‾x−1)31x3−2+1−4x+4x−1)31x3−2+11−+4x+−4‾x−1)31x3−2+11+1×\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \\ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \\ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 1}31x^3-2}11x^2 - 15x \\ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \\ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \\ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \\ \phantom{{x - 1)}31x^3-2+11+1}\times \end{array}

⇒ 3x3 + 8x2 - 15x + 4 = (x - 1)(3x2 + 11x - 4)

= (x - 1)[3x2 + 12x - x - 4]

= (x - 1)[3x(x + 4) - 1(x + 4)]

= (x - 1)(3x - 1)(x + 4).

Hence, 3x3 + 8x2 - 15x + 4 = (x - 1)(3x - 1)(x + 4).