# RD Sharma Solutions Class 9 Factorization Of Polynomials Exercise 6.4

## RD Sharma Solutions Class 9 Chapter 6 Ex 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)

Q1. $f(x) = x^{3} – 6x^{2} +11x – 6 , g(x) = x – 3$

Sol :

Here , $f(x) = x^{3} – 6x^{2} +11x – 6$

g(x) = x – 3

To prove that g(x) is the factor of f(x) ,

we should show => f(3) = 0

here , x – 3 = 0

=> x = 3

Substitute the value of x in f(x)

$f(3) = 3^{3} – 6*(3)^{2} +11(3) – 6$

= 27 – (6*9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q2. $f(x) = 3x^{4} + 17x^{3} + 9x^{2} – 7x – 10 , g(x) = x + 5$

Sol :

Here , $f(x) = 3x^{4} + 17x^{3} + 9x^{2} – 7x – 10$

g(x) = x + 5

To prove that g(x) is the factor of f(x) ,

we should show => f(-5) = 0

here , x + 5 = 0

=> x = -5

Substitute the value of x in f(x)

$f(-5) = 3(-5)^{4} + 17(-5)^{3}+ 9(-5)^{2} – 7(-5) – 10$

= (3 * 625) + (12 * (-125)) +(9*25) + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 2135 – 2135

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q3. $f(x) = x^{5} + 3x^{4} – x^{3} – 3x^{2} + 5x + 15 , g(x)= x + 3$

Sol :

Here , $f(x) = x^{5} + 3x^{4} – x^{3} – 3x^{2} + 5x + 15$

g(x) = x + 3

To prove that g(x) is the factor of f(x) ,

we should show => f(-3) = 0

here , x + 3 = 0

=> x = -3

Substitute the value of x in f(x)

$f(-3) = (-3)^{5} + 3(-3)^{4} – (-3)^{3} – 3(-3)^{2} + 5(-3) + 15$

= -243 + 243 + 27 – 27 – 15 + 15

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q4. $f(x) = x^{3} – 6x^{2} – 19x + 84 , g(x) = x – 7$

Sol :

Here , $f(x) = x^{3} – 6x^{2} – 19x + 84$

g(x) = x – 7

To prove that g(x) is the factor of f(x) ,

we should show => f(7) = 0

here , x – 7 = 0

=> x = 7

Substitute the value of x in f(x)

$f(7) = 7^{3} – 6(7)^{2} – 19(7) + 84$

= 343 – (6*49) – (19*7) + 84

= 342 – 294 – 133 + 84

= 427 – 427

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q5. $f(x) = 3x^{3} + x^{2} – 20x + 12 , g(x) = 3x – 2$

Sol :

Here , $f(x) = 3x^{3} + x^{2} – 20x + 12$

g(x) = 3x – 2

To prove that g(x) is the factor of f(x) ,

we should show => f($\frac{2}{3}$) = 0

here , 3x – 2 = 0

=> 3x = 2

=> x = $\frac{2}{3}$

Substitute the value of x in f(x)

$f(\frac{2}{3}$) = 3($\frac{2}{3}$)^{3} + ($\frac{2}{3})^{2}$ – 20($\frac{2}{3}$) + 12

= $3(\frac{8}{27}) + \frac{4}{9} – \frac{40}{3} + 12$

= $\frac{8}{9} + \frac{4}{9} – \frac{40}{3} + 12$

= $\frac{12}{9} – \frac{40}{3} + 12$

Taking L.C.M

= $\frac{12 – 120 + 108}{9}$

= $\frac{120 – 120 }{9}$

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q6. $f(x) = 2x^{3} – 9x^{2} + x + 13 , g(x) = 3 – 2x$

Sol :

Here , $f(x) = 2x^{3} – 9x^{2} + x + 13$

g(x) = 3 – 2x

To prove that g(x) is the factor of f(x) ,

To prove that g(x) is the factor of f(x) ,

we should show => f($\frac{3}{2}$) = 0

here , 3 – 2x = 0

=> -2x = -3

=> 2x = 3

=> x = $\frac{3}{2}$

Substitute the value of x in f(x)

f($\frac{3}{2}$) = 2($\frac{3}{2})^{3}$ – 9($\frac{3}{2})^{2}$ + ($\frac{3}{2}$) + 13

