# RD Sharma Solutions Class 9 Chapter 6 Factorization of Polynomials

Read RD Sharma Solutions Class 9 Chapter 6 Factorization of Polynomials below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

Exercise 6.1

Question 1: Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:
(i) 3x2 – 4x + 15
(ii) y2 + 2√3
(iii) 3√x + √2 x
(iv) x – 4/x
(v) x12 + y3 + t50
Solution:
(i) 3x2 – 4x + 15 is an expression having only non-negative integral powers of x. So it is a polynomial.

(ii) y2 + 2√3 is an expression having only non-negative integral power of y. So it is a polynomial.

(iii) 3√x + √2 x is an expression in which one term, namely 3√x we can write it as 3x(1/2) has, rational power of x. So, it is not a polynomial.

(iv) x – 4/x is an expression in which one term, namely – 4/x we can write it as 4x-1 has, which is not a positive term. So, it is not a polynomial.

(v) x12 + y3 + t50 is an expression having only non-negative integral powers of its variables which is x, y and t. So it is a polynomial.

Question 2: Write the coefficient of x2 in each of the following:
(i) 17 – 2x + 7x2
(ii) 9 – 12x + x3
(iii) π/6 x2 – 3x + 4
(iv) √3 x – 7
Solution:
(i) 17 – 2x + 7x2 the coefficient of x2 is 7

(ii) 9 – 12x + x3 the coefficient of x2 is 0

(iii) π/6x2 – 3x + 4 the coefficient of x2 is π/6

(iv) √3 x – 7 the coefficient of x2 is

Question 3: Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y – √2
(iv) 7
(v) 0
Solution:
As we know, Degree is the highest power in the polynomial
(i) 7x3 + 4x2 – 3x + 12, Degree is the highest power in the polynomial. So the Degree of the given polynomial is 3.

(ii) 12 – x + 2x3 Degree is the highest power in the polynomial. So the Degree of the given polynomial is 3.

(iii) 5y – √2 Degree is the highest power in the polynomial. So the Degree of the given polynomial is 1.

(iv) 7 Degree is the highest power in the polynomial. So the Degree of the given polynomial is 0

(v) 0 Degree is the highest power in the polynomial. So, the Degree of the polynomial is undefined

Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:
(i) x + x2 + 4
(ii) 3x – 2
(iii) 2x + x2
(iv) 3y
(v) t2 + 1
(vi) 7t4 + 4t3 + 3t – 2
Solution:
(i) x + x2 + 4
The degree of the given polynomial is 2. Hence, it is a quadratic polynomial.

(ii) 3x – 2
The degree of the given polynomial is 1. Hence, it is a linear polynomial.

(iii) 2x + x2
The degree of the given polynomial is 2. Hence, it is a quadratic polynomial.

(iv) 3y
The degree of the given polynomial is 1. Hence, it is a linear polynomial.

(v) t2+ 1
The degree of the given polynomial is 2. Hence, it is a quadratic polynomial.

(vi) 7t4 + 4t3 + 3t – 2
The degree of the given polynomial is 4. Hence, it is a biquadratic polynomial.

Exercise 6.2
Question 1: If f(x)=2x3  – 13x+17x+ 12, find
(i) f(2)
(ii) f(-3)
(iii) f(0)
Solution:
It is given that,
f(x)=2x3  – 13x3+17x+ 12

(i) f(2)
f(2)= 2(2)3 – 13(2)2 + 17(2) + 12
f(2)= 2 x 8 – 13 x 4 + 17 x 2 + 12
f(2)= 16 – 52 + 34 + 12
f(2)= 62 – 52
f(2)= 10

(ii) f(-3)
f(-3)= 2(-3)3 – 13(-3)2 + 17 x (-3) + 12
f(-3)=2×(-27)–13×9+17×(-3)+12
f(-3)=-54 –117-51+12
f(-3)=-222+12
f(-3)=-210

