Long Questions

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

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

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

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

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

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

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

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

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