Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths

RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q14. Find the values of a and b, if x2 – 4 is a factor of ax4+2x3–3x2+bx–4

Sol :

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


Q15. Find α,β if (x + 1) and (x + 2) are the factors of x3+3x2−2αx+β

Sol:

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

From factor theorem, if they are tha 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


Q16. Find the values of p and q so that x4+px3+2x2−3x+q is divisible by (x2 – 1)

Sol :

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


Q17. Find the values of a and b so that (x + 1) and (x – 1) are the factors of x4+ax3–3x2+2x+b

Sol :

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


Q18. If x3+ax2–bx+10 is divisible by x3–3x+2, find the values of a and b

Sol :

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


Q19. If both (x + 1) and (x – 1) are the factors of ax3+x2−2x+b , Find the values of a and b

Sol:

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


Q20. What must be added to x3–3x2–12x+19 so that the result is exactly divisible by x2+x–6

Sol :

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 wiil 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


Q21. What must be added to x3–6x2–15x+80 so that the result is exactly divisible by x2+x–12

Sol :

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) wiil 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


Q22. What must be added to 3x3+x2–22x+9 so that the result is exactly divisible by 3x2+7x–6

Sol :

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

RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Equate to zero

⇒ RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ 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


Q23. 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

Sol :

(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 = 3/2


Q24. 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

Sol :

(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= −1/3

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


Q25. 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

Sol :

(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

The document RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the importance of factorization of polynomials in mathematics?
Ans. Factorization of polynomials is important in mathematics as it helps us simplify complex expressions and solve equations easily. It also helps in finding the roots of polynomials, which are essential in various mathematical applications such as engineering, physics, and economics.
2. How can factorization of polynomials be used to solve equations?
Ans. Factorization of polynomials can be used to solve equations by setting each factor equal to zero and solving for the variable. This is based on the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be equal to zero. By finding the solutions for each factor, we can find the solutions for the entire equation.
3. What are the different methods of factorizing polynomials?
Ans. There are several methods for factorizing polynomials. Some common methods include: - Factorization by grouping: This method involves grouping the terms of the polynomial in pairs and then factoring out the common factor from each pair. - Factorization by using common factors: This method involves factoring out the common factors from the terms of the polynomial. - Factorization by using the difference of squares: This method is used when the polynomial is a perfect square, and it involves writing the polynomial as the difference of two squares and then factoring accordingly. - Factorization by using the sum or difference of cubes: This method is used when the polynomial is a perfect cube, and it involves writing the polynomial as the sum or difference of cubes and then factoring accordingly.
4. Can all polynomials be factored?
Ans. No, not all polynomials can be factored. There are certain polynomials, known as irreducible polynomials, that cannot be factored into simpler polynomials with integer coefficients. These polynomials have prime factors or irreducible quadratic factors that cannot be further factored.
5. How is factorization of polynomials related to finding the roots of a polynomial equation?
Ans. Factorization of a polynomial equation is directly related to finding its roots. The roots of a polynomial equation are the values of the variable that make the equation equal to zero. By factoring the polynomial equation, we can rewrite it in the form of a product of factors, where one or more factors are equal to zero. By setting each factor equal to zero and solving for the variable, we can find the roots of the polynomial equation.
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