(iii) f(0)
f(0)= 2 x (0)3 – 13(0)2 + 17 x 0 + 12
f(0)=0 - 0 + 0 + 12
f(0)=12

Question 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) = x2 – 1, x = 1,−1
(iii) g(x) = 3x2 – 2 , x = 2/√3 , -2/√3
(iv) p(x) = x3 – 6x2 + 11x – 6 , x = 1, 2, 3
(v) f(x) = 5x – π, x = 4/5
(vi) f(x) = x2 , x = 0
(vii) f(x) = lx + m, x = −m/l
(viii) f(x) = 2x + 1, x = 1/2
Solution:
(i) It is given that,
f(x) = 3x + 1
and x = −1/3
f(x) = 3x + 1 ………………….(1)
Put the value of x = − 1/3 in equation (1)
f(− 1/3) = 3 (− 1/3) + 1
f(− 1/3) = -1 + 1
f(− 1/3) = 0
The result is 0, Hence, x = -1/3 is the root of 3x + 1.

(ii) It is given that,
f(x) = x2 – 1
and x = 1,−1
f(x) = x2 – 1 ………………….(1)
Put the value of x = 1 in equation (1)
f(1) = (1)2 – 1
f(1) = 1 – 1
f(1) = 0
Now, put the value of x as -1 in equation (1)
f(-1) = (−1)2 – 1
f(-1) = 1 – 1
f(-1) = 0
The result is 0 with both the values, Hence, x = 1,-1 is the root of x2 – 1.

(iii) It is given that,
g(x) = 3x2 – 2
and x = 2/√3 , −2/√3
g(x) = 3x2 – 2………………….(1)
Put the value of x = 2/(√3) in equation (1)
g(2/(√3)) = 3(2/(√3))2 – 2
g(2/(√3)) = 3(4/3) – 2
g(2/(√3))  = 4 – 2
g(2/(√3)) = 2
Now, Put the value of x = -2/(√3) in equation (1)
g(-2/(√3)) = 3(-2/(√3))2–2
g(-2/(√3))  = 3(4/3)–2
g(-2/(√3))  = 4 – 2
g(-2/(√3))  = 2
The result is not 0 with both the values, Hence, x = (2/(√3)) , (-2/(√3)) is the not root of 3x2 – 2.

(iv) It is given that,
p(x) = x3 – 6x2 + 11x – 6
and x = 1, 2, 3
p(1) = 13 – 6(1)2 + 11x 1 – 6
p(1) = 1 – 6 + 11 – 6
p(1) = 0

p(2) = 23 – 6(2)2 + 11×2 – 6
p(2) = 8 – 24 + 22 – 6
p(2) = 0

p(3) = 33 – 6(3)2 + 11×3 – 6
p(3) = 27 – 54 + 33 – 6
p(3) = 0
The result is 0 with all the given values, Hence, x = 1,2,3 is the root of x3 – 6x2 + 11x – 6.

(v) It is given that,
f(x) = 5x – π
and x =4/5
f(4/5) = 5 x 4/5 – π
f(4/5) = 4 – π
The result is not 0 with the given values, Hence, x = 4/5 is not the root of 5x – π.

(vi) We have f(x) = x2, x = 0
f(0) = 0
f(0) = 0
The result is 0 with the given values, Hence, x = 0 is the root of x2.

(vii) It is given that,
f(x) = lx + m……………………..(1)
and x = -m/l
Put the value of x in equation (1)
f(−m/l) = l x −m/l + m
f(−m/l) = -m + m
f(−m/l) = 0
The result is 0 with the given values, Hence, x = 0 is the root of lx + m.

(viii) It is given that,
f(x) = 2x + 1 ……………………..(1)
and x = 1/2
Put the value of x in equation (1)
f(1/2) = 2×1/2 + 1
f(1/2) = 1 + 1
f(1/2) = 2
The result is not 0 with the given value, Hence, x = 1/2 is not the root of 2x + 1.

