Chapter 6: Factorization of Polynomials Exercise – 6.4

Question: 1

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = x³ − 6x² + 11x − 6, g(x) = x − 3

Solution:

Here, f(x) = x³ − 6x² + 11x - 6

g(x) = x - 3

To prove that g(x) is the factor of f(x),

we should show ⟹ f(3) = 0

here, x - 3 = 0

⟹ x = 3

Substitute the value of x in f(x)

f(3) = 3³ - 6∗(3)² + 11(3) - 6

= 27 - (6*9) + 33 - 6

= 27 - 54 + 33 - 6

= 60 - 60

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 2

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = 3x⁴ + 17x³ + 9x² − 7x − 10, g(x) = x + 5

Solution:

Here, f(x) = 3x⁴ + 17x³ + 9x² − 7x − 10

g(x) = x + 5

To prove that g(x) is the factor of f(x),

we should show ⟹ f(-5) = 0

here, x + 5 = 0

⟹ x = - 5

Substitute the value of x in f(x)

f(−5) = 3(−5)⁴ + 17(−5)³ + 9(−5)² - 7(−5) - 10

= (3 * 625) + (12 * (-125)) + (9*25) + 35 - 10

= 1875 - 2125 + 225 + 35 - 10

= 2135 - 2135

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 3

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = x⁵ + 3x⁴ − x³ − 3x² + 5x + 15, g(x) = x + 3

Solution:

Here , f(x) = x⁵ + 3x⁴ − x³ − 3x² + 5x + 15

g(x) = x + 3

To prove that g(x) is the factor of f(x),

we should show ⟹ f(-3) = 0

here, x + 3 = 0

⟹ x = -3

Substitute the value of x in f(x)

f(-3) = (-3)⁵ + 3(-3)⁴ - (-3)³ - 3(-3)² + 5(-3) + 15

= – 243 + 243 + 27 - 27 - 15 + 15

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 4

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = x³ − 6x² − 19x + 84, g(x) = x − 7

Solution:

Here, f(x) = x³ − 6x² − 19x + 84

g(x) = x - 7

To prove that g(x) is the factor of f(x),

we should show ⟹ f(7) = 0

here, x - 7 = 0

⟹ x = 7

Substitute the value of x in f(x)

f(7) = 7³ - 6(7)² - 19(7) + 84

= 343 - (6 * 49) - (19 * 7) + 84

= 342 - 294 - 133 + 84

= 427 - 427

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 5

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = 3x³ + x² − 20x + 12, g(x) = 3x − 2

Solution:

Here, f(x) = 3x³ + x² − 20x + 12

g(x) = 3x - 2

To prove that g(x) is the factor of f(x),

we should show ⟹ f(2/3) = 0

here, 3x - 2 = 0

⟹ 3x = 2

⟹ x = 2/3

Substitute the value of x in f(x)

f(2/3) = 3(2/3)³ + (2/3)² - 20(2/3) + 12

= 3(8/27) + 4/9 − 40/3 + 12

= 8/9 + 4/9 − 40/3 + 12

= 12/9 − 40/3 + 12

Taking L.C.M

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 6

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = 2x³ − 9x² + x + 13, g(x) = 3 − 2x

Solution:

Here, f(x) = 2x³ − 9x² + x + 13

g(x) = 3 - 2x

To prove that g(x) is the factor of f(x),

To prove that g(x) is the factor of f(x),

we should show ⟹ f(3/2) = 0

here, 3 - 2x = 0

⟹ -2x = -3

⟹ 2x = 3

⟹ x = 3/2

Substitute the value of x in f(x)

f(3/2) = 2(3/2)³ - 9(3/2)² + (3/2) + 13

= 2(27/8) − 9(9/4) + 3/2 + 12

= (27/4) − (81/4) + 3/2 + 12

Taking L.C.M

= 0

Since, the result is 0 g(x) is the factor of f(x)

 

Question: 7

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not:

f(x) = x³ − 6x² + 11x − 6, g(x) = x² − 3x + 2

Solution:

