In Class 9 mathematics, understanding the factorization of polynomials is crucial for solving quadratic equations and simplifying algebraic expressions. Factorisation involves expressing a polynomial as a product of its factors which simplifies the polynomial and aids in solving the various problems. Exercise 6.4 from RD Sharma’s book focuses on applying factorization techniques to the different types of polynomial problems. This article provides solutions to Set 2 of Exercise 6.4 demonstrating the practical application of these techniques.
Factorization of Polynomials
The Factorisation of polynomials is the process of breaking down a polynomial into polynomials that when multiplied together give the original polynomial. This technique is fundamental in algebra and helps in solving polynomial equations simplifying the expressions and understanding the polynomial functions. The Common methods of factorization include:
- Factor by Grouping: The Rearranging and grouping terms to find common factors.
- Factoring Quadratics: Using methods like completing the square or applying the quadratic formula.
- Using Identities: Applying algebraic identities such as the difference of squares or perfect square trinomials.
- Polynomial Division: Dividing polynomials to simplify and find factors.
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
According to factor theorem
If x - 3 is the factor of f(x) then f(3) = 0
⇒ x - 3 = 0
⇒ x = 3
On substituting the value of x in f(x), we get
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 k = 1/27
When k = 0 and 1/27, (x - 3) will be the factor of f(x)
Question 14. Find the value of a and b, if x2 - 4 is a factor of f(x) = ax4 + 2x3 - 3x2 + bx - 4
Solution:
Given: f(x) = ax4 + 2x3 - 3x2 + bx - 4, g(x) = x2 - 4
We need to find the factors of g(x)
⇒ x2 - 4 = 0
⇒ x2 = 4
⇒ x = √4
⇒ x = ±2
(x - 2) and (x + 2) are the factors
According to factor theorem
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
On substituting the value of x in f(x), we get
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 the value of f(2) to zero
⇒ 16a + 2b = 0
⇒ 2(8a + b) = 0
⇒ 8a + b = 0 -(1)
Let, x + 2 = 0
x = -2
On substituting the value of x in f(x), we get
f(-2) = a(-2)4 + 2(-2)3- 3(-2)2 + b(-2) - 4
= 16a + 2(-8) - 3(4) - 2b - 4
= 16a - 16 - 12 - 2b - 4
= 16a - 2b - 32
Equate the value of f(-2) to zero
⇒ 16a - 2b - 32 = 0
⇒ 16a - 2b - 32 = 0
⇒ 8a - b = 16 -(2)
On solving equation (1) and (2)
8a + b = 0
8a - b = 16
16a = 16
a = 1
On substituting the value of a in eq (1), we get
8(1) + b = 0
b = -8
The values are a = 1 and b = -8
Question 15. Find \alpha, \beta if (x + 1) and (x + 2) are factors of x^3+3x^2-2\alpha x+\beta
Solution:
Given:
f(x)=x^3+3x^2-2\alpha x+\beta and factors are (x + 1) and (x + 2)According to 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
On substituting the value of x in f(x), we get
f(x)=x^3+3x^2-2\alpha x+\beta =
(-1)^3+3(-1)^2-2\alpha (-1)+\beta =
-1+3+2\alpha +\beta = 2\alpha +\beta +2 -(1)
Let,
⇒ x + 2 = 0
⇒ x = -2
On substituting the value of x in f(x), we get
f(x)=x^3+3x^2-2\alpha x+\beta
=(-2)^3+3(-2)^2-2\alpha (-2)+\beta =
-8+12+4\alpha +\beta
=4\alpha +\beta +4 -(2)On solving eq(1) and (2)
⇒
2\alpha +\beta+2 -(4\alpha +\beta +4)=0 ⇒
-2\alpha -2 =0 ⇒
2\alpha=-2 ⇒
\alpha =-1 On substituting
\alpha =-1 in equation(1)⇒
2(-1)+\beta=-2 ⇒
\beta=-2+2 ⇒
\beta=0 The values are
\alpha =-1 and\beta=0
Question 16. Find the values of p and q so that x4 + px3 + 2x2 - 3x + q is divisible by x2 - 1
Solution:
Given: 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)