Exercise 6.3

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) = x3 + 4x2 – 3x + 10, g(x) = x + 4
Solution:
It is given that,
f(x) = x3 + 4x2 – 3x + 10
g(x) = x + 4
Let, g(x) =0
x + 4 = 0
x = - 4
Put the value of x = - 4 in f(x)
f(-4) = (-4)3 + 4(-4)2 – 3(-4) + 10
f(-4) = -64 + 64 + 12 + 10
f(-4) = 22
Alternate Method:- (Division Method)

Question 2: f(x) = 4x4 – 3x3 – 2x2 + x – 7, g(x) = x – 1
Solution:
It is given that,
f(x) = 4x4–3x3–2x2+x–7
g(x) = x – 1
Let, g(x) = 0
x – 1 = 0
x = 1
Put the value of x = 1 in f(x)
f(1) = 4(1)4 – 3(1)3 – 2(1)2 + (1) – 7
f(1) = 4 – 3 – 2 + 1 – 7
f(1) = 4 – 5 + 1 – 7
f(1) = 5 – 5 – 7
f(1) = – 7
Alternate Method:- (Division Method)

Question 3: f(x) = 2x4 – 6x3 + 2x2 – x + 2, g(x) = x + 2
Solution:
It is given that,
f(x) = 2x4 – 6x3 + 2x2 – x + 2
g(x) = x + 2
Let g(x) = 0
x + 2 = 0
x = -2
Put the value of x = -2 in f(x)
f(-2) = 2(-2)4 – 6(-2)3 + 2(-2)2 – (-2) + 2
f(-2) = 2(16) – 6(-8) + 2(4) – (-2) + 2
f(-2) = 32 - (-48) + 8 + 2 + 2
f(-2) = 32 + 48 + 8 + 2 + 2
f(-2) = 92
Alternate Method:- (Division Method)

Question 4: f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
Solution:
It is given that,
f(x) = 4x3 – 12x2 + 14x – 3
g(x) = 2x – 1
Let g(x) = 0
2x - 1 = 0
x = 1/2
Put the value of x = 1/2 in f(x)
f(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2)–3
f(1/2) = 4(1/8) – 12(1/4) + 14(1/2)–3
f(1/2) = 1/2 – 3 + 7 – 3
f(1/2) = 1/2  + 7 – 6
f(1/2) = 1/2  + 1
f(1/2) = 3/2

Alternate Method:- (Division Method)

Question 5: f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x
Solution:
It is given that,
f(x) = x3 – 6x2 + 2x – 4
g(x) = 1 – 2x
Let g(x) = 0
1 – 2x = 0
x = 1/2
Put the value of x = 1/2 in f(x)
f(1/2) = (1/2)3 – 6(1/2)2 + 2(1/2) – 4
f(1/2) = 1/8 – 6(1/4) + 2(1/2) – 4
f(1/2) = 1/8 – 3/2 + 1 – 4
f(1/2)  = -35/8

Alternate Method:- (Division Method)

Question 6: f(x) = x4 – 3x2 + 4, g(x) = x – 2
Solution:
It is given that,
f(x) = x4 – 3x2 + 4
g(x) = x – 2
Let g(x) = 0
x – 2 = 0
x = 2
Put the value of x = 2 in f(x)
f(2) = (2)4 – 3(2)2 + 4
f(2) = 16 – 12 + 4
f(2) = 8

Alternate Method:- (Division Method)

Question 7:  f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Solution:
f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Let g(x) = 0
x – 2/3 = 0
x = 2/3
Put the value of x = 2/3 in f(x)
f(2/3) = 9(2/3)3 – 3(2/3)2+ (2/3) – 5
f(2/3) = 9(8/27) – 3(4/9) + (2/3) – 5
f(2/3)  = 8/3 – 4/3 + 2/3 – 5/1
f(2/3) = -3
Alternate Method:- (Division Method)

Exercise 6.4

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) = x3 – 6x2 + 11x – 6; g(x) = x – 3
Solution:
It is given that,
f(x) = x3 – 6x2 + 11x – 6
Let, g(x) = 0
x -3 = 0
x = 3
Put the value of x in given equation.
f(3) = (3)3 – 6(3)2 +11 x 3 – 6
f(3) = 27 – 54 + 33 – 6
f(3) = 60 – 60
f(3) = 0
Hence, g(x) is a factor of f(x)

Question 2: f(x) = 3x4 + 17x3 + 9x2 – 7x – 10; g(x) = x + 5
Solution:
It is given that,
f(x) = 3x4 + 17x3 + 9x2 – 7x – 10
Let, g(x) = 0
x + 5 = 0
x = -5
Put the value of x in given equation.
f(3) = 3(-5)4 + 17(-5)3 + 9(-5)2 – 7(-5) – 10
f(3) = 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10
f(3) = 1875 -2125 + 225 + 35 – 10
f(3) = 0
Hence, g(x) is a factor of f(x).

