(3x + 5) is a factor of the polynomial (a - 1)x 3 + (a + 1)x 2 - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

3x + 5 = 0 ⇒ x = - Since, (3x + 5) is a factor of the polynomial (a - 1)x 3 + (a + 1)x 2 - (2a + 1)x - 15. Substituting x = - in (a - 1)x 3 + (a + 1)x 2 - (2a + 1)x - 15, remainder = 0. Substituting a = 4 in (a - 1)x 3 + (a + 1)x 2 - (2a + 1)x - 15, ⇒ (4 - 1)x 3 + (4 + 1)x 2 - (2(4) + 1)x - 15 ⇒ 3x 3 + 5x 2 - 9x - 15 ⇒ x 2 (3x + 5) - 3(3x + 5) ⇒ (x 2 - 3)(3x + 5) ⇒ Hence, a = 4 and 3x 3 + 5x 2 - 9x - 15 = .

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Mathematics

(3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Factorisation

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Mathematics

(3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Factorisation

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Mathematics

(3x + 5) is a factor of the polynomial (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Factorisation

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(3x + 5) is a factor of the polynomial (a - 1)x³ + (a + 1)x² - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Find the value of 'm', if mx³ + 2x² - 3 and x² - mx + 4 leave the same remainder when divided by x - 2.

The polynomial px³ + 4x² - 3x + q is completely divisible by x² - 1; find the values of p and q. Also for these values of p and q factorize the given polynomial completely.

When the polynomial x³ + 2x² - 5ax - 7 is divided by (x - 1), the remainder is A and when the polynomial x³ + ax² - 12x + 16 is divided by (x + 2), the remainder is B. Find the value of 'a' if 2A + B = 0.

Using remainder theorem, find the value of k if on dividing 2x³ + 3x² - kx + 5 by (x - 2) leaves a remainder 7.