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**Q14. Find the values of a and b, if x ^{2} â€“ 4 is a factor of ax4+2x^{3}â€“3x^{2}+bxâ€“4**

**Sol :**

Given , f(x) = ax^{4}+2x^{3}â€“3x^{2}+bxâ€“4

g(x) = x^{2} â€“ 4

first we need to find the factors of g(x)

â‡’ x^{2} â€“ 4

â‡’ x^{2} = 4

â‡’ x = âˆš4

â‡’ x = Â±2

(x â€“ 2) and (x + 2) are the factors

By factor therorem if (x â€“ 2) and (x + 2) are the factors of f(x) the result of f(2) and f(-2) should be zero

Let , x â€“ 2 = 0

â‡’ x = 2

Substitute the value of x in f(x)

f(2) = a(2)^{4}+2(2)^{3}â€“3(2)^{2}+b(2)â€“4

= 16a + 2(8) â€“ 3(4) + 2b â€“ 4

= 16a + 2b + 16 â€“ 12 â€“ 4

= 16a + 2b

Equate f(2) to zero

â‡’ 16a + 2b = 0

â‡’ 2(8a + b) = 0

â‡’ 8a + b = 0 â€”â€”â€”- 1

Let , x + 2 = 0

â‡’ x = -2

Substitute the value of x in f(x)

f(-2) = a(âˆ’2)^{4}+2(âˆ’2)^{3}â€“3(âˆ’2)^{2}+b(âˆ’2)â€“4

= 16a + 2(-8) â€“ 3(4) â€“ 2b â€“ 4

= 16a â€“ 2b â€“ 16 â€“ 12 â€“ 4

= 16a â€“ 2b â€“ 32

= 16a â€“ 2b â€“ 32

Equate f(2) to zero

â‡’ 16a â€“ 2b â€“ 32 = 0

â‡’ 2(8a â€“ b) = 32

â‡’ 8a â€“ b = 16 â€”â€”â€”â€” 2

Solve equation 1 and 2

8a + b = 0

8a â€“ b = 16

16a = 16

a = 1

substitute a value in eq 1

8(1) + b = 0

â‡’ b = -8

The values are a = 1 and b = -8

**Q15. Find Î±,Î² if (x + 1) and (x + 2) are the factors of x ^{3}+3x^{2}âˆ’2Î±x+Î²**

**Sol:**

Given, f(x) = x^{3}+3x^{2}âˆ’2Î±x+Î² 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

â‡’ x = -1

Substitute value of x in f(x)

f(-1) = (âˆ’1)^{3}+3(âˆ’1)^{2}âˆ’2Î±(âˆ’1)+Î²

=âˆ’1+3+2Î±+Î²

= 2Î±+Î² + 2 â€”â€”â€”â€” 1

Let , x + 2 = 0

â‡’ x = -2

Substitute value of x in f(x)

f(-2) = (âˆ’2)^{3}+3(âˆ’2)^{2}âˆ’2Î±(âˆ’2)+Î²

=âˆ’8+12+4Î±+Î²

=4Î±+Î² + 4 â€”â€”â€”â€”â€“ 2

Solving 1 and 2 i.e (1 â€“ 2)

â‡’2Î±+Î²+2â€“(4Î±+Î² + 4) = 0

â‡’ âˆ’2Î±â€“2 = 0

â‡’ 2Î±=âˆ’2

â‡’ Î± = âˆ’1

Substitute Î± = -1 in equation 1

â‡’ 2(âˆ’1)+Î² = -2

â‡’ Î² = -2 + 2

â‡’ Î² = 0

The values are Î± = âˆ’1 and Î² = 0

**Q16. Find the values of p and q so that x ^{4}+px^{3}+2x^{2}âˆ’3x+q is divisible by (x^{2} â€“ 1)**

**Sol :**

Here , f(x) = x^{4}+px^{3}+2x^{2}âˆ’3x+q

g(x) = x^{2}â€“1

first, we need to find the factors of x^{2}â€“1

â‡’ x^{2}â€“1 = 0

â‡’ x^{2} = 1

â‡’ x = Â±1

â‡’ (x + 1) and (x â€“ 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

â‡’ x + 1 = 0

â‡’ x = -1

Substitute the value of x in f(x)

f(-1) = (âˆ’1)^{4}+p(âˆ’1)^{3}+2(âˆ’1)^{2}âˆ’3(âˆ’1)+q

= 1 â€“ p + 2 + 3 + q

= -p + q + 6 â€”â€”â€”- 1

Let us take , x â€“ 1

â‡’ x â€“ 1 = 0

â‡’ x = 1

Substitute the value of x in f(x)

f(1) = (1)^{4}+p(1)^{3}+2(1)^{2}âˆ’3(1)+q

= 1 + p + 2 â€“ 3 + q

= p + q â€”â€”â€”- 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 â€“ 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 â€“ 1) are the factors of x ^{4}+ax^{3}â€“3x^{2}+2x+b**