Question 3: f(x) = x5 + 3x4 – x3 – 3x2 + 5x + 15, g(x) = x + 3
Solution:
It is given that,
Let, g(x) = 0
x + 3 = 0
x = -3
Put the value of x in given equation.
f(-3) = (-3)5 + 3(-3)4 – (-3)3 – 3(-3)2 + 5(-3) + 15
f(-3) = -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15
f(-3) = -243 +243 + 27-27- 15 + 15
f(-3) = 0
Hence, g(x) is a factor of f(x).

Question 4: f(x) = x3 – 6x2 – 19x + 84, g(x) = x – 7
Solution:
It is given that,
f(x) = x3 – 6x2 – 19x + 84
Let, g(x) = 0,
x – 7 = 0
x = 7
Put the value of x in given equation.
f(7) = (7)3 – 6(7)2 – 19 x 7 + 84
f(7) = 343 – 294 – 133 + 84
f(7) = 343 + 84 – 294 – 133
f(7) = 0
Therefore, g(x) is a factor of f(x).

Question 5: f(x) = 3x3 + x2 – 20x + 12, g(x) = 3x – 2
Solution:
It is given that,
f(x) = 3x3 + x2 – 20x + 12
Let, g(x) = 0
3x – 2 = 0
x = 2/3
Put the value of x in given equation.
f(2/3) = 3(2/3)3 + (2/3)2 – 20(2/3) + 12
f(2/3)  = 3 x  8/27 +  4/9 –  40/3 + 12
f(2/3) =  8/9 +  4/9 –  40/3 + 12
f(2/3)  =  0/9
f(2/3)  = 0
Hence, g(x) is a factor of f(x).

Question 6: f(x) = 2x3 – 9x2 + x + 12, g(x) = 3 – 2x
Solution:
It is given that,
f(x) = 2x3 – 9x2 + x + 12
Let, g(x) = 0
3 – 2x = 0
x = 3/2
Put the value of x in given equation.
f(3/2) = 2(3/2)3 – 9(3/2)2 + (3/2) + 12
f(3/2) = 2(27/8) – 9(9/4) + (3/2) + 12
f(3/2) = 27/4 – 81/4 + 3/2 + 12
f(3/2) = 0/4
f(3/2) = 0
Hence, g(x) is a factor of f(x).

Question 7: f(x) = x3 – 6x2 + 11x – 6, g(x) = x2 – 3x + 2
Solution:
It is given that,
f(x) = x3 – 6x2 + 11x – 6
Let, g(x) = 0
x2 – 3x + 2 = 0
x2 – x – 2x + 2 = 0
x(x – 1) – 2(x – 1) = 0
(x – 1) (x – 2) = 0
x = 1
x = 2
Now,
f(1) = (1)3 – 6(1)2 + 11(1) – 6
f(1) = 1 – 6 + 11 – 6
f(1) = 12 - 12
f(1) = 0
and
f(2) = (2)3 – 6(2)2 + 11(2) – 6
f(2) = 8 – 24 + 22 – 6
f(2) = 30 – 30
f(2) = 0
f(1) = 0 and f(2) = 0
Hence, g(x) is factor of f(x).