Here, f(x) = x³ − 6x² + 11x − 6

g(x) = x² − 3x + 2

First we need to find the factors of x² − 3x + 2

⟹ x² − 2x - x + 2

⟹ x(x - 2) -1(x - 2)

⟹ (x - 1) and (x - 2) are the factors

To prove that g(x) is the factor of f(x),

The results of f(1) and f(2) should be zero

Let, x – 1 = 0

x = 1

substitute the value of x in f(x)

f(1) = 1³ – 6(1)² + 11(1) – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

Let, x – 2 = 0

x = 2

substitute the value of x in f(x)

f(2) = 2³ – 6(2)² + 11(2) – 6

= 8 – (6 * 4) + 22 – 6

= 8 – 24 + 22 - 6

= 30 – 30

= 0

Since, the results are 0 g(x) is the factor of f(x)

 

Question: 8

Show that (x – 2), (x + 3) and (x – 4) are the factors of x³ − 3x² − 10x + 24

Solution:

Here, f(x) = x³ − 3x² − 10x + 24

The factors given are (x - 2), (x + 3) and (x - 4)

To prove that g(x) is the factor of f(x),

The results of f(2), f(-3) and f(4) should be zero

Let, x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

f(2) = 2³ - 3(2)² - 10(2) + 24

= 8 - (3 * 4) - 20 + 24

= 8 - 12 - 20 + 24

= 32 - 32

= 0

Let, x + 3 = 0

⟹ x = -3

Substitute the value of x in f(x)

f(-3) = (−3)³ - 3(−3)² - 10(−3) + 24

= -27 - 3(9) + 30 + 24

= -27 - 27 + 30 + 24

= 54 - 54

= 0

Let, x - 4 = 0

⟹ x = 4

Substitute the value of x in f(x)

f(4) = (4)³ - 3(4)² - 10(4) + 24

= 64 - (3 * 16) - 40 + 24

= 64 - 48 - 40 + 24

= 84 - 84

= 0

Since, the results are 0 g(x) is the factor of f(x)

 

Question: 9

Show that (x + 4), (x - 3) and (x - 7) are the factors of x³ − 6x² - 19x + 84

Solution:

Here, f(x) = x³ − 6x² − 19x + 84

The factors given are (x + 4), (x - 3) and (x - 7)

To prove that g(x) is the factor of f(x),

The results of f(- 4) , f(3) and f(7) should be zero

Let, x + 4 = 0

⟹ x = - 4

Substitute the value of x in f(x)

f(-4) = (−4)³ - 6(−4)² - 19(−4) + 84

= -64 - (6 * 16) - (19 * (- 4)) + 84

= - 64 - 96 + 76 + 84

= 160 - 160

= 0

Let, x - 3 = 0

⟹ x = 3

Substitute the value of x in f(x)

f(3) = (3)³ – 6(3)² – 19(3) + 84

= 27 – (6 * 9) – (19 * 3) + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0

Let, x - 7 = 0

⟹ x = 7

Substitute the value of x in f(x)

f(7) = (7)³ – 6(7)² – 19(7) + 84

= 343 – (6 * 49) – (19 * 7) + 84

= 343 – 294 - 133 + 84

= 427 – 427

= 0

Since, the results are 0 g(x) is the factor of f(x)

 

Question: 10

For what value of a is (x – 5) a factor of x³ − 3x² + ax − 10

Solution:

Here, f(x) = x³ − 3x² + ax − 10

By factor theorem

If (x - 5) is the factor of f(x) then, f(5) = 0

⟹ x - 5 = 0

⟹ x = 5

Substitute the value of x in f(x)

f(5) = 5³ - 3(5)² + a(5) − 10

= 125 - (3 * 25) + 5a - 10

= 125 - 75 + 5a - 10

= 5a + 40

Equate f(5) to zero

f(5) = 0

⟹ 5a + 40 = 0

⟹ 5a = - 40

⟹ a = − 40/5

= - 8

When a = - 8, (x - 5) will be factor of f(x)

