RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials
Ananya Gupta6 Apr, 2024
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RD Sharma Solutions Class 9 Maths Chapter 6: In RD Sharma Class 9 Maths Chapter 6, students learn about factorising polynomials. This means breaking down complicated polynomial expressions into simpler parts. The chapter teaches different methods to do this effectively.
By using the solutions provided, students can understand how to factorise polynomials step by step. Practicing with these solutions helps students improve their understanding of polynomial factorisation and become better at solving related problems.CBSE Class 9 Science Syllabus 2024-25
RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials PDF
You can access the PDF for RD Sharma Class 9 Solutions Maths Chapter 6 on Factorisation of Polynomials by clicking the link provided below. This PDF contains comprehensive solutions to all the exercises in the chapter, helping students understand the concept of polynomial factorisation better and improve their problem-solving skills. Accessing the PDF is convenient and allows students to study the solutions at their own pace, reinforcing their learning and preparation for exams.RD Sharma Solutions Class 9 Maths Chapter 6 PDF
RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials
RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials Exercise 6.1 Page No: 6.2
Question 1: Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer:
(i) 3x 2 – 4x + 15
(ii) y 2 + 2√3
(iii) 3√x + √2x
(iv) x – 4/x
(v) x 12 + y 3 + t 50
Solution:
(i) 3x 2 – 4x + 15
It is a polynomial of x.(ii) y 2 + 2√3
It is a polynomial of y.(iii) 3√x + √2x
It is not a polynomial since the exponent of 3√x is a rational term.(iv) x – 4/x
It is not a polynomial since the exponent of – 4/x is not a positive term.(v) x 12 + y 3 + t 50
It is a three-variable polynomial, x, y and t.Question 2: Write the coefficient of x 2 in each of the following:
(i) 17 – 2x + 7x 2
(ii) 9 – 12x + x 3
(iii) ∏/6 x 2 – 3x + 4
(iv) √3x – 7
Solution:
(i) 17 – 2x + 7x 2
Coefficient of x 2 = 7(ii) 9 – 12x + x 3
Coefficient of x 2 =0(iii) ∏/6 x 2 – 3x + 4
Coefficient of x 2 = ∏/6(iv) √3x – 7
Coefficient of x 2 = 0Question 3: Write the degrees of each of the following polynomials:
(i) 7x 3 + 4x 2 – 3x + 12
(ii) 12 – x + 2x 3
(iii) 5y – √2
(iv) 7
(v) 0
Solution :
As we know, degree is the highest power in the polynomial(i) Degree of the polynomial 7x 3 + 4x 2 – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x 3 is 3
(iii) Degree of the polynomial 5y – √2 is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is undefined.
CBSE Class 10 Result 2024 Expected To Be Out Soon
Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:
(i) x + x 2 + 4
(ii) 3x – 2
(iii) 2x + x 2
(iv) 3y
(v) t 2 + 1
(vi) 7t 4 + 4t 3 + 3t – 2
Solution:
(i) x + x 2 + 4: It is a quadratic polynomial as its degree is 2 .
(ii) 3x – 2 : It is a linear polynomial as its degree is 1 .
(iii) 2x + x 2 : It is a quadratic polynomial as its degree is 2 .
(iv) 3y: It is a linear polynomial as its degree is 1 .
(v) t 2 + 1: It is a quadratic polynomial as its degree is 2 .
(vi) 7t 4 + 4t 3 + 3t – 2: It is a biquadratic polynomial as its degree is 4 .
RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials Exercise 6.2 Page No: 6.8
Question 1: If f(x) = 2x 3 – 13x 2 + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x 3 – 13x 2 + 17x + 12(i) f(2) = 2(2) 3 – 13(2) 2 + 17(2) + 12
= 2 x 8 – 13 x 4 + 17 x 2 + 12 = 16 – 52 + 34 + 12 = 62 – 52 = 10(ii) f(-3) = 2(-3) 3 – 13(-3) 2 + 17 x (-3) + 12
= 2 x (-27) – 13 x 9 + 17 x (-3) + 12 = -54 – 117 -51 + 12 = -222 + 12 = -210(iii) f(0) = 2 x (0) 3 – 13(0) 2 + 17 x 0 + 12
= 0-0 + 0+ 12 = 12Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:
(i) f(x) = 3x + 1, x = −1/3
(ii) f(x) = x 2 – 1, x = 1,−1
(iii) g(x) = 3x 2 – 2 , x = 2/√3 , −2/√3
(iv) p(x) = x 3 – 6x 2 + 11x – 6 , x = 1, 2, 3
(v) f(x) = 5x – π, x = 4/5
(vi) f(x) = x 2 , x = 0
(vii) f(x) = lx + m, x = −m/l
(viii) f(x) = 2x + 1, x = 1/2
Solution:
(i) f(x) = 3x + 1, x = −1/3
f(x) = 3x + 1 Substitute x = −1/3 in f(x) f( −1/3) = 3(−1/3) + 1 = -1 + 1 = 0 Since, the result is 0, so x = −1/3 is the root of 3x + 1(ii) f(x) = x 2 – 1, x = 1,−1
f(x) = x 2 – 1 Given that x = (1 , -1) Substitute x = 1 in f(x) f(1) = 1 2 – 1 = 1 – 1 = 0 Now, substitute x = (-1) in f(x) f(-1) = (−1) 2 – 1 = 1 – 1 = 0 Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x 2 – 1(iii) g(x) = 3x 2 – 2 , x = 2/√3 , −2/√3
g(x) = 3x 2 – 2 Substitute x = 2/√3 in g(x) g(2/√3) = 3(2/√3) 2 – 2 = 3(4/3) – 2 = 4 – 2 = 2 ≠ 0 Now, Substitute x = −2/√3 in g(x) g(2/√3) = 3(-2/√3) 2 – 2 = 3(4/3) – 2 = 4 – 2 = 2 ≠ 0 The results when x = 2/√3 and x = −2/√3) are not 0. Therefore, (2/√3 , −2/√3 ) are not zeros of 3x 2 –2.(iv) p(x) = x 3 – 6x 2 + 11x – 6 , x = 1, 2, 3
p(1) = 1 3 – 6(1) 2 + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0 p(2) = 2 3 – 6(2) 2 + 11×2 – 6 = 8 – 24 + 22 – 6 = 0 p(3) = 3 3 – 6(3) 2 + 11×3 – 6 = 27 – 54 + 33 – 6 = 0 Therefore, x = 1, 2, 3 are zeros of p(x).(v) f(x) = 5x – π, x = 4/5
f(4/5) = 5 x 4/5 – π = 4 – π ≠ 0 Therefore, x = 4/5 is not a zeros of f(x).(vi) f(x) = x 2 , x = 0
f(0) = 0 2 = 0 Therefore, x = 0 is a zero of f(x).(vii) f(x) = lx + m, x = −m/l
f(−m/l) = l x −m/l + m = -m + m = 0 Therefore, x = −m/l is a zero of f(x).(viii) f(x) = 2x + 1, x = ½
f(1/2) = 2x 1/2 + 1 = 1 + 1 = 2 ≠ 0 Therefore, x = ½ is not a zero of f(x).RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials Exercise 6.3 Page No: 6.14
In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 – 8)
Question 1: f(x) = x 3 + 4x 2 – 3x + 10, g(x) = x + 4
Solution:
f(x) = x 3 + 4x 2 – 3x + 10, g(x) = x + 4 Put g(x) =0 ⇒ x + 4 = 0 or x = -4 Remainder = f(-4) Now, f(-4) = (-4) 3 + 4(-4) 2 – 3(-4) + 10 = -64 + 64 + 12 + 10 = 22Actual Division:
Question 2: f(x) = 4x 4 – 3x 3 – 2x 2 + x – 7, g(x) = x – 1
Solution:
f(x) = 4x 4 – 3x 3 – 2x 2 + x – 7 Put g(x) =0 ⇒ x – 1 = 0 or x = 1 Remainder = f(1) Now, f(1) = 4(1) 4 – 3(1) 3 – 2(1) 2 + (1) – 7 = 4 – 3 – 2 + 1 – 7 = -7Actual Division:
Question 3: f(x) = 2x 4 – 6X 3 + 2x 2 – x + 2, g(x) = x + 2
Solution:
f(x) = 2x 4 – 6X 3 + 2x 2 – x + 2, g(x) = x + 2 Put g(x) = 0 ⇒ x + 2 = 0 or x = -2 Remainder = f(-2) Now, f(-2) = 2(-2) 4 – 6(-2) 3 + 2(-2) 2 – (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92Actual Division:
Question 4: f(x) = 4x 3 – 12x 2 + 14x – 3, g(x) = 2x – 1
Solution:
f(x) = 4x 3 – 12x 2 + 14x – 3, g(x) = 2x – 1 Put g(x) =0 ⇒ 2x -1 =0 or x = 1/2 Remainder = f(1/2) Now, f(1/2) = 4(1/2) 3 – 12(1/2) 2 + 14(1/2) – 3 = ½ – 3 + 7 – 3 = 3/2Actual Division:
Question 5: f(x) = x 3 – 6x 2 + 2x – 4, g(x) = 1 – 2x
Solution:
f(x) = x 3 – 6x 2 + 2x – 4, g(x) = 1 – 2x Put g(x) = 0 ⇒ 1 – 2x = 0 or x = 1/2 Remainder = f(1/2) Now, f(1/2) = (1/2) 3 – 6(1/2) 2 + 2(1/2) – 4 = 1 + 1/8 – 4 – 3/2 = -35/8Actual Division:
Question 6: f(x) = x 4 – 3x 2 + 4, g(x) = x – 2
Solution:
f(x) = x 4 – 3x 2 + 4, g(x) = x – 2 Put g(x) = 0 ⇒ x – 2 = 0 or x = 2 Remainder = f(2) Now, f(2) = (2) 4 – 3(2) 2 + 4 = 16 – 12 + 4 = 8Actual Division:
Question 7: f(x) = 9x 3 – 3x 2 + x – 5, g(x) = x – 2/3
Solution:
f(x) = 9x 3 – 3x 2 + x – 5, g(x) = x – 2/3 Put g(x) = 0 ⇒ x – 2/3 = 0 or x = 2/3 Remainder = f(2/3) Now, f(2/3) = 9(2/3) 3 – 3(2/3) 2 + (2/3) – 5 = 8/3 – 4/3 + 2/3 – 5/1 = -3Actual Division:
RD Sharma Class 9 Solutions Maths Chapter 6 Factorisation of Polynomials Exercise 6.4 Page No: 6.24
In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)
Question 1: f(x) = x 3 – 6x 2 + 11x – 6; g(x) = x – 3Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = x -3 = 0 or x = 3 Remainder = f(3) Now, f(3) = (3) 3 – 6(3) 2 +11 x 3 – 6 = 27 – 54 + 33 – 6 = 60 – 60 = 0 Therefore, g(x) is a factor of f(x)Question 2: f(x) = 3X 4 + 17x 3 + 9x 2 – 7x – 10; g(x) = x + 5
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = x + 5 = 0, then x = -5 Remainder = f(-5) Now, f(3) = 3(-5) 4 + 17(-5) 3 + 9(-5) 2 – 7(-5) – 10 = 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10 = 1875 -2125 + 225 + 35 – 10 = 0 Therefore, g(x) is a factor of f(x).Question 3: f(x) = x 5 + 3x 4 – x 3 – 3x 2 + 5x + 15, g(x) = x + 3
Solution :
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = x + 3 = 0, then x = -3 Remainder = f(-3) Now, f(-3) = (-3) 5 + 3(-3) 4 – (-3) 3 – 3(-3) 2 + 5(-3) + 15 = -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15 = -243 +243 + 27-27- 15 + 15 = 0 Therefore, g(x) is a factor of f(x).Question 4: f(x) = x 3 – 6x 2 – 19x + 84, g(x) = x – 7
Solution :
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = x – 7 = 0, then x = 7 Remainder = f(7) Now, f(7) = (7) 3 – 6(7) 2 – 19 x 7 + 84 = 343 – 294 – 133 + 84 = 343 + 84 – 294 – 133 = 0 Therefore, g(x) is a factor of f(x).Question 5: f(x) = 3x 3 + x 2 – 20x + 12, g(x) = 3x – 2
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = 3x – 2 = 0, then x = 2/3 Remainder = f(2/3) Now, f(2/3) = 3(2/3) 3 + (2/3) 2 – 20(2/3) + 12 = 3 x 8/27 + 4/9 – 40/3 + 12 = 8/9 + 4/9 – 40/3 + 12 = 0/9 = 0 Therefore, g(x) is a factor of f(x).Question 6: f(x) = 2x 3 – 9x 2 + x + 12, g(x) = 3 – 2x
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = 3 – 2x = 0, then x = 3/2 Remainder = f(3/2) Now, f(3/2) = 2(3/2) 3 – 9(3/2) 2 + (3/2) + 12 = 2 x 27/8 – 9 x 9/4 + 3/2 + 12 = 27/4 – 81/4 + 3/2 + 12 = 0/4 = 0 Therefore, g(x) is a factor of f(x).Question 7: f(x) = x 3 – 6x 2 + 11x – 6, g(x) = x 2 – 3x + 2
Solution:
If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. g(x) = 0 or x 2 – 3x + 2 = 0 x 2 – x – 2x + 2 = 0 x(x – 1) – 2(x – 1) = 0 (x – 1) (x – 2) = 0 Therefore x = 1 or x = 2 Now, f(1) = (1) 3 – 6(1) 2 + 11(1) – 6 = 1-6+11-6= 12- 12 = 0 f(2) = (2) 3 – 6(2) 2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 30 – 30 = 0 ⇒ f(1) = 0 and f(2) = 0 Which implies g(x) is factor of f(x).Question 8: Show that (x – 2), (x + 3) and (x – 4) are factors of x 3 – 3x 2 – 10x + 24.
