# Long Questions

2x3 - 9x2 + 7x + 6 = 0

Solution,
Let, p(x) = 2x3 - 9x2 + 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) = 2x2 - 5x - 3
= 2x2 - (6 - 1)x - 3
= 2x2 - 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: 3x3 = 7x2 - 4

Solution,
Let p(x) = 3x3 - 7x2 + 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) = 3x2 - 4x - 4
= 3x2 - (6 - 2)x - 4
= 3x2 - 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

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

Solve: y3 - 6y2 + 11y - 6 = 0

Solution,
Let, p(y) = y3 - 6y2 + 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) = y2 - 5y + 6
= y2 - 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: 6x3 + x2 - 19x + 6 = 0

Solution,
Let, p(x) = 6x3 + x2 - 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) = 6x2 - 11x + 3
= 6x2 - (9 + 2)x + 3
= 6x2 - 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: 2x3 + 6 - 3x2 - 11x = 0

Solution,
Let, p(x) = 2x3 + 6 - 3x2 - 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) = 2x2 - 7x + 3
= 2x2 - (6 + 1)x + 3
= 2x2 - 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 = -2, 1/2, 3

Solve: x3 - 3x2 - 10x + 24 = 0

Solution,
Let, p(x) = x3 - 3x2 - 10x + 24 = 0 ---- (a)

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

When x = 2,
p(2) = 23 - 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) = x2 - x - 12
= x2 - (4 - 3)x - 12
= x2 - 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

Hence, x = -3, 2, 4

Solve: x3 = 7x2 - 36

Solution,
Let, p(x) = x3 - 7x2 + 0x + 36 = 0 ---- (a)

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

When x = 3,
p(3) = 33 - 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) = x2 - 4x - 12
= x2 - (6 - 2)x - 12
= x2 - 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

Hence, x = -2, 3, 6

Solve: 6x3 = 4 - 13x2

Solution,
Let, p(x) = 6x3 + 13x2 + 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) = 6x2 + x - 2
= 6x2 + 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: 2x3 - 4x2 - 7x + 14 =0

Solution,
Let, p(x) = 2x3 - 4x2 - 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) = 2x2 + 0x - 7
= 2x2 - 7 ---- (c)

Now, we know,
p(x) = q(x).d(x)
= (2x2 - 7)(x - 2) [from (b) and (c)]

Again, from (a),
(2x2 - 7)(x - 2) = 0
Either, x2 = 7/2
or, x = ± $\sqrt{\frac 72}$
OR, x = 2

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