 

Question: 11

Find the value of a such that (x - 4) is a factor of 5x³ − 7x² - ax - 28

Solution:

Here, f(x) = 5x³ − 7x² - ax - 28

By factor theorem

If (x - 4) is the factor of f(x) then, f(4) = 0

⟹ x – 4 = 0

⟹ x = 4

Substitute the value of x in f(x)

f(4) = 5(4)³ – 7(4)² – a(4) – 28

= 5(64) – 7(16) – 4a – 28

= 320 – 112 – 4a – 28

= 180 – 4

Equate f(4) to zero, to find a

f(4) = 0

⟹ 180 – 4a = 0

⟹ -4a = -180

⟹ 4a = 180

⟹ a = 180/4

⟹ a = 45

When a = 45, (x - 4) will be factor of f(x)

 

Question: 12

Find the value of a, if (x + 2) is a factor of 4x⁴ + 2x³ − 3x² + 8x + 5a

Solution:

Here, f(x) = 4x⁴ + 2x³ − 3x² + 8x + 5a

By factor theorem

If (x + 2) is the factor of f(x) then, f(-2) = 0

⟹ x + 2 = 0

⟹ x = -2

Substitute the value of x in f(x)

f(-2) = 4(−2)⁴ + 2(−2)³ - 3(−2)² + 8(−2) + 5a

= 4(16) + 2(-8) - 3(4) - 16 + 5a

= 64 - 16 - 12 - 16 + 5a

= 5a + 20

equate f(-2) to zero

f(-2) = 0

⟹ 5a + 20 = 0

⟹ 5a = - 20

⟹ a = -20/5

⟹ a = - 4

When a = - 4, (x + 2) will be factor of f(x)

 

Question: 13

Find the value of k if x - 3 is a factor of k²x³ − kx² + 3kx − k

Solution:

Let f(x) = k²x³ − kx² + 3kx − k

From factor theorem if x - 3 is the factor of f(x) then f(3) = 0

⟹ x - 3 = 0

⟹ x = 3

Substitute the value of x in f(x)

f(3) = k²(3)³ - k(3)² + 3k(3) − k

= 27k² − 9k + 9k - k

= 27k² - k

= k(27k - 1)

Equate f(3) to zero, to find k

⟹ f(3) = 0

⟹ k(27k - 1) = 0

⟹ k = 0 and 27k - 1 = 0

⟹ k = 0 and 27k = 1

⟹ k = 0 and k = 1/27

When k = 0 and 1/27, (x - 3) will be the factor of f(x)

 

Question: 14

Find the values of a and b, if x² - 4 is a factor of ax⁴ + 2x³ − 3x² + bx − 4

Solution:

Given, f(x) = ax⁴ + 2x³ − 3x² + bx − 4

g(x) = x² - 4

first we need to find the factors of g(x)

⟹ x² - 4

⟹ x² = 4

⟹ x = √4

⟹ x = ± 2

(x - 2) and (x + 2) are the factors

By factor therorem if (x - 2) and (x + 2) are the factors of f(x) the result of f(2) and f(-2) should be zero

Let, x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

f(2) = a(2)⁴ + 2(2)³ - 3(2)² + b(2) − 4

= 16a + 2(8) - 3(4) + 2b - 4

= 16a + 2b + 16 - 12 - 4

= 16a + 2b

Equate f(2) to zero

⟹ 16a + 2b = 0

⟹ 2(8a + b) = 0

⟹ 8a + b = 0 ..... 1

Let, x + 2 = 0

⟹ x = -2

Substitute the value of x in f(x)

f(-2) = a(- 2)⁴ + 2(- 2)³ - 3(-2)² + b(−2) - 4

= 16a + 2(-8) - 3(4) - 2b - 4

= 16a - 2b - 16 - 12 - 4

= 16a - 2b - 32

= 16a - 2b - 32

Equate f(2) to zero

⟹ 16a - 2b - 32 = 0

⟹ 2(8a - b) = 32

⟹ 8a - b = 16 .... 2

Solve equation 1 and 2

8a + b = 0

8a - b = 16

16a = 16

a = 1

substitute a value in eq 1

8(1) + b = 0

⟹ b = - 8

The values are a = 1 and b = - 8

 