**Sol :**

Here, f(x) = x^{4}+ax^{3}â€“3x^{2}+2x+b

The factors are (x + 1) and (x â€“ 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

â‡’ x + 1 = 0

â‡’ x = -1

Substitute value of x in f(x)

f(-1) = (âˆ’1)^{4}+a(âˆ’1)^{3}â€“3(âˆ’1)^{2}+2(âˆ’1)+b

= 1 â€“ a â€“ 3 â€“ 2 + b

= -a + b â€“ 4 â€”â€”- 1

Let , us take x â€“ 1

â‡’ x â€“ 1 = 0

â‡’ x = 1

Substitute value of x in f(x)

f(1) = (1)4+a(1)^{3}â€“3(1)^{2}+2(1)+b

= 1 + a â€“ 3 + 2 + b

= a + b â€”â€”- 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 x ^{3}+ax^{2}â€“bx+10 is divisible by x^{3}â€“3x+2, find the values of a and b**

**Sol :**

Here , f(x) = x^{3}+ax^{2}â€“bx+10

g(x) = x^{3}â€“3x+2

first, we need to find the factors of g(x)

g(x) = x^{3}â€“3x+2

= x^{3}â€“2xâ€“x+2

= x(x â€“ 2) -1( x â€“ 2)

= ( x â€“ 1) and ( x â€“ 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 â€“ 1

â‡’ x â€“ 1 = 0

â‡’ x = 1

Substitute the value of x in f(x)

f(1) = 13+a(1)^{2}â€“b(1)+10

= 1 + a â€“ b + 10

= a â€“ b + 11 â€”â€”- 1

Let, us take x â€“ 2

â‡’ x â€“ 2 = 0

â‡’ x = 2

Substitute the value of x in f(x)

f(2) = 2^{3}+a(2)^{2}â€“b(2)+10

= 8 + 4a â€“ 2b + 10

= 4a â€“ 2b + 18

Equate f(2) to zero

â‡’ 4a â€“ 2b + 18 = 0

â‡’ 2(2a â€“ b + 9) = 0

â‡’ 2a â€“ b + 9 â€”â€”â€”- 2

Solve 1 and 2

a â€“ b = -11

2a â€“ b = -9

(-) (+) (+)

-a = -2

a = 2

substitute a value in eq 1

â‡’ 2 â€“ b = -11

â‡’ â€“ b = -11 â€“ 2

â‡’ -b = -13

â‡’ b = 13

The values are a = 2 and b = 13

**Q19. If both (x + 1) and (x â€“ 1) are the factors of ax ^{3}+x^{2}âˆ’2x+b , Find the values of a and b**

**Sol:**

Here, f(x) = ax^{3}+x^{2}âˆ’2x+b

(x + 1) and (x â€“ 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 â€“ 1= 0

â‡’ x = -1

Substitute x value in f(x)

f(1) = a(1)^{3}+(1)^{2}âˆ’2(1)+b

= a + 1 â€“ 2 + b

= a + b â€“ 1 â€”â€”â€”- 1

Let, x + 1= 0

â‡’ x = -1

Substitute x value in f(x)

f(-1) = a(âˆ’1)^{3}+(âˆ’1)^{2}âˆ’2(âˆ’1)+b

= -a + 1 + 2 + b

= -a + b + 3 â€”â€”â€”- 2

Solve equations 1 and 2

a + b = 1

-a + b = -3

2b = -2

â‡’ b = -1

substitute b value in eq 1

â‡’ a â€“ 1 = 1

â‡’ a = 1 + 1

â‡’ a = 2

The values are a= 2 and b = -1

**Q20. What must be added to x ^{3}â€“3x^{2}â€“12x+19 so that the result is exactly divisible by x^{2}+xâ€“6**

**Sol :**

Here , p(x) = x^{3}â€“3x^{2}â€“12x+19

g(x) = x^{2}+xâ€“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)

â‡’ f(x) = p(x) + r(x)

= x^{3}â€“3x^{2}â€“12x+19 + ax + b

f(x) = x^{3}â€“3x^{2}+x(aâ€“12)+19 + b

We know that , g(x) = x^{2}+xâ€“6

First, find the factors for g(x)

g(x) = x^{2}+3xâ€“2xâ€“6

= x(x + 3) -2(x + 3)

= (x + 3) ( x â€“ 2) are the factors

From, factor theorem when (x + 3) and (x â€“ 2) are the factors of f(x) the f(-3) = 0 and f(2) = 0