Question 8: Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24.
Solution:
It is assumed that,
f(x) = x3 – 3x2 – 10x + 24
x – 2 = 0
x = 2,
x + 3 = 0
x = -3,
x – 4 = 0
x = 4
Put the value of x in given equation.
f(2) = (2)3 – 3(2)2 – 10 x 2 + 24
f(2) = 8 – 12 – 20 + 24
f(2) = 32 – 32
f(2) = 0
and
f(-3) = (-3)3 – 3(-3)2 – 10 (-3) + 24
f(-3) = -27 -27 + 30 + 24
f(-3) = -54 + 54
f(-3) = 0
and
f(4) = (4)3 – 3(4)2 – 10 x 4 + 24
f(4) = 64-48 -40 + 24
f(4) = 88 – 88
f(4) = 0
Therefore, (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 x3 – 6x2 – 19x + 84.
Solution:
Let f(x) = x3 – 6x2 – 19x + 84
x + 4 = 0
x = -4
and
x – 3 = 0
x = 3
and
x – 7 = 0
x = 7
Put the value of x in given equations.
f(-4) = (-4)3 – 6(-4)2 – 19(-4) + 84
f(-4) = -64 – 96 + 76 + 84
f(-4) = 160 – 160
f(-4) = 0
and
f(3) = (3)3 – 6(3)2 – 19 x 3 + 84
f(3) = 27 – 54 – 57 + 84
f(3) = 111 -111
f(3) = 0
and
f(7) = (7)3 – 6(7)2 – 19 x 7 + 84
f(7) = 343 – 294 – 133 + 84
f(7) = 427 – 427
f(7) = 0
Thus, (x + 4), (x – 3), (x – 7) are the factors of f(x).

Exercise 6.5
Using factor theorem, factorize each of the following polynomials:

Question 1: x3 + 6x2 + 11x + 6
Solution:
It is given that,
f(x) = x3 + 6x2 + 11x + 6
The Factors of Constant term 6 are ±1,±2,±3,±6
It is assumed that
x + 1 = 0
x = -1
Put the value of x in given equation.
f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6
f(-1) = -1 + 6 -11 + 6
f(-1) = 12 – 12
f(-1) = 0
Hence, (x + 1) is the factor of given equation.
It is assumed that,
x + 2 = 0
x = -2
Put the value of x in given equation.
f(-2) = (−2)3 + 6(−2)2 + 11(−2) + 6
f(-2) = -8 + 24 – 22 + 6
f(-2) = 0
Hence, (x + 2) is the factor of given equation
It is assumed that,
x + 3 = 0
x = -3
Put the value of x in given equation.
f(-3) = (−3)3 + 6(−3)2 + 11(−3) + 6
f(-3) = -27 + 54 – 33 + 6
f(-3) = 0
Hence, (x + 3) is the factor of given equation

Question 2: x3 + 2x2 – x – 2
Solution:
It is given that,
f(x) = x3 + 2x2 – x – 2
The Factors of constant term -2 are ±1,±2
It is assumed that,
x – 1 = 0
x = 1
Put the value of x in given equation.
f(1) = (1)3 + 2(1)2 – 1 – 2
f(1) = 1 + 2 – 1 – 2
f(1) = 0
Hence, (x – 1) is factor of given equation.
It is assumed that,
x + 1 = 0
x = -1
Put the value of x in given equation.
f(-1) = (-1)3 + 2(-1)2 – 1 – 2
f(-1) = -1 + 2 + 1 – 2
f(-1) = 0
Hence, (x + 1) is a factor of given equation.
It is assumed that,
x + 2 = 0
x = -2
Put the value of x in given equation.
f(-2) = (-2)3 + 2(-2)2 – (-2) – 2
f(-2) = -8 + 8 + 2 – 2
f(-2) = 0
Hence, (x + 2) is a factor of given equation.
It is assumed that,
x – 2 = 0
x = 2
Put the value of x in given equation.
f(2) = (2)3 + 2(2)2 – 2 – 2
f(2) = 8 + 8 – 2 – 2
f(2) = 12 ≠ 0
Hence, (x – 2) is not a factor of given equation.

Question 3: x3 – 6x2 + 3x + 10
Solution:
It is given that,
f(x) = x3 – 6x2 + 3x + 10
The Factors of Constant term 10 are±1, ±2, ±5, ±10

Let
x + 1 = 0
x = -1
f(-1) = (-1)3 – 6(-1)2 + 3(-1) + 10
f(-1) = 10 – 10 = 0
f(-1) = 0

Let
x + 2 = 0
x = -2
f(-2) = (-2)3 – 6(-2)2 + 3(-2) + 10
f(-2) = -8 – 24 – 6 + 10
f(-2) = -28
f(-2) ≠ 0