According to 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 = 0
x = -1
On substituting the value of x in f(x), we get
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 = 0
x = 1
On substituting the value of x in f(x), we get
f(1) = (1)4 + p(1)3 + 2(1)2 - 3(1) + q
= 1 + p + 2 - 3 + q
= p + q -(2)
On solving eq(1) and (2), we get
-p + q = -6
p + q = 0
2q = -6
q = -3
On substituting q value in eq(2), we get
p + q = 0
p - 3 = 0
p = 3
The value of 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:
Given: f(x) = x4 + ax3 - 3x2 + 2x + b
The factors are (x + 1) and (x - 1)
According to 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
On substituting the value of x in f(x), we get
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
On substituting the value of x in f(x), we get
f(1) = (1)4 + a(1)3 - 3(1)2 + 2(1) + b
= 1 + a - 3 + 2 + b
= a + b -(2)
On solving equation (1) and (2)
-a + b = 4
a + b = 0
2b = 4
b = 2
On substituting value of b in eq (2), we get
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:
Given: 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)(x - 2) are the factors
Let us take (x - 1)
⇒ x - 1 = 0
⇒ x = 1
On substituting the value of x in f(x), we get
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
On substituting the value of x in f(x), we get
f(2) = 23 + a(2)2 - b(2) + 10
= 8 + 4a - 2b + 10
= 4a - 2b + 18
Equating f(2) to zero
⇒ 4a - 2b +18 = 0
⇒ 2a - b + 9 = 0 -(2)
On solving eq(1) and (2), we get
a - b = -11
2a - b = -9
a = 2
On substituting the value of a in equation (1), we get
⇒ 2 - b = - 11
⇒ -b = -11 - 2
⇒ b = 13
The value is 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:
Given: f(x) = ax3 + x2 - 2x + b, (x + 1) and (x - 1) are the factors
According to factor theorem,
If x = -1 & 1 are factors of f(x) then f(1) = 0 and f(-1) = 0
Let, x - 1 = 0
⇒ x = 1
On substituting the value of x in f(x), we get
f(1) = a(1)3 + (1)2 - 2(1) + b
= a +1 - 2 + b
= a + b - 1 -(1)
Let, x + 1 = 0
⇒ x = -1
On substituting the value of x in f(x), we get
f(-1) = a(-1)3 + (-1)2 - 2(-1) + b
= -a + 1 + 2 + b
= -a + b + 3 -(2)
On solving equation (1) and (2), we get
⇒ a + b = 1
⇒ -a + b = -3
⇒ 2b = -2
⇒ b = -1
On substituting the b in eq (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:
Given: p(x) = x3 - 3x2 - 12x + 19, g(x) = x2 + x - 6
According to division algorithm when p(x) is divided by g(x),
the remainder will be the linear expression in x
Let, r(x) = ax + b is added to p(x)
⇒ f(x) = p(x) + r(x)
= f(x) = x3 - 3x2 - 12x + 19 + ax + b
We know that, g(x) = x2 + x - 6
First, we find the factors of g(x)
⇒ g(x) = x2 + x - 6
= x2 + 3x - 2x - 6
= x(x + 3) - 2(x + 3)
= (x - 2)(x + 3)
According to factor theorem
If (x - 2) & (x + 3) are factors of f(x) then f(-3) = 0 and f(2) = 0
Let, x + 3 = 0
⇒ x = -3
On substituting the value of x in f(x), we get
f(-3) = (-3)3 - 3(-3)2 - 12(-3) + 19 + a(-3) + b
= -27 - 27 - 3a + 24 + 19 + b
= -3a + b +1 -(1)
Let, x - 2 = 0
⇒ x = 2
On substituting the value of x in f(x), we get
f(2) = (2)3 - 3(2)2 - 12(2) + 19 + a(2) + b
= 8 - 12 + 2a - 24 + b
= 2a + b - 9 -(2)
On solving eq(1) and eq (2), we get
⇒ -3a + b = -1
⇒ 2a + b = 9
⇒ -5a = -10
⇒ a = 2
On substituting the value of a in eq(1)
⇒ -3(2) + b = -1
⇒ -6 + b = -1
⇒ b = 5
Therefore, r(x) = ax + b
= 2x + 5
Hence, x3 - 3x2 - 12x + 19 is divisible 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
According to 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 - 3)(x + 4)
Clearly, (x - 3) and (x + 4) are factors of q(x)
So, f(x) is divisible by q(x) if (x - 3) and (x + 4) are factors of q(x)
According to 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