Let, x + 3 = 0

â‡’ x = -3

Substitute the value of x in f(x)

f(-3) = (âˆ’3)^{3}â€“3(âˆ’3)^{2}+(âˆ’3)(aâ€“12)+19 + b

= -27 â€“ 27 â€“ 3a + 24 + 19 + b

= -3a + b + 1 â€”â€”â€” 1

Let, x â€“ 2 = 0

â‡’ x = 2

Substitute the value of x in f(x)

f(2) = (2)^{3}â€“3(2)^{2}+(2)(aâ€“12)+19 + b

= 8 â€“ 12 + 2a â€“ 24 + b

= 2a + b â€“ 9 â€”â€”â€” 2

Solve equations 1 and 2

-3a + b = -1

2a + b = 9

(-) (-) (-)

-5a = â€“ 10

a = 2

substitute the value of a in eq 1

â‡’ -3(2) + b = -1

â‡’ -6 + b = -1

â‡’ b = -1 + 6

â‡’ b = 5

âˆ´ r(x) = ax + b

= 2x + 5

âˆ´ x^{3}â€“3x^{2}â€“12x+19 is divided by x^{2}+xâ€“6 when it is added by 2x + 5

**Q21. What must be added to x ^{3}â€“6x^{2}â€“15x+80 so that the result is exactly divisible by x^{2}+xâ€“12**

**Sol :**

Let, p(x) = x^{3}â€“6x^{2}â€“15x+80

q(x) = x^{2}+xâ€“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) â€“ q(x) is divisible by q(x)

let f(x) = p(x) â€“ q(x)

q(x) = x^{2}+xâ€“12

= x^{2}+4xâ€“3xâ€“12

= x(x + 4) (-3)(x + 4)

= (x+4) , (x â€“ 3)

clearly, (x â€“ 3) and (x + 4) are factors of q(x)

so, f(x) wiil be divisible by q(x) if (x â€“ 3) and (x + 4) are factors of q(x)

from , factor theorem

f(-4) = 0 and f(3) = 0

â‡’ f(3) = 3^{3}â€“6(3)^{2}â€“3(a+15)+80 â€“ b = 0

= 27 â€“ 54 -3a -45 + 80 â€“b

= -3a â€“b + 8 â€”â€”â€” 1

Similarly,

f(-4) = 0

â‡’ f(-4) â‡’ (âˆ’4)^{3}â€“6(âˆ’4)^{2}â€“(âˆ’4)(a+15)+80 â€“ b = 0

â‡’ -64 â€“ 96 -4a + 60 + 80 â€“b = 0

â‡’ 4a â€“ b â€“ 20 = 0 â€”â€”â€”- 2

Substract eq 1 and 2

â‡’ 4a â€“ b â€“ 20 â€“ 8 + 3a + b = 0

â‡’ 7a â€“ 28 = 0

â‡’ a = 28/7

â‡’ a= 4

Put a = 4 in eq 1

â‡’ -3(4) â€“ b = -8

â‡’ -b â€“ 12 = -8

â‡’ -b = -8 + 12

â‡’ b = -4

Substitute a and b values in r(x)

â‡’ r(x) = ax + b

= 4x â€“ 4

Hence, p(x) is divisible by q(x) , if r(x) = 4x â€“ 4 is subtracted from it

**Q22. What must be added to 3x ^{3}+x^{2}â€“22x+9 so that the result is exactly divisible by 3x^{2}+7xâ€“6**

**Sol :**

Let, p(x) = 3x^{3}+x^{2}â€“22x+9 and q(x) = 3x^{2}+7xâ€“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)

â‡’ f(x) = 3x^{3}+x^{2}â€“22x+9(ax+b)

â‡’ = 3x^{3}+x^{2}+x(aâ€“22)+b+9

We know that,

q(x) = 3x^{2}+7xâ€“6

= 3x^{2}+9xâ€“2xâ€“6

= 3x(x+3) â€“ 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 â€“ 2 = 0

3x = 2

x = 2/3

â‡’ f(2/3) = 3(2/3)^{3}+(2/3)^{2} + (2/3)(a â€“ 22) + b + 9

Equate to zero

â‡’

â‡’ 6a + 9b â€“ 39 = 0

â‡’ 3(2a + 3b â€“ 13) = 0

â‡’ 2a + 3b â€“ 13 = 0 â€”â€”â€”- 1

Similarly,

Let, x + 3 = 0

â‡’ x = -3

â‡’ f(-3) = 3(âˆ’3)^{3}+(âˆ’3)^{2}+(âˆ’3)(aâ€“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 â€”â€”â€“ 2

Substact eq 1 from 2

â‡’ -9a + 3b + 9 -2a â€“ 3b + 13 = 0

â‡’ -11a + 22 = 0

â‡’ -11a = -22

â‡’ a = 22/11

â‡’ a = 2

Substitute a value in eq 1

â‡’ -3(2) + b = -3

â‡’ -6 + b = -3

â‡’ b = -3 + 6

â‡’ 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 â€“ 2 is a factor of each of the following two polynomials , find the value of a in each case :**