Let
x – 2 = 0
x = 2
f(2) = (2)3 – 6(2)2 + 3(2) + 10
f(2) = 8 – 24 + 6 + 10
f(2) = 0
x – 5 = 0
x = 5
f(5) = (5)3 – 6(5)2 + 3(5) + 10
f(5) = 125 – 150 + 15 + 10
f(5) = 0
Hence, (x + 1), (x – 2) and (x - 5) are factors of f(x)

Question 4: x4 – 7x3 + 9x2 + 7x- 10
Solution:
It is given that,
f(x) = x4 – 7x3 + 9x2 + 7x- 10
The Factors of Constant term -10 are ±1, ±2, ±5, ±10

Let
x – 1 = 0
x = 1
f(1) = (1)4 – 7(1)3 + 9(1)2 + 7(1) – 10
f(1) = 1 – 7 + 9 + 7 -10
f(1) = 0

Let
x + 1 = 0
x = -1
f(-1) = (-1)4 – 7(-1)3 + 9(-1)2 + 7(-1) – 10
f(-1) = 1 + 7 + 9 – 7 -10
f(-1) = 0

Let
x – 2 = 0
x = 2
f(2) = (2)4 – 7(2)3 + 9(2)2 + 7(2) – 10
f(2) = 16 – 56 + 36 + 14 – 10
f(2) = 0

Let
x – 5 = 0
x = 5
f(5) = (5)4 – 7(5)3 + 9(5)2 + 7(5) – 10
f(5) = 625 – 875 + 225 + 35 – 10
f(5) = 0
Hence, (x – 1), (x + 1), (x – 2) and (x – 5) are factors of f(x)

Question 5:   x4 – 2x3 – 7x2 + 8x + 12
Solution:
It is given that,
f(x) = x4 – 2x3 – 7x2 + 8x + 12
The Factors of Constant term 12 are ±1, ±2, ±3, ±4, ±6, ±12

Let
x – 1 = 0
x = 1
f(1) = (1)4 – 2(1)3 – 7(1)2 + 8(1) + 12
f(1) = 1 – 2 – 7 + 8 + 12
f(1) = 12
f(1) ≠ 0
(x – 1) is not a factor of given equation.

Let
x + 1 = 0
x = -1
f(-1) = (-1)4 – 2(-1)3 – 7(-1)2 + 8(-1) + 12
f(-1) = 1 + 2 – 7 – 8 + 12
f(-1) = 0

Let
x +2 = 0
x = -2
f(-2) = (-2)4 – 2(-2)3 – 7(-2)2 + 8(-2) + 12
f(-2) = 16 + 16 – 28 – 16 + 12
f(-2) = 0

Let
x – 2 = 0
x = 2
f(2) = (2)4 – 2(2)3 – 7(2)2 + 8(2) + 12
f(2) = 16 – 16 – 28 + 16 + 12
f(2) = 0

Let
x – 3 = 0
x = 3
f(3) = (3)4 – 2(3)3 – 7(3)2 + 8(3) + 12
f(3) = 81 - 2×27 - 7×9 + 24 +12
f(3) = 81 - 54 - 63 + 24 +12
f(3) = 0
Hence, (x + 1), (x + 2), (x – 2) and (x-3) are factors of f(x)

Question 6: x4 + 10x3 + 35x2 + 50x + 24
Solution:
It is given that,
f(x) = x4 + 10x3 + 35x2 + 50x + 24
The Factors of Constant term 24 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

Let
x + 1 = 0
x = -1
f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24
f(-1) = 1 – 10 + 35 – 50 + 24
f(-1) = 0
(x + 1) is a factor of given equation.

Let
x + 2 = 0
x = -2
f(-2) = (-2)4 + 10(-2)3 + 35(-2)2 + 50(-2) + 24
f(-2) = 16 + 10(-8) + 35(4) + 50(-2) + 24
f(-2) = 16 - 80 + 140 - 100 + 24
f(-2) = 0
(x + 2) is a factor of given equation.

Let
x + 3 = 0
x = -3
f(-3) = (-3)4 + 10(-3)3 + 35(-3)2 + 50(-3) + 24
f(-3) = 81 + 10(-27) + 35(9) + 50(-3) + 24
f(-3) = 81 - 270 + 315 - 150 + 24
f(-3) = 0
(x + 3) is a factor of given equation.