= $2(\frac{27}{8}) – 9(\frac{9}{4}) + \frac{3}{2} + 12$

= $(\frac{27}{4}) – (\frac{81}{4}) + \frac{3}{2} + 12$

Taking L.C.M

= $\frac{21 – 81 + 6 + 48}{4}$

= $\frac{81 – 81}{4}$

= 0

Since, the result is 0 g(x) is the factor of f(x)

Q7. $f(x) = x^{3} – 6x^{2} + 11x – 6 , g(x) = x^{2} – 3x + 2$

Sol :

Here , $f(x) = x^{3} – 6x^{2} + 11x – 6$

$g(x) = x^{2} – 3x + 2$

First we need to find the factors of $x^{2} – 3x + 2$

=>$x^{2} – 2x – x + 2$

=> x(x – 2) -1(x – 2)

=> (x – 1) and (x – 2) are the factors

To prove that g(x) is the factor of f(x) ,

The results of f(1) and f(2) should be zero

Let , x – 1 = 0

x = 1

substitute the value of x in f(x)

$f(1) = 1^{3} – 6(1)^{2} + 11(1) – 6$

= 1 – 6 + 11 – 6

= 12 – 12

= 0

Let , x – 2 = 0

x = 2

substitute the value of x in f(x)

$f(2) = 2^{3} – 6(2)^{2} + 11(2) – 6$

= 8 – (6 * 4) + 22 – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

Since, the results are 0 g(x) is the factor of f(x)

Q8. Show that (x – 2) , (x + 3) and (x – 4) are the factors of $x^{3} – 3x^{2} – 10x + 24$

Sol :

Here , f(x) = $x^{3} – 3x^{2} – 10x + 24$

The factors given are (x – 2) , (x + 3) and (x – 4)

To prove that g(x) is the factor of f(x) ,

The results of f(2) , f(-3) and f(4) should be zero

Let , x – 2 = 0

=> x = 2

Substitute the value of x in f(x)

f(2) = $2^{3} – 3(2)^{2} – 10(2) + 24$

= 8 – (3 * 4) – 20 + 24

= 8 – 12 – 20 + 24

= 32 – 32

= 0

Let , x + 3 = 0

=> x = -3

Substitute the value of x in f(x)

f(-3) = $(-3)^{3} – 3(-3)^{2} – 10(-3) + 24$

= -27 – 3(9) + 30 + 24

= -27 – 27 + 30 + 24

= 54 – 54

= 0

Let , x – 4 = 0

=> x = 4

Substitute the value of x in f(x)

f(4) = $(4)^{3} – 3(4)^{2} – 10(4) + 24$

= 64 – (3*16) – 40 + 24

= 64 – 48 – 40 + 24

= 84 – 84

= 0

Since, the results are 0 g(x) is the factor of f(x)

Q9. Show that (x + 4) , (x – 3) and (x – 7) are the factors of $x^{3} – 6x^{2} – 19x + 84$

Sol :

Here , f(x) = $x^{3} – 6x^{2} – 19x + 84$

The factors given are (x + 4) , (x – 3) and (x – 7)

To prove that g(x) is the factor of f(x) ,

The results of f(-4) , f(3) and f(7) should be zero

Let, x + 4 = 0

=> x = -4

Substitute the value of x in f(x)

f(-4) = $(-4)^{3} – 6(-4)^{2} – 19(-4) + 84$

= -64 – (6 * 16) – ( 19 * (-4)) + 84

= -64 – 96 + 76 + 84

= 160 – 160

= 0

Let, x – 3 = 0

=> x = 3

Substitute the value of x in f(x)

f(3) = $(3)^{3} – 6(3)^{2} – 19(3) + 84$

= 27 – (6 * 9) – ( 19 * 3) + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0

Let, x – 7 = 0

=> x = 7

Substitute the value of x in f(x)

f(7) = $(7)^{3} – 6(7)^{2} – 19(7) + 84$

= 343 – (6 * 49) – ( 19 * 7) + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

Since, the results are 0 g(x) is the factor of f(x)

Q10. For what value of a is (x – 5) a factor of $x^{3} – 3x^{2} + ax – 10$

Sol :

