# Long Questions

2x^{3} - 9x^{2} + 7x + 6 = 0

Let, p(x) = 2x

^{3}- 9x

^{2}+ 7x + 6 = 0 ---- (a)

Factor of 6 = ±1, ±2, ±3, ±4, ±5, ±6

When x = 2,

p(2) = 2(2)^{3} - 9(2)^{2} + 7(2) + 6

= 16 - 36 + 14 + 6

= 0

So, factor of p(x) = d(x) = x - 2 ---- (b)

Using synthetic division,

From above,

q(x) = 2x^{2} - 5x - 3

= 2x^{2} - (6 - 1)x - 3

= 2x^{2} - 6x + x - 3

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

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

Now, we know,

p(x) = q(x).d(x)

= (2x + 1)(x - 3)(x - 2) [from 'b' and 'c']

Again, from (a),

(2x + 1)(x - 3)(x - 2) = 0

Either, x = -1/2

OR, x = 3

OR, x = 2

Hence, x = -1/2, 2, 3

Solve: 3x^{3} = 7x^{2} - 4

Let p(x) = 3x

^{3}- 7x

^{2}+ 0x + 4 = 0 ---- (a)

Factor of 4 = ±1, ±2, ±3, ±4

When x = 1, f(1) = 3 - 7 + 4 = 0

So, factor of p(x) = d(x) = x - 1 ---- (b)

Using synthetic division,

From above,

q(x) = 3x^{2} - 4x - 4

= 3x^{2} - (6 - 2)x - 4

= 3x^{2} - 6x + 2x - 4

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

= (x - 2)(3x + 2) ---- (c)

Now, we know,

p(x) = q(x).d(x)

= (x - 2)(3x + 2)(x - 1) [from 'b' and 'c']

Again, from (a),

(x - 2)(3x + 2)(x - 1) = 0

Either, x = 2

OR, x = -2/3

OR, x = 1

Solve: y^{3} - 6y^{2} + 11y - 6 = 0

Let, p(y) = y

^{3}- 6y

^{2}+ 11y - 6 = 0 ---- (a)

Factor of 6 = ±1, ±2, ±3, ±4, ±5, ±6

When y = 1, p(1) = 1 - 6 + 11 - 6 = 0

So, d(y) = y - 1 ---- (b)

Using synthetic division,

From above,

q(y) = y^{2} - 5y + 6

= y^{2} - 2y - 3y + 6

= y(y - 2) - 3(y - 2)

= (y - 2)(y - 3) ---- (c)

Now, we know,

p(y) = q(y).d(y)

= (y - 2)(y - 3)(y - 1) [from 'b' and 'c']

Again, from (a),

(y - 2)(y - 3)(y - 1) = 0

Either, y = 1

OR, y = 2

OR, y = 3

Hence, y = 1, 2, 3

Solve: 6x^{3} + x^{2} - 19x + 6 = 0

Let, p(x) = 6x

^{3}+ x

^{2}- 19x + 6 = 0 ---- (a)

Factor of 6 = ±1, ±2, ±3, ±4, ±5, ±6

When x = -2,

p(-2) = 6(-2)^{3} + (-2)^{2} - 19(-2) + 6

= -48 + 4 + 38 + 6

= 0

So, factor of p(x) = d(x) = x + 2 ---- (b)

Using synthetic division,

From above,

q(x) = 6x^{2} - 11x + 3

= 6x^{2} - (9 + 2)x + 3

= 6x^{2} - 9x - 2x + 3

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

= (2x - 3)(3x - 1) ---- (c)

Now, we know,

p(x) = q(x).d(x)
= (2x - 3)(3x - 1)(x + 2) [from 'b' and 'c']

Again, from (a),

(2x - 3)(3x - 1)(x + 2) = 0

Either, x = 3/2

OR, x = 1/3

OR, x = -2

Hence, x = -2, 1/3, 3/2

Solve: 2x^{3} + 6 - 3x^{2} - 11x = 0

Let, p(x) = 2x

^{3}+ 6 - 3x

^{2}- 11x = 0 ---- (a)

Factor of 6 = ±1, ±2, ±3, ±4, ±5, ±6

When x = -2,

p(-2) = 2(-2)^{3} + 6 - 3(-2)^{2} - 11(-2)
= -16 + 6 - 12 + 22

= 0

So, factor of p(x) = d(x) = x + 2 ---- (b)