Let
x + 4 = 0
x = -4
f(-4) = (-4)4 + 10(-4)3 + 35(-4)2 + 50(-4) + 24
f(-4) = 256 + 10(-64) + 35(16) + 50(-4) + 24
f(-4) = 256 - 640 + 560 - 200 + 24
f(-4) = 0
(x + 4) is a factor of given equation.
Hence (x + 1) (x + 2) (x + 3) (x + 4) are the factor of x4 + 10x3 + 35x2 + 50x + 24.

Question 7: 2x4 – 7x3 – 13x2 + 63x – 45
Solution:
It is given that,
f(x) = 2x4 – 7x3 – 13x2 + 63x – 45
The Factors of Constant term -45 are ±1, ±3, ±5, ±9, ±15, ±45.
The coefficient of x4 is 2.
So possible rational roots of f(x) are ±1,±3,±5,±9,±15,±45,±1/2,±3/2,±5/4,±9/2,±15/2,±45/2

Let
x – 1 = 0
x = 1
f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45
f(1) = 2 – 7 – 13 + 63 – 45
f(1) = 0

Let
x – 3 = 0
x = 3
f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3) – 45
f(3) = 162 – 189 – 117 + 189 – 45
f(3) = 0
(x – 1) and (x – 3) are the factors of given equation.
(x – 1) (x – 3)
x2 - 3x - 1x + 3
x2 - 4x + 3
So, x2 - 4x + 3 is root of 2x4 – 7x3 – 13x2 + 63x – 45.
By Dividing

By splitting the middle term method we get,
2x2 + x – 15
2x2 + 6x – 5x – 15
2x(x – 3) + 5(x – 3)
(2x + 5) (x – 3)
Hence, the factors of 2x4 – 7x3 – 13x2 + 63x – 45 are (x – 1) (x – 3) (2x + 5) (x – 3).

Exercise VSAQs : ->

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) = 8x3 + ax2 – 4x + 2, find the value of a.
Solution:
It is given that,
f(x) = 8x3 + ax2 – 4x + 2
x = 1/2 is a zero of the polynomial f(x),
so f(1/2) = 0
Put the value of x = 1/2 in the given equation.
f(1/2)= 8(1/2)3+ a(1/2)2– 4(1/2) + 2 = 0
f(1/2)= 8× 1/8 + a/4 – 2 + 2 = 0
f(1/2)= 1 +  a/4 = 0
f(1/2)= (4 + a)/4 = 0
f(1/2)= 4+a=0
f(1/2)= a=-4

Question 3: Write the remainder when the polynomial f(x) = x3 + x2 – 3x + 2 is divided by x + 1.
Solution:
It is given that,
f(x) = x3 + x2 – 3x + 2
x + 1 = 0
x = -1
Put the value of x in given equation.
f(-1) = (-1)3 + (-1)2 – 3(-1) + 2
f(-1) = -1 + 1 + 3 + 2
f(-1) = 5
Hence, 5 is the remainder.

Question 4: Find the remainder when x3 + 4x2 + 4x - 3 if divided by x
Solution:
It is given that,
f(x) = x3 + 4x2 + 4x - 3
x = 0
f(0) is the remainder.
f(0) = x3 + 4x2 + 4x - 3
f(0) = 03 + 4(0)2 + 4(0) -3
f(0) = -3
Hence, the remainder is -3.

Question 5: If x + 1 is a factor of x3 + a, then write the value of a.
Solution:
It is given that,
f(x) = x3 + a
Let
x + 1 = 0
x = -1
Put the value of value of x in given equation.
f(-1) = x3 + a
(-1)3 + a = 0
-1 + a = 0
a = 1

Question 6: If f(x) = x4 – 2x3 + 3x2 – ax – b when divided by x – 1, the remainder is 6, then find the value of a + b.
Solution:
It is given that,
f(x) = x4 – 2x3 + 3x2 – ax – b
The remainder is 6
Here,
x -1 = 0
x = 1
(1)4 – 2(1)3 + 3(1)2 – a(1) – b = 6
1 – 2 + 3 – a – b = 6
2 – a – b = 6
– a – b = 6 - 2
– (a + b) = 4
a + b = -4
Hence, the value of a + b is -4.