Solution:
Let f(x) = x 3 – 3x 2 – 10x + 24 If x – 2 = 0, then x = 2, If x + 3 = 0 then x = -3, and If x – 4 = 0 then x = 4 Now, f(2) = (2) 3 – 3(2) 2 – 10 x 2 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0 f(-3) = (-3) 3 – 3(-3) 2 – 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0 f(4) = (4) 3 – 3(4) 2 – 10 x 4 + 24 = 64-48 -40 + 24 = 88 – 88 = 0 f(2) = 0 f(-3) = 0 f(4) = 0 Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)Question 9: Show that (x + 4), (x – 3) and (x – 7) are factors of x 3 – 6x 2 – 19x + 84.
Solution:
Let f(x) = x 3 – 6x 2 – 19x + 84 If x + 4 = 0, then x = -4 If x – 3 = 0, then x = 3 and if x – 7 = 0, then x = 7 Now, f(-4) = (-4) 3 – 6(-4) 2 – 19(-4) + 84 = -64 – 96 + 76 + 84 = 160 – 160 = 0 f(-4) = 0 f(3) = (3) 3 – 6(3) 2 – 19 x 3 + 84 = 27 – 54 – 57 + 84 = 111 -111=0 f(3) = 0 f(7) = (7) 3 – 6(7) 2 – 19 x 7 + 84 = 343 – 294 – 133 + 84 = 427 – 427 = 0 f(7) = 0 Hence (x + 4), (x – 3), (x – 7) are the factors of f(x).RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials Exercise 6.5 Page No: 6.32
Using factor theorem, factorize each of the following polynomials:
Question 1: x 3 + 6x 2 + 11x + 6
Solution :
Let f(x) = x 3 + 6x 2 + 11x + 6 Step 1: Find the factors of constant term Here constant term = 6 Factors of 6 are ±1, ±2, ±3, ±6 Step 2: Find the factors of f(x) Let x + 1 = 0 ⇒ x = -1 Put the value of x in f(x) f(-1) = (−1) 3 + 6(−1) 2 + 11(−1) + 6 = -1 + 6 -11 + 6 = 12 – 12 = 0 So, (x + 1) is the factor of f(x) Let x + 2 = 0 ⇒ x = -2 Put the value of x in f(x) f(-2) = (−2) 3 + 6(−2) 2 + 11(−2) + 6 = -8 + 24 – 22 + 6 = 0 So, (x + 2) is the factor of f(x) Let x + 3 = 0 ⇒ x = -3 Put the value of x in f(x) f(-3) = (−3) 3 + 6(−3) 2 + 11(−3) + 6 = -27 + 54 – 33 + 6 = 0 So, (x + 3) is the factor of f(x) Hence, f(x) = (x + 1)(x + 2)(x + 3)Question 2: x 3 + 2x 2 – x – 2
Solution :
Let f(x) = x 3 + 2x 2 – x – 2 Constant term = -2 Factors of -2 are ±1, ±2 Let x – 1 = 0 ⇒ x = 1 Put the value of x in f(x) f(1) = (1) 3 + 2(1) 2 – 1 – 2 = 1 + 2 – 1 – 2 = 0 So, (x – 1) is factor of f(x) Let x + 1 = 0 ⇒ x = -1 Put the value of x in f(x) f(-1) = (-1) 3 + 2(-1) 2 – 1 – 2 = -1 + 2 + 1 – 2 = 0 (x + 1) is a factor of f(x) Let x + 2 = 0 ⇒ x = -2 Put the value of x in f(x) f(-2) = (-2) 3 + 2(-2) 2 – (-2) – 2 = -8 + 8 + 2 – 2 = 0 (x + 2) is a factor of f(x) Let x – 2 = 0 ⇒ x = 2 Put the value of x in f(x) f(2) = (2) 3 + 2(2) 2 – 2 – 2 = 8 + 8 – 2 – 2 = 12 ≠ 0 (x – 2) is not a factor of f(x) Hence f(x) = (x + 1)(x- 1)(x+2)Question 3: x 3 – 6x 2 + 3x + 10
Solution :