**1. x ^{3}â€“2ax^{2}+axâ€“1**

2. x^{5}â€“3x^{4}â€“ax^{3}+3ax^{2}+2ax+4

**Sol :**

(1) let f(x) = x^{3}â€“2ax^{2}+axâ€“1

from factor theorem

if (x â€“ 2) is the factor of f(x) the f(2) = 0

let , x â€“ 2 = 0

â‡’ x = 2

Substitute x value in f(x)

f(2) = 2^{3}â€“2a(2)^{2}+a(2)â€“1

= 8 â€“ 8a + 2a â€“ 1

= -6a + 7

Equate f(2) to zero

â‡’ -6a + 7 = 0

â‡’ -6a = -7

â‡’ a= 76

When , (x â€“ 2) is the factor of f(x) then a= 76

(2) Let, f(x) = x^{5}â€“3x^{4}â€“ax^{3}+3ax^{2}+2ax+4

from factor theorem

if (x â€“ 2) is the factor of f(x) the f(2) = 0

let , x â€“ 2 = 0

â‡’ x = 2

Substitute x value in f(x)

f(2) = 2^{5}â€“3(2)^{4}â€“a(2)^{3}+3a(2)^{2}+2a(2)+4

= 32 â€“ 48 â€“ 8a + 12 + 4a + 4

= 8a â€“ 12

Equate f(2) to zero

â‡’ 8a â€“ 12 = 0

â‡’ 8a = 12

â‡’ a = 12/8

= 3/2

So, when (x â€“ 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 â€“ a) is a factor :**

**1. x ^{6}â€“ax^{5}+x^{4}â€“ax^{3}+3xâ€“a+2**

2. x^{5}â€“a^{2}x^{3}+2x+a+1

**Sol :**

(1) x^{6}â€“ax^{5}+x^{4}â€“ax^{3}+3xâ€“a+2

let , f(x) = x^{6}â€“ax^{5}+x^{4}â€“ax^{3}+3xâ€“a+2

here , x â€“ a = 0

â‡’ x = a

Substitute the value of x in f(x)

f(a) = a^{6}â€“a(a)^{5}+(a)^{4}â€“a(a)^{3}+3(a)â€“a+2

= a^{6}â€“a^{6}+(a)^{4}â€“a^{4}+3(a)â€“a+2

= 2a + 2

Equate to zero

â‡’ 2a + 2 = 0

â‡’ 2(a + 1) = 0

â‡’ a = -1

So, when (x â€“ a) is a factor of f(x) then a = -1

(2) x^{5}â€“a^{2}x^{3}+2x+a+1

let, f(x) = x^{5}â€“a^{2}x^{3}+2x+a+1

here , x â€“ a = 0

â‡’ x = a

Substitute the value of x in f(x)

f(a) = a^{5}â€“a^{2}a^{3}+2(a)+a+1

= a^{5}â€“a^{5}+2a+a+1

= 3a + 1

Equate to zero

â‡’ 3a + 1 = 0

â‡’ 3a = -1

â‡’ a= âˆ’1/3

So, when (x â€“ a) is a factor of f(x) then a = âˆ’1/3

**Q25. In each of the following two polynomials , find the value of a, if (x + a) is a factor :**

**1. x ^{3}+ax^{2}â€“2x+a+4**

**2. x ^{4}â€“a^{2}x^{2}+3xâ€“a**

**Sol :**

(1) x^{3}+ax^{2}â€“2x+a+4

let, f(x) = x^{3}+ax^{2}â€“2x+a+4

here , x + a = 0

â‡’ x = -a

Substitute the value of x in f(x)

f(-a) = (âˆ’a)^{3}+a(âˆ’a)^{2}â€“2(âˆ’a)+a+4

= (âˆ’a)^{3}+a^{3}â€“2(âˆ’a)+a+4

= 3a + 4

Equate to zero

â‡’ 3a + 4 = 0

â‡’ 3a = -4

â‡’ a = âˆ’4/3

So, when (x + a) is a factor of f(x) then a = âˆ’4/3

(2) x^{4}â€“a^{2}x^{2}+3xâ€“a

let, f(x) = x^{4}â€“a^{2}x^{2}+3xâ€“a

here , x + a = 0

â‡’ x = -a

Substitute the value of x in f(x)

f(-a) = (âˆ’a)^{4}â€“a^{2}(âˆ’a)^{2}+3(âˆ’a)â€“a

= a^{4}â€“a^{4}â€“3(a)â€“a

= -4a

Equate to zero

â‡’ -4a = 0

â‡’ a = 0

So, when (x + a) is a factor of f(x) then a = 0