Factorise completely using factor theorem : 6x 3 - 11x 2 - 3x + 2.

Substituting x = 2 in above equation, we get : ⇒ 6(2) 3 - 11(2) 2 - 3(2) + 2 ⇒ 6(8) - 11(4) - 6 + 2 ⇒ 48 - 44 - 4 ⇒ 0. ∴ x - 2 is the factor of 6x 3 - 11x 2 - 3x + 2. On dividing 6x 3 - 11x 2 - 3x + 2 by (x - 2), we get : ∴ 6x 3 - 11x 2 - 3x + 2 = (x - 2)(6x 2 + x - 1) = (x - 2)(6x 2 + 3x - 2x - 1) = (x - 2)[3x(2x + 1) -1(2x + 1)] = (x - 2)(3x - 1)(2x + 1). Hence, 6x 3 - 11x 2 - 3x + 2 = (x - 2)(3x - 1)(2x + 1).

Prove that : tan² A + cot² A + 2 = sec² A. cosec² A

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A(8, 6), B(10, -10) and C(4, -4) are the vertices of triangle ABC. If P is the mid-point of side AB and Q is mid-point of side AC, show that PQ is parallel to side BC also show 2 × PQ = BC.

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In the given figure, triangle ABC is circumscribed by the given circle. Use the given information to find the value of x.

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Mathematics

Factorise completely using factor theorem :
6x³ - 11x² - 3x + 2.

Factorisation

Related Questions

Prove that : tan² A + cot² A + 2 = sec² A. cosec² A

A(8, 6), B(10, -10) and C(4, -4) are the vertices of triangle ABC. If P is the mid-point of side AB and Q is mid-point of side AC, show that PQ is parallel to side BC also show 2 × PQ = BC.

In the given figure, triangle ABC is circumscribed by the given circle. Use the given information to find the value of x.

Mathematics

Factorise completely using factor theorem :6x3 - 11x2 - 3x + 2.

Factorisation

Related Questions

Prove that : tan2 A + cot2 A + 2 = sec2 A. cosec2 A

A(8, 6), B(10, -10) and C(4, -4) are the vertices of triangle ABC. If P is the mid-point of side AB and Q is mid-point of side AC, show that PQ is parallel to side BC also show 2 × PQ = BC.

In the given figure, triangle ABC is circumscribed by the given circle. Use the given information to find the value of x.

Factorise completely using factor theorem :
6x³ - 11x² - 3x + 2.

Prove that : tan² A + cot² A + 2 = sec² A. cosec² A