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

RD Sharma Solutions for Class 9 Mathematics

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Class 9 : RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths Class 9 Notes | EduRev

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

Equate to zero

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

⇒ 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

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