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