Question: 15

Find α, β if (x + 1) and (x + 2) are the factors of x³ + 3x² − 2αx + β

Solution:

Given, f(x) = x³ + 3x² − 2αx + β and the factors are (x + 1) and (x + 2)

From factor theorem, if they are the factors of f(x) then results of f(-2) and f(-1) should be zero

Let, x + 1 = 0

⟹ x = -1

Substitute value of x in f(x)

f(-1) = (−1)³ + 3(−1)² − 2α(−1) + β

= −1 + 3 + 2α + β

= 2α + β + 2 ... 1

Let, x + 2 = 0

⟹ x = -2

Substitute value of x in f(x)

f(-2) = (−2)³ + 3(−2)² − 2α(−2) + β

= −8 + 12 + 4α + β

= 4α + β + 4 .... 2

Solving 1 and 2 i.e (1 - 2)

⟹ 2α + β + 2 - (4α + β + 4) = 0

⟹ −2α - 2 = 0

⟹ 2α = −2

⟹ α = −1

Substitute α= -1 in equation 1

⟹ 2(−1) + β = -2

⟹ β = -2 + 2

⟹ β = 0

The values are α = −1 and β = 0

 

Question: 16

Find the values of p and q so that x⁴ + px³ + 2x² − 3x + q is divisible by (x² - 1)

Solution:

Here, f(x) = x⁴ + px³ + 2x² − 3x + q

g(x) = x² − 1

first, we need to find the factors of x² − 1

⟹ x² − 1 = 0

⟹ x² = 1

⟹ x = ±1

⟹ (x + 1) and (x - 1)

From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let us take, x + 1

⟹ x + 1 = 0

⟹ x = -1

Substitute the value of x in f(x)

f(-1) = (−1)⁴ + p(−1)³ + 2(−1)² − 3(−1) + q

= 1 - p + 2 + 3 + q

= -p + q + 6 .... 1

Let us take, x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute the value of x in f(x)

f(1) = (1)⁴ + p(1)³ + 2(1)² − 3(1) + q

= 1 + p + 2 - 3 + q

= p + q  ..... 2

Solve equations 1 and 2

- p + q = - 6

p + q = 0

2q = - 6

q = - 3

substitute q value in equation 2

p + q = 0

p - 3 = 0

p = 3

the values of are p = 3 and q = – 3

 

Question: 17

Find the values of a and b so that (x + 1) and (x - 1) are the factors of x⁴ + ax³ − 3x² + 2x + b

Solution:

Here, f(x) = x⁴ + ax³ − 3x² + 2x + b

The factors are (x + 1) and (x - 1)

From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let, us take x + 1

⟹ x + 1 = 0

⟹ x = -1

Substitute value of x in f(x)

f(-1) = (−1)⁴ + a(−1)³ - 3(−1)² + 2(−1) + b

= 1 - a - 3 - 2 + b

= -a + b - 4 ... 1

Let, us take x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute value of x in f(x)

f(1) = (1)⁴ + a(1)³ - 3(1)² + 2(1) + b

= 1 + a - 3 + 2 + b

= a + b .... 2

Solve equations 1 and 2

- a + b = 4

a + b = 0

2b = 4

b = 2

substitute value of b in eq 2

a + 2 = 0

a = - 2

the values are a = -2 and b = 2

 

Question: 18

If x³ + ax² − bx + 10 is divisible by x³ − 3x + 2, find the values of a and b

Solution:

Here, f(x) = x³ + ax² − bx + 10

g(x) = x³ − 3x + 2

first, we need to find the factors of g(x)

g(x) = x³ − 3x + 2

= x³ − 2x - x + 2

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

= (x - 1) and (x - 2) are the factors

From factor theorem, if x = 1, 2 are the factors of f(x) then f(1) = 0 and f(2) = 0

Let, us take x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute the value of x in f(x)

f(1) = 1³ + a(1)² - b(1) + 10

= 1 + a - b + 10

= a - b + 11  .... 1

Let, us take x - 2

⟹ x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

f(2) = 2³ + a(2)² - b(2) + 10

= 8 + 4a - 2b + 10

= 4a - 2b + 18

Equate f(2) to zero

⟹ 4a - 2b + 18 = 0

⟹ 2(2a - b + 9) = 0

⟹ 2a - b + 9 ..... 2

Solve 1 and 2

a - b = -11

2a - b = -9

(-) (+) (+)

- a = - 2

a = 2

substitute a value in eq 1

⟹ 2 - b = -11

⟹ - b = - 11 - 2

⟹ - b = - 13

=> b = 13

The values are a = 2 and b = 13

 

Question: 19

 If both (x + 1) and (x - 1) are the factors of ax³ + x² − 2x + b, Find the values of a and b

Solution:

Here, f(x) = ax³ + x² − 2x + b

(x + 1) and (x - 1) are the factors

From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let, x - 1= 0

⟹ x = -1

Substitute x value in f(x)

f(1) = a(1)³ + (1)² − 2(1) + b

= a + 1 - 2 + b

= a + b - 1 ..... 1

Let, x + 1 = 0

⟹ x = -1

Substitute x value in f(x)

f(-1) = a(−1)³ + (−1)² − 2(−1) + b

= -a + 1 + 2 + b

= -a + b + 3 ..... 2

Solve equations 1 and 2

a + b = 1

-a + b = -3

2b = -2

⟹ b = -1

substitute b value in eq 1

⟹ a - 1 = 1

⟹ a = 1 + 1

⟹ a = 2

The values are a= 2 and b = -1

 

Question: 20

What must be added to x³ − 3x² − 12x + 19 so that the result is exactly divisible by x² + x − 6

Solution:

Here, p(x) = x³ − 3x² − 12x + 19

g(x) = x² + x − 6

by division algorithm, when p(x) is divided by g(x) , the remainder will be a linear expression in x

Let, r(x) = ax + b is added to p(x)

⟹ f(x) = p(x) + r(x)

= x³ − 3x² − 12x + 19 + ax + b

f(x) = x³ − 3x² + x(a − 12) + 19 + b

We know that, g(x) = x² + x − 6

First, find the factors for g(x)

g(x) = x² + 3x - 2x − 6

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

= (x + 3) (x - 2) are the factors

From, factor theorem when (x + 3) and (x - 2) are the factors of f(x) the f(-3) = 0 and f(2) = 0

Let, x + 3 = 0

⟹ x = -3

Substitute the value of x in f(x)

f(-3) = (−3)³ - 3(−3)² + (−3)(a − 12) + 19 + b

= -27 - 27 - 3a + 24 + 19 + b

= -3a + b + 1 ...... 1

Let, x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

f(2) = (2)³ - 3(2)² + (2)(a − 12) + 19 + b

= 8 - 12 + 2a - 24 + b

= 2a + b - 9 .... 2

Solve equations 1 and 2

- 3a + b = -1

2a + b = 9

(-) (-) (-)

- 5a = - 10

a = 2

substitute the value of a in eq 1

⟹ -3(2) + b = -1

⟹ - 6 + b = - 1

⟹ b = - 1 + 6

⟹ b = 5

∴ r(x) = ax + b

= 2x + 5

∴  x³ − 3x² − 12x + 19 is divided by x² + x − 6 when it is added by 2x + 5

 

Question: 21

What must be added to x³ − 6x² − 15x + 80 so that the result is exactly divisible by x² + x - 12

Solution:

Let, p(x) = x³ − 6x² − 15x + 80

q(x) = x² + x - 12

by division algorithm, when p(x) is divided by q(x) the remainder is a linear expression in x.

so, let r(x) = ax + b is subtracted from p(x), so that p(x) - q(x) is divisible by q(x) let f(x) = p(x) - q(x)

q(x) = x² + x - 12

= x² + 4x − 3x - 12

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

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

clearly, (x - 3) and (x + 4) are factors of q(x)

so, f(x) will be divisible by q(x) if (x - 3) and (x + 4) are factors of q(x)

from, factor theorem

f(-4) = 0 and f(3) = 0

⟹ f(3) = 33 - 6(3)² - 3(a + 15) + 80 - b = 0

= 27 - 54 - 3a - 45 + 80 - b

= -3a - b + 8 .... 1

Similarly,

f(- 4) = 0

⟹ f(- 4)

⟹ (- 4)³ - 6(- 4)² - (- 4)(a + 15) + 80 - b = 0

⟹ - 64 - 96 - 4a + 60 + 80 - b = 0

⟹ 4a - b - 20 = 0  .... 2

Substract eq 1 and 2

⟹ 4a - b - 20 - 8 + 3a + b = 0

⟹ 7a - 28 = 0

⟹ a = 28/7

⟹ a = 4

Put a = 4 in eq 1

⟹ - 3(4) - b = - 8

⟹ - b - 12 = - 8

⟹ - b = - 8 + 12

⟹ b = - 4

Substitute a and b values in r(x)

⟹ r(x) = ax + b

= 4x - 4

Hence, p(x) is divisible by q(x), if r(x) = 4x - 4 is subtracted from it

 

Question: 22

What must be added to 3x³ + x² − 22x + 9 so that the result is exactly divisible by 3x² + 7x − 6

Solution:

Let, p(x) = 3x³ + x² − 22x + 9 and q(x) = 3x² + 7x − 6

By division theorem, when p(x) is divided by q(x), the remainder is a linear equation in x.

Let, r(x) = ax + b is added to p(x), so that p(x) + r(x) is divisible by q(x)

f(x) = p(x) + r(x)

⟹ f(x) = 3x³ + x² − 22x + 9(ax + b)

⟹ = 3x³ + x² + x(a - 22) + b + 9

We know that,

q(x) = 3x² + 7x − 6

= 3x² + 9x - 2x − 6

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

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

So, f(x) is divided by q(x) if (3x - 2) and (x + 3) are the factors of f(x)

From, factor theorem

f(2/3) = 0 and f(-3) = 0

let, 3x - 2 = 0

3x = 2

x = 2/3

⟹ f(2/3) = 3(2/3)³ + (2/3)² + (2/3)(a - 22) + b + 9

= 3(8/27) + 4/9 + 2/3a − 44/3 + b + 9

= 12/9 + 2/3a − 44/3 + b + 9

Equate to zero

⟹ 6a + 9b - 39 = 0

⟹ 3(2a + 3b - 13) = 0

⟹ 2a + 3b - 13 = 0 .... 1

Similarly,

Let, x + 3 = 0

⟹ x = - 3

⟹ f(-3) = 3(- 3)³ + (-3)² + (-3)(a - 22) + b + 9

= – 81 + 9 - 3a + 66 + b + 9

= – 3a + b + 3

Equate to zero

- 3a + b + 3 = 0

Multiply by 3

- 9a + 3b + 9 = 0 ... 2

Substact eq 1 from 2

⟹ -9a + 3b + 9 -2a - 3b + 13 = 0

⟹ -11a + 22 = 0

⟹ -11a = -22

⟹ a = 22/11

⟹ a = 2

Substitute a value in eq 1

⟹ - 3(2) + b = - 3

⟹ - 6 + b = - 3

⟹ b = - 3 + 6

⟹ b = 3

Put the values in r(x)

r(x) = ax + b

= 2x + 3

Hence, p(x) is divisible by q(x), if r(x) = 2x + 3 is added to it

 