Let f(x) = x 3 – 6x 2 + 3x + 10 Constant term = 10 Factors of 10 are ±1, ±2, ±5, ±10 Let x + 1 = 0 or x = -1 f(-1) = (-1) 3 – 6(-1) 2 + 3(-1) + 10 = 10 – 10 = 0 f(-1) = 0 Let x + 2 = 0 or x = -2 f(-2) = (-2) 3 – 6(-2) 2 + 3(-2) + 10 = -8 – 24 – 6 + 10 = -28 f(-2) ≠ 0 Let x – 2 = 0 or x = 2 f(2) = (2) 3 – 6(2) 2 + 3(2) + 10 = 8 – 24 + 6 + 10 = 0 f(2) = 0 Let x – 5 = 0 or x = 5 f(5) = (5) 3 – 6(5) 2 + 3(5) + 10 = 125 – 150 + 15 + 10 = 0 f(5) = 0 Therefore, (x + 1), (x – 2) and (x-5) are factors of f(x) Hence f(x) = (x + 1) (x – 2) (x-5)Question 4: x 4 – 7x 3 + 9x 2 + 7x- 10
Solution:
Let f(x) = x 4 – 7x 3 + 9x 2 + 7x- 10 Constant term = -10 Factors of -10 are ±1, ±2, ±5, ±10 Let x – 1 = 0 or x = 1 f(1) = (1) 4 – 7(1) 3 + 9(1) 2 + 7(1) – 10 = 1 – 7 + 9 + 7 -10 = 0 f(1) = 0 Let x + 1 = 0 or x = -1 f(-1) = (-1) 4 – 7(-1) 3 + 9(-1) 2 + 7(-1) – 10 = 1 + 7 + 9 – 7 -10 = 0 f(-1) = 0 Let x – 2 = 0 or x = 2 f(2) = (2) 4 – 7(2) 3 + 9(2) 2 + 7(2) – 10 = 16 – 56 + 36 + 14 – 10 = 0 f(2) = 0 Let x – 5 = 0 or x = 5 f(5) = (5) 4 – 7(5) 3 + 9(5) 2 + 7(5) – 10 = 625 – 875 + 225 + 35 – 10 = 0 f(5) = 0 Therefore, (x – 1), (x + 1), (x – 2) and (x-5) are factors of f(x) Hence f(x) = (x – 1) (x + 1) (x – 2) (x-5)Question 5: x 4 – 2x 3 – 7x 2 + 8x + 12
Solution :
f(x) = x 4 – 2x 3 – 7x 2 + 8x + 12 Constant term = 12 Factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12 Let x – 1 = 0 or x = 1 f(1) = (1) 4 – 2(1) 3 – 7(1) 2 + 8(1) + 12 = 1 – 2 – 7 + 8 + 12 = 12 f(1) ≠ 0 Let x + 1 = 0 or x = -1 f(-1) = (-1) 4 – 2(-1) 3 – 7(-1) 2 + 8(-1) + 12 = 1 + 2 – 7 – 8 + 12 = 0 f(-1) = 0 Let x +2 = 0 or x = -2 f(-2) = (-2) 4 – 2(-2) 3 – 7(-2) 2 + 8(-2) + 12 = 16 + 16 – 28 – 16 + 12 = 0 f(-2) = 0 Let x – 2 = 0 or x = 2 f(2) = (2) 4 – 2(2) 3 – 7(2) 2 + 8(2) + 12 = 16 – 16 – 28 + 16 + 12 = 0 f(2) = 0 Let x – 3 = 0 or x = 3 f(3) = (3) 4 – 2(3) 3 – 7(3) 2 + 8(3) + 12 = 0 f(3) = 0 Therefore, (x + 1), (x + 2), (x – 2) and (x-3) are factors of f(x) Hence f(x) = (x + 1)(x + 2) (x – 2) (x-3)Question 6: x 4 + 10x 3 + 35x 2 + 50x + 24
Solution :
Let f(x) = x 4 + 10x 3 + 35x 2 + 50x + 24 Constant term = 24 Factors of 24 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 Let x + 1 = 0 or x = -1 f(-1) = (-1) 4 + 10(-1) 3 + 35(-1) 2 + 50(-1) + 24 = 1 – 10 + 35 – 50 + 24 = 0 f(1) = 0 (x + 1) is a factor of f(x) Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x) Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4)Question 7: 2x 4 – 7x 3 – 13x 2 + 63x – 45
Solution:
Let f(x) = 2x 4 – 7x 3 – 13x 2 + 63x – 45 Constant term = -45 Factors of -45 are ±1, ±3, ±5, ±9, ±15, ±45 Here coefficient of x^4 is 2. So possible rational roots of f(x) are ±1, ±3, ±5, ±9, ±15, ±45, ±1/2,±3/2,±5/2,±9/2,±15/2,±45/2 Let x – 1 = 0 or x = 1 f(1) = 2(1) 4 – 7(1) 3 – 13(1) 2 + 63(1) – 45 = 2 – 7 – 13 + 63 – 45 = 0 f(1) = 0 f(x) can be written as, f(x) = (x-1) (2x 3 – 5x 2 -18x +45) or f(x) =(x-1)g(x) …(1) Let x – 3 = 0 or x = 3 f(3) = 2(3) 4 – 7(3) 3 – 13(3) 2 + 63(3) – 45 = = 162 – 189 – 117 + 189 – 45= 0 f(3) = 0 Now, we are available with 2 factors of f(x), (x – 1) and (x – 3) Here g(x) = 2x 2 (x-3) + x(x-3) -15(x-3) Taking (x-3) as common = (x-3)(2x 2 + x – 15) = (x-3)(2x 2 +6x – 5x -15) = (x-3)(2x-5)(x+3) = (x-3)(x+3)(2x-5) ….(2) From (1) and (2) f(x) =(x-1) (x-3)(x+3)(2x-5)RD Sharma Solutions Class 9 Maths Chapter 6 Factorisation of Polynomials Exercise VSAQs Page No: 6.33
Question 1: Define zero or root of a polynomial
Solution :
zero or root, is a solution to the polynomial equation, f(y) = 0. It is that value of y that makes the polynomial equal to zero.Question 2: If x = 1/2 is a zero of the polynomial f(x) = 8x 3 + ax 2 – 4x + 2, find the value of a.
Solution:
If x = 1/2 is a zero of the polynomial f(x), then f(1/2) = 0 8(1/2) 3 + a(1/2) 2 – 4(1/2) + 2 = 0 8 x 1/8 + a/4 – 2 + 2 = 0 1 + a/4 = 0 a = -4Question 3: Write the remainder when the polynomial f(x) = x 3 + x 2 – 3x + 2 is divided by x + 1.
Solution:
Using factor theorem, Put x + 1 = 0 or x = -1 f(-1) is the remainder. Now, f(-1) = (-1) 3 + (-1) 2 – 3(-1) + 2 = -1 + 1 + 3 + 2 = 5 Therefore 5 is the remainder.Question 4: Find the remainder when x 3 + 4x 2 + 4x-3 if divided by x
Solution:
Using factor theorem, Put x = 0 f(0) is the remainder. Now, f(0) = 0 3 + 4(0) 2 + 4×0 -3 = -3 Therefore -3 is the remainder.Question 5: If x+1 is a factor of x 3 + a, then write the value of a.
Solution:
Let f(x) = x 3 + a If x+1 is a factor of x 3 + a then f(-1) = 0 (-1) 3 + a = 0 -1 + a = 0 or a = 1Question 6: If f(x) = x^4 – 2x 3 + 3x 2 – ax – b when divided by x – 1, the remainder is 6, then find the value of a+b.
Solution:
From the statement, we have f(1) = 6 (1)^4 – 2(1) 3 + 3(1) 2 – a(1) – b = 6 1 – 2 + 3 – a – b = 6 2 – a – b = 6 a + b = -4| CBSE Class 9 Maths Syllabus | CBSE Class 9 Science Syllabus |
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RD Sharma Solutions Class 9 Maths Chapter 6 FAQs
What is factorisation of polynomials?
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What are the different methods of factorising polynomials?
How can I determine the factors of a polynomial expression?
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