⇒ -64 - 96 - 4a + 60 + 80 - b = 0
⇒ 4a - b - 20 = 0
On subtracting eq (1) and eq (2), we get
4a - b - 20 = 0 -(2)
⇒ 7a - 28 = 0
⇒ a = 28/7
⇒ a = 4
On Putting a = 4 in eq (1), we get
⇒ -3(4) - b = -8
⇒ -b - 12 = -8
⇒ -b = -8 + 12
⇒ b = -4
On substituting 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
According to divisible theorem, when p(x) is divided by q(x), the reminder is 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 divisible 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(\frac23)=3(\frac23)^3+(\frac23)^2+\frac23(a-22)+b+9 =
3(\frac8{27})+\frac49+\frac23a-\frac{44}3+b+9 =
\frac{12}9+\frac23a-\frac{44}3+b+9 =
\frac{12+6a-132+9b+81}{9} Equate to zero
⇒
\frac{12+6a-132+9b+81}{9}=0 ⇒ 6a + 9b - 39 = 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 the given equation by 3
⇒ -9a + 3b + 9 = 0 -(2)
On Subtracting eq (1) from eq(2)
⇒ -9a + 3b + 9 - 2a - 3b + 13 = 0
⇒-11a + 22 = 0
⇒-11a = -22
⇒ a = 2
On substituting value of a in eq (1)
⇒ -3(2) + b = -3
⇒ -6 + b = -3
⇒ 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:
(i) x3 - 2ax2 + ax - 1
(ii) x5 - 3x4 - ax3 + 3ax2 + 2ax + 4
Solution:
(i) Let f(x) = x3 - 2ax2 + ax - 1
According to factor theorem
If (x-2) is a factor of f(x) then f(2) = 0
Let, x - 2 = 0
⇒ x = 2
On substituting the value of x in f(x), we get
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 = 7/6
So, (x - 2) is the factor of f(x)
(ii) Let f(x) = x5 - 3x4 - ax3 + 3ax2 + 2ax + 4
According to factor theorem
If (x - 2) is a factor of f(x) then f(2) = 0
Let, x - 2 = 0
⇒ x = 2
On substituting the value of x in f(x), we get
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
⇒ a = 12/8
⇒ a = 3/2
So, (x - 2) is a factor of f(x)
Question 24. In each of the following two polynomials, find the value of a, if (x - a) is a factor:
(i) x6 - ax5 + x4 - ax3 + 3x - a + 2
(ii) x5 - a2x3 + 2x + a + 1
Solution:
(i) Let, f(x) = x6 - ax5 + x4 - ax3 + 3x - a + 2
Here, x - a = 0
⇒ x = a
On substituting the value of x in f(x), we get
f(a) = a6 - a(a)5 + (a)4 - a(a)3 + 3(a) - a + 2
= a6 - a6 + a4 - a4 + 3a - a + 2
= 2a+2
Equate to zero
⇒ 2a + 2 = 0
⇒ 2(a + 1) = 0
⇒ a = -1
So, (x - a) is a factor of f(x)
(ii) Let, f(x) = x5 - a2x3 + 2x + a + 1
Here, x - a = 0
⇒ x = a
On substituting the value of x in f(x), we get
f(a) = a5 - a2(a)3 + 2(a) + a + 1
= a5 - a5 + 2a + a + 1
= 3a + 1
Equate to zero
⇒ 3a + 1 = 0
⇒ 3a = -1
⇒ a = -1/3
So, (x - a) is a factor of f(x)
Question 25. In each of the following two polynomials, find the value of a, if (x + a) is a factor:
(i) x3 + ax2 - 2x + a +4
(ii) x4 - a2x2 + 3x - a
Solution:
(i) Let, f(x) = x3 + ax2 - 2x + a + 4
Here, x + a = 0
⇒ x = - a
On substituting the value of x in f(x), we get
f(-a) = (-a)3+ a(-a)2- 2(-a) + a + 4
= 3a + 4
Equate to zero
⇒ 3a + 4 = 0
⇒ 3a = -4
⇒ a = -4/3
So, (x + a) is a factor of f(x)
(ii) Let, f(x) = x4- a2x2 + 3x - a
Here, x + a = 0
⇒ x = -a
On substituting the value of x in f(x), we get
f(-a) = (-a)4 - a2(-a)2 + 3(-a) - a
= a4 - a4 - 3a - a
= -4a
Equate to zero
⇒ -4a = 0
⇒ a = 0
So, (x + a) is a factor of f(x)
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Conclusion
The Factorisation of polynomials is a key concept in algebra that simplifies the complex polynomial expressions and solves equations. The solutions provided for the Exercise 6.4 | Set 2 demonstrate the various factorisation techniques including the splitting the middle term factoring by grouping and using the algebraic identities. Mastery of these methods is essential for the advancing in algebra and solving more complex mathematical problems.