Here, f(x) = $x^{3} – 3x^{2} + ax – 10$

By factor theorem

If (x – 5) is the factor of f(x) then , f(5) = 0

=> x – 5 = 0

=> x = 5

Substitute the value of x in f(x)

f(5) = $5^{3} – 3(5)^{2} + a(5) – 10$

= 125 – (3 * 25) + 5a – 10

= 125 – 75 + 5a – 10

= 5a + 40

Equate f(5) to zero

f(5) = 0

=> 5a + 40 = 0

=> 5a = -40

=> a = $\frac{-40}{5}$

= -8

When a= -8 , (x – 5) will be factor of f(x)

Q11. Find the value of a such that (x – 4) is a factor of $5x^{3} – 7x^{2} – ax – 28$

Sol :

Here, f(x) = $5x^{3} – 7x^{2} – ax – 28$

By factor theorem

If (x – 4) is the factor of f(x) then , f(4) = 0

=> x – 4 = 0

=> x = 4

Substitute the value of x in f(x)

f(4) = $5(4)^{3} – 7(4)^{2} – a(4) – 28$

= 5(64) – 7(16) – 4a – 28

= 320 – 112 – 4a – 28

= 180 – 4

Equate f(4) to zero, to find a

f(4) = 0

=> 180 – 4a = 0

=> -4a = -180

=> 4a = 180

=> a = $\frac{180}{4}$

=> a = 45

When a = 45 , (x – 4) will be factor of f(x)

Q12. Find the value of a, if (x + 2) is a factor of $4x^{4} + 2x^{3} – 3x^{2} + 8x + 5a$

Sol :

Here, f(x) = $4x^{4} + 2x^{3} – 3x^{2} + 8x + 5a$

By factor theorem

If (x + 2) is the factor of f(x) then , f(-2) = 0

=> x + 2 = 0

=> x = -2

Substitute the value of x in f(x)

f(-2) = $4(-2)^{4} + 2(-2)^{3} – 3(-2)^{2} + 8(-2) + 5a$

= 4(16) + 2(-8) – 3( 4) – 16 + 5a

= 64 – 16 – 12 – 16 + 5a

= 5a + 20

equate f(-2) to zero

f(-2) = 0

=> 5a + 20 = 0

=> 5a = -20

=> a = $\frac{-20}{5}$

=> a = -4

When a = -4 , (x + 2) will be factor of f(x)

Q13. Find the value of k if x – 3 is a factor of $k^{2}x^{3} – kx^{2} + 3kx – k$

Sol :

Let f(x) = $k^{2}x^{3} – kx^{2} + 3kx – k$

From factor theorem if x – 3 is the factor of f(x) then f(3) = 0

=> x – 3 = 0

=> x = 3

Substitute the value of x in f(x)

f(3) = $k^{2}(3)^{3} – k(3)^{2} + 3k(3) – k$

= $27k^{2} – 9k + 9k$ – k

= $27k^{2} – k$

= k(27k – 1)

Equate f(3) to zero, to find k

=> f(3) = 0

=> k(27k – 1) = 0

=> k = 0 and 27k – 1 = 0

=> k = 0 and 27k = 1

=> k = 0 and k = $\frac{1}{27}$

When k = 0 and $\frac{1}{27}$ , (x – 3) will be the factor of f(x)

Q14. Find the values of a and b, if $x^{2}$ – 4 is a factor of $ax^{4} + 2x^{3} – 3x^{2} + bx – 4$

Sol :

Given , f(x) = $ax^{4} + 2x^{3} – 3x^{2} + bx – 4$

g(x) = $x^{2}$ – 4

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

=> $x^{2}$ – 4

=> $x^{2}$ = 4

=> x = $\sqrt{4}$

=> 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 $\alpha\; ,\; \beta$ if (x + 1) and (x + 2) are the factors of $x^{3}+3x^{2}-2\alpha x+\beta$

Sol:

Given, f(x) = $x^{3}+3x^{2}-2\alpha x+\beta$ 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) = $(-1)^{3}+3(-1)^{2}-2\alpha (-1)+\beta$