Using synthetic division,

From above,

q(x) = 2x^{2} - 7x + 3

= 2x^{2} - (6 + 1)x + 3

= 2x^{2} - 6x - x + 3

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

= (2x - 1)(x - 3) ---- (c)

Now, we know,

p(x) = q(x).d(x)

= (2x - 1)(x - 3)(x + 2) [from (b) and (c)]

Again, from (a)

(2x - 1)(x - 3)(x + 2) = 0

Either, x = 1/2

OR, x = 3

OR, x = -2

Solve: x^{3} - 3x^{2} - 10x + 24 = 0

Let, p(x) = x

^{3}- 3x

^{2}- 10x + 24 = 0 ---- (a)

Factor of 24 = ±1, ±2, ±3, ±4, ±5, ±6, ... ..., ±24

When x = 2,

p(2) = 2^{3} - 3(2)^{2} - 10(2) + 24

= 8 - 12 - 20 + 24

= 0

So, factor of p(x) = d(x) = x - 2 ---- (b)

Using synthetic division,

From above,

q(x) = x^{2} - x - 12

= x^{2} - (4 - 3)x - 12

= x^{2} - 4x + 3x - 12

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

= (x - 4)(x + 3) ---- (c)

Now, we know,

p(x) = q(x).d(x)

= (x - 4)(x + 3)(x - 2) [from (b) and (c)]

Again, from (a),

(x - 4)(x + 3)(x - 2) = 0

Either, x = 4

OR, x = -3

OR, x = 2

Solve: x^{3} = 7x^{2} - 36

Let, p(x) = x

^{3}- 7x

^{2}+ 0x + 36 = 0 ---- (a)

Factor of 36 = ±1, ±2, ±3, ±4, ±5, ±6, ... ..., ±36

When x = 3,

p(3) = 3^{3} - 7(3)^{2} + 36

= 27 - 63 + 36

= 0

So, factor of p(x) = d(x) = x - 3 ---- (b)

Using synthetic division,

From above,

q(x) = x^{2} - 4x - 12

= x^{2} - (6 - 2)x - 12

= x^{2} - 6x + 2x - 12

= x(x - 6) + 2(x - 6)

= (x - 6)(x + 2) ---- (c)

Now, we know,

p(x) = q(x).d(x)

= (x - 6)(x + 2)(x - 3) [from 'b' and 'c']

Again, from (a),

(x - 6)(x + 2)(x - 3) = 0

Either, x = 6

OR, x = -2

OR, x = 3

Solve: 6x^{3} = 4 - 13x^{2}

Let, p(x) = 6x

^{3}+ 13x

^{2}+ 0x - 4 = 0 ---- (a)

Factor of 4 = ±1, ±2, ±3, ±4

When x = -2,

p(-2) = 6(-2)^{3} + 13(-2)^{2} + 0(-2) - 4

= -48 + 52 - 4

= 0

So, factor of p(x) = d(x) = x + 2 ---- (b)

Using synthetic division,

From above,

q(x) = 6x^{2} + x - 2

= 6x^{2} + 4x - 3x - 2

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

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

Now, we know,

p(x) = q(x).d(x)

= (2x - 1)(3x + 2)(x + 2) [from 'b' and 'c']

Again, from (a),

(2x - 1)(3x + 2)(x + 2) = 0

Either, x = 1/2

OR, x = -2/3

OR, x = -2

Hence, x = -2, -2/3, 1/2

Solve: 2x^{3} - 4x^{2} - 7x + 14 =0

Let, p(x) = 2x

^{3}- 4x

^{2}- 7x + 14 = 0 ---- (a)

Factor of 14 = ±1, ±2, ±3, ±4, ±5, ±6, ... ..., ±14

When x = 2,

p(2) = 2(2)^{3} - 4(2)^{2} - 7(2) + 14

= 16 - 16 - 14 + 14

= 0

So, factor of p(x) = d(x) = x - 2 ---- (b)

Using synthetic division,

From above,

q(x) = 2x^{2} + 0x - 7

= 2x^{2} - 7 ---- (c)

Now, we know,

p(x) = q(x).d(x)

= (2x^{2} - 7)(x - 2) [from (b) and (c)]

Again, from (a),

(2x^{2} - 7)(x - 2) = 0

Either, x^{2} = 7/2

or, x = ± $\sqrt{\frac 72}$

OR, x = 2

Hence, x = $\sqrt{\frac 72}$, 2