Question: 23

If x - 2 is a factor of each of the following two polynomials, find the value of a in each case:

1. x³ − 2ax² + ax − 1

2. x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4

Solution:

(1) Let f(x) = x³ − 2ax² + ax - 1

from factor theorem

if (x - 2) is the factor of f(x) the f(2) = 0

let, x - 2 = 0

⟹ x = 2

Substitute x value in f(x)

f(2) = 2³ − 2a(2)² + a(2) - 1

= 8 - 8a + 2a - 1

= - 6a + 7

Equate f(2) to zero

⟹ - 6a + 7 = 0

⟹ - 6a = -7

⟹ a= 76

When, (x - 2) is the factor of f(x) then a = 76

(2) Let, f(x) = x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4

from factor theorem

if (x - 2) is the factor of f(x) the f(2) = 0

let, x - 2 = 0

⟹ x = 2

Substitute x value in f(x)

f(2) = 2⁵ - 3(2)⁴ - a(2)³ + 3a(2)² + 2a(2) + 4

= 32 - 48 - 8a + 12 + 4a + 4

= 8a - 12

Equate f(2) to zero

⟹ 8a - 12 = 0

⟹ 8a = 12

⟹ a = 12/8

= 3/2

So, when (x - 2) is a factor of f(x) then a = 32

 

Question: 24

In each of the following two polynomials, find the value of a, if (x - a) is a factor:

1. x⁶ − ax⁵ + x⁴ − ax³ + 3x − a + 2

2. x⁵ − a²x³ + 2x + a + 1

Solution:

(1) x⁶ − ax⁵ + x⁴ − ax³ + 3x − a + 2

let, f(x) = x⁶ − ax⁵ + x⁴ − ax³ + 3x − a + 2

here, x - a = 0

⟹ x = a

Substitute the value of x in f(x)

f(a) = a⁶ - a(a)⁵ + (a)⁴ - a(a)³ + 3(a) − a + 2

= a⁶ - a⁶ + (a)⁴ - a⁴ + 3(a) − a + 2

= 2a + 2

Equate to zero

⟹ 2a + 2 = 0

⟹ 2(a + 1) = 0

⟹ a = -1

So, when (x - a) is a factor of f(x) then a = -1

(2) x⁵ − a²x³ + 2x + a + 1

let, f(x) = x⁵ − a²x³ + 2x + a + 1

here, x - a = 0

⟹ x = a

Substitute the value of x in f(x)

f(a) = a⁵ − a²a³ + 2(a) + a + 1

= a⁵ − a⁵ + 2a + a + 1

= 3a + 1

Equate to zero

⟹ 3a + 1 = 0

⟹ 3a = -1

⟹ a = −13

So, when (x - a) is a factor of f(x) then a = −1/3

 

Question: 25

In each of the following two polynomials, find the value of a, if (x + a) is a factor:

1. x³ + ax² − 2x + a + 4

2. x⁴ − a²x² + 3x − a

Solution:

1. x³ + ax² − 2x + a + 4

let, f(x) = x³ + ax² − 2x + a + 4

here, x + a = 0

⟹ x = - a

Substitute the value of x in f(x)

f(-a) = (−a)³ + a(−a)² - 2(−a) + a + 4

= (−a)³ + a³ - 2(−a) + a + 4

= 3a + 4

Equate to zero

⟹ 3a + 4 = 0

⟹ 3a = -4

⟹ a = − 4/3

So, when (x + a) is a factor of f(x) then a = −4/3

2. x⁴ − a²x² + 3x − a

let, f(x) = x⁴ − a²x² + 3x − a

here, x + a = 0

⟹ x = - a

Substitute the value of x in f(x)

f(-a) = (−a)⁴ − a²(−a)² + 3(−a) − a

= a⁴ − a⁴ − 3(a) − a

= -4a

Equate to zero

⟹ -4a = 0

⟹ a = 0

So, when (x + a) is a factor of f(x) then a = 0