$= -1 + 3 + 2\alpha + \beta$

$= 2\alpha + \beta$ + 2 ———— 1

Let , x + 2 = 0

=> x = -2

Substitute value of x in f(x)

f(-2) = $(-2)^{3}+3(-2)^{2}-2\alpha (-2)+\beta$

$= -8 + 12 + 4\alpha + \beta$

$= 4\alpha + \beta$ + 4 ————– 2

Solving 1 and 2 i.e ( 1 – 2)

$=> 2\alpha + \beta + 2 – (4\alpha + \beta$ + 4 ) = 0

$=> -2\alpha – 2 = 0$

$=> 2\alpha = -2$

$=> \alpha = -1$

Substitute $\alpha$= -1 in equation 1

$=> 2(-1) + \beta$ = -2

$=> \beta$= -2 + 2

$=> \beta$= 0

The values are $\alpha = -1 and \beta = 0$

Q16. Find the values of p and q so that $x^{4} + px^{3} + 2x^{2} -3x + q$ is divisible by ($x^{2}$ – 1)

Sol :

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

g(x) = $x^{2} – 1$

first, we need to find the factors of $x^{2} – 1$

=> $x^{2} – 1$ = 0

=> $x^{2} = 1$

=> 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 $x^{4} + ax^{3} – 3x^{2} + 2x + b$

Sol :

Here, f(x) = $x^{4} + ax^{3} – 3x^{2} + 2x + b$

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 $x^{3} + ax^{2} – bx + 10$ is divisible by $x^{3} – 3x + 2$, find the values of a and b

Sol :

Here , f(x) = $x^{3} + ax^{2} – bx + 10$

g(x) = $x^{3} – 3x + 2$

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

g(x) = $x^{3} – 3x + 2$

= $x^{3} – 2x – x + 2$

= 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) = $1^{3} + a(1)^{2} – b(1) + 10$

= 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) = $2^{3} + a(2)^{2} – b(2) + 10$

= 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 $ax^{3} + x^{2} -2x + b$ , Find the values of a and b

Sol:

Here, f(x) = $ax^{3} + x^{2} -2x + b$

(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 $x^{3} – 3x^{2} – 12x + 19$ so that the result is exactly divisible by $x^{2} + x – 6$

Sol :

Here , p(x) = $x^{3} – 3x^{2} – 12x + 19$

g(x) = $x^{2} + x – 6$

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)

= $x^{3} – 3x^{2} – 12x + 19$ + ax + b

f(x) = $x^{3} – 3x^{2} + x(a – 12) + 19$ + b

We know that , g(x) = $x^{2} + x – 6$

First, find the factors for g(x)

g(x) = $x^{2} + 3x – 2x – 6$

= 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

$∴$ $x^{3} – 3x^{2} – 12x + 19$ is divided by $x^{2} + x – 6$ when it is added by 2x + 5

Q21. What must be added to $x^{3} – 6x^{2} – 15x + 80$ so that the result is exactly divisible by $x^{2} + x – 12$

Sol :

Let, p(x) = $x^{3} – 6x^{2} – 15x + 80$

q(x) = $x^{2} + x – 12$

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) = $x^{2} + x – 12$

= $x^{2} + 4x – 3x – 12$

= 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) = $3^{3} – 6(3)^{2} – 3(a+15) + 80$ – 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 = $\frac{28}{7}$

=> 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 $3x^{3} + x^{2} – 22x + 9$ so that the result is exactly divisible by $3x^{2} + 7x – 6$

Sol :

Let, p(x) = $3x^{3} + x^{2} – 22x + 9$ and q(x) = $3x^{2} + 7x – 6$

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) = $3x^{3} + x^{2} – 22x + 9(ax + b)$

=> = $3x^{3} + x^{2} + x(a – 22) + b + 9$

We know that,

q(x) = $3x^{2} + 7x – 6$

= $3x^{2} + 9x – 2x – 6$

= 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($\frac{2}{3}$) = 0 and f(-3) = 0

let , 3x – 2 = 0

3x = 2

x = $\frac{2}{3}$

=> f($\frac{2}{3}$) = $3(\frac{2}{3})^{3} + (\frac{2}{3}$)^{2} + ($\frac{2}{3}$)(a – 22) + b + 9

= $3(\frac{8}{27})+\frac{4}{9}+\frac{2}{3}a-\frac{44}{3}+b+9$

=$\frac{12}{9}+\frac{2}{3}a-\frac{44}{3}+b+9$

= $\frac{12+6a-132+9b+81}{9}$

Equate to zero

=> $\frac{12+6a-132+9b+81}{9}$ = 0

=> 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 = $\frac{22}{11}$

=> 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. $x^{3} – 2ax^{2} + ax – 1$
2. $x^{5} – 3x^{4} – ax^{3} + 3ax^{2} + 2ax + 4$

Sol :

(1) let f(x) = $x^{3} – 2ax^{2} + ax – 1$

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) = $2^{3} – 2a(2)^{2} + a(2) – 1$

= 8 – 8a + 2a – 1

= -6a + 7

Equate f(2) to zero

=> -6a + 7 = 0

=> -6a = -7

=> a= $\frac{7}{6}$

When , (x – 2 ) is the factor of f(x) then a= $\frac{7}{6}$

(2) Let, f(x) = $x^{5} – 3x^{4} – ax^{3} + 3ax^{2} + 2ax + 4$

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) = $2^{5} – 3(2)^{4} – a(2)^{3} + 3a(2)^{2} + 2a(2) + 4$

= 32 – 48 – 8a + 12 + 4a + 4

= 8a – 12

Equate f(2) to zero

=> 8a – 12 = 0

=> 8a = 12

=> a = $\frac{12}{8}$

= $\frac{3}{2}$

So, when (x – 2) is a factor of f(x) then a= $\frac{3}{2}$

Q24. In each of the following two polynomials , find the value of a, if (x – a) is a factor :

1. $x^{6} – ax^{5} + x^{4} – ax^{3} + 3x – a + 2$
2. $x^{5} – a^{2}x^{3} + 2x + a + 1$

Sol :

(1) $x^{6} – ax^{5} + x^{4} – ax^{3} + 3x – a + 2$

let , f(x) = $x^{6} – ax^{5} + x^{4} – ax^{3} + 3x – a + 2$

here , x – a = 0

=> x = a

Substitute the value of x in f(x)

f(a) = $a^{6} – a(a)^{5} + (a)^{4} – a(a)^{3} + 3(a) – a + 2$

= $a^{6} – a^{6} + (a)^{4} – a^{4} + 3(a) – a + 2$

= 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) $x^{5} – a^{2}x^{3} + 2x + a + 1$

let, $f(x) = x^{5} – a^{2}x^{3} + 2x + a + 1$

here , x – a = 0

=> x = a

Substitute the value of x in f(x)

f(a) = $a^{5} – a^{2}a^{3} + 2(a) + a + 1$

= $a^{5} – a^{5} + 2a + a + 1$

= 3a + 1

Equate to zero

=> 3a + 1 = 0

=> 3a = -1

=> a= $\frac{-1}{3}$

So, when (x – a) is a factor of f(x) then a = $\frac{-1}{3}$

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

1. $x^{3} + ax^{2} – 2x + a + 4$
2. $x^{4} – a^{2}x^{2} + 3x – a$

Sol :

(1) $x^{3} + ax^{2} – 2x + a + 4$

let, f(x) = $x^{3} + ax^{2} – 2x + a + 4$

here , x + a = 0

=> x = -a

Substitute the value of x in f(x)

f(-a) = $(-a)^{3} + a(-a)^{2} – 2(-a) + a + 4$

= $(-a)^{3} + a^{3} – 2(-a) + a + 4$

= 3a + 4

Equate to zero

=> 3a + 4 = 0

=> 3a = -4

=> a = $\frac{-4}{3}$

So, when (x + a) is a factor of f(x) then a = $\frac{-4}{3}$

(2) $x^{4} – a^{2}x^{2} + 3x – a$

let, $f(x) = x^{4} – a^{2}x^{2} + 3x – a$

here , x + a = 0

=> x = -a

Substitute the value of x in f(x)

f(-a) = $(-a)^{4} – a^{2}(-a)^{2} + 3(-a) – a$

= $a^{4} – a^{4} – 3(a) – a$

= -4a

Equate to zero

=> -4a = 0

=> a = 0

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

#### Practise This Question

Which of the following figures has/have both line symmetry and rotational symmetry?

(i) Square                               (iii) Pentagon

(ii) Rectangle                          (iv) Isosceles triangle