If 2x³ + ax² + bx - 6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2). Calculate the values of a and b hence factorise the given expression completely.
Factorisation
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Question

If 2x³ + ax² + bx - 6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2). Calculate the values of a and b hence factorise the given expression completely.

KnowledgeBoat · Accepted Answer

By factor theorem, A polynomial f(x) has a factor (x - a), if and only if, f(a) = 0. ⇒ 2x + 1 = 0 ⇒ 2x = -1 ⇒ x = -. Since, (2x + 1) is a factor of 2x³ + ax² + bx - 6. By remainder theorem, If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a). ⇒ x + 2 = 0 ⇒ x = -2. Given, 2x³ + ax² + bx - 6 leaves a remainder 12 when divided by (x + 2). ∴ 2(-2)³ + a(-2)² + b(-2) - 6 = 12 ⇒ -2(8) + 4a - 2b - 6 = 12 ⇒ -16 + 4a - 2b - 6 = 12 ⇒ 4a - 2b = 12 + 16 + 6 ⇒ 4a - 2b = 34 ⇒ 2(2a - b) = 34 ⇒ 2a - b = 17 Substituting value of a in above equation from Eq. 1 : ⇒ 2(2b + 25) - b = 17 ⇒ 4b + 50 - b = 17 ⇒ 3b = 17 - 50 ⇒ 3b = -33 ⇒ b = -11. Substituting value of b in equation (1), we get : ⇒ a = 2b + 25 ⇒ a = 2(-11) + 25 = -22 + 25 = 3. Substituting value of a and b in 2x³ + ax² + bx - 6, we get : ⇒ 2x³ + (3)x² + (-11)x - 6 ⇒ 2x³ + 3x² - 11x - 6 Since, (2x + 1) is the factor. On dividing 2x³ + 3x² - 11x - 6 by (2x + 1), we get : ⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x² + x - 6) ⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x² + 3x - 2x - 6) ⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)[x(x + 3) - 2(x + 3)] ⇒ 2x³ + 3x² - 11x - 6 = (2x + 1)(x - 2)(x + 3). Hence, a = 3, b = -11 and 2x³ + 3x² - 11x - 6 = (2x + 1)(x - 2)(x + 3).

Mathematics

If 2x³ + ax² + bx - 6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2). Calculate the values of a and b hence factorise the given expression completely.

Factorisation

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Mathematics

If 2x3 + ax2 + bx - 6 has a factor (2x + 1) and leaves a remainder 12 when divided by (x + 2). Calculate the values of a and b hence factorise the given expression completely.

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