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**In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) , or not : (1 â€“ 7)**

**Q1. f(x) = x ^{3}â€“6x^{2}+11xâ€“6,g(x) = xâ€“3**

**Sol :**

Here , f(x) = x^{3}â€“6x^{2}+11xâ€“6

g(x) = x â€“ 3

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

we should show â‡’ f(3) = 0

here , x â€“ 3 = 0

â‡’ x = 3

Substitute the value of x in f(x)

f(3) = 3^{3}â€“6âˆ—(3)^{2}+11(3)â€“6

= 27 â€“ (6*9) + 33 â€“ 6

= 27 â€“ 54 + 33 â€“ 6

= 60 â€“ 60

= 0

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

**Q2. f(x) = 3x ^{4}+17x^{3}+9x^{2}â€“7xâ€“10,g(x) = x+5**

**Sol :**

Here , f(x) = 3x^{4}+17x^{3}+9x^{2}â€“7xâ€“10

g(x) = x + 5

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

we should show â‡’ f(-5) = 0

here , x + 5 = 0

â‡’ x = -5

Substitute the value of x in f(x)

f(âˆ’5)=3(âˆ’5)^{4}+17(âˆ’5)^{3}+9(âˆ’5)^{2}â€“7(âˆ’5)â€“10

= (3 * 625) + (12 * (-125)) +(9*25) + 35 â€“ 10

= 1875 â€“ 2125 + 225 + 35 â€“ 10

= 2135 â€“ 2135

= 0

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

**Q3. f(x) = x ^{5}+3x^{4}â€“x^{3}â€“3x^{2}+5x+15,g(x) = x+3**

**Sol :**

Here , f(x) = x^{5}+3x^{4}â€“x^{3}â€“3x^{2}+5x+15

g(x) = x + 3

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

we should show â‡’ f(-3) = 0

here , x + 3 = 0

â‡’ x = -3

Substitute the value of x in f(x)

f(âˆ’3)=(âˆ’3)^{5}+3(âˆ’3)^{4}â€“(âˆ’3)^{3}â€“3(âˆ’3)^{2}+5(âˆ’3)+15

= -243 + 243 + 27 â€“ 27 â€“ 15 + 15

= 0

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

**Q4. f(x) = x ^{3}â€“6x^{2}â€“19x+84,g(x)=xâ€“7**

**Sol :**

Here , f(x) = x^{3}â€“6x^{2}â€“19x+84

g(x) = x â€“ 7

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

we should show â‡’ f(7) = 0

here , x â€“ 7 = 0

â‡’ x = 7

Substitute the value of x in f(x)

f(7) = 73â€“6(7)^{2}â€“19(7)+84

= 343 â€“ (6*49) â€“ (19*7) + 84

= 342 â€“ 294 â€“ 133 + 84

= 427 â€“ 427

= 0

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

**Q5. f(x) = 3x ^{3}+x^{2}â€“20x+12,g(x) = 3xâ€“2**

**Sol :**

Here , f(x)=3x^{3}+x^{2}â€“20x+12

g(x) = 3x â€“ 2

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

we should show â‡’

here , 3x â€“ 2 = 0

â‡’ 3x = 2

â‡’ x = 2/3

Substitute the value of x in f(x)

= 0

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

**Q6. f(x) = 2x ^{3}â€“9x^{2}+x+13,g(x) = 3â€“2x**

**Sol :**

Here , f(x) = 2x^{3}â€“9x^{2}+x+13

g(x) = 3 â€“ 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 â‡’ f(3/2) = 0

here , 3 â€“ 2x = 0

â‡’ -2x = -3

â‡’ 2x = 3

â‡’ x = 3/2

Substitute the value of x in f(x)

Taking L.C.M

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

**Q7. f(x) = x ^{3}â€“6x^{2}+11xâ€“6,g(x) = x^{2}â€“3x+2**

**Sol :**

Here , f(x) = x^{3}â€“6x^{2}+11xâ€“6

g(x) = x^{2}â€“3x+2

First we need to find the factors of x^{2}â€“3x+2

â‡’ x^{2}â€“2xâ€“x+2

â‡’ x(x â€“ 2) -1(x â€“ 2)

â‡’ (x â€“ 1) and (x â€“ 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 â€“ 1 = 0

x = 1

substitute the value of x in f(x)

f(1) = 1^{3}â€“6(1)^{2}+11(1)â€“6

= 1 â€“ 6 + 11 â€“ 6

= 12 â€“ 12

= 0

Let , x â€“ 2 = 0

x = 2

substitute the value of x in f(x)

f(2) = 2^{3}â€“6(2)^{2}+11(2)â€“6

= 8 â€“ (6 * 4) + 22 â€“ 6

= 8 â€“ 24 + 22 â€“ 6

= 30 â€“ 30

= 0

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

**Q8. Show that (x â€“ 2) , (x + 3) and (x â€“ 4) are the factors of x ^{3}â€“3x^{2}â€“10x+24**

**Sol :**

Here , f(x) = x^{3}â€“3x^{2}â€“10x+24

The factors given are (x â€“ 2) , (x + 3) and (x â€“ 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 â€“ 2 = 0

â‡’ x = 2

Substitute the value of x in f(x)

f(2) = 2^{3}â€“3(2)^{2}â€“10(2)+24

= 8 â€“ (3 * 4) â€“ 20 + 24

= 8 â€“ 12 â€“ 20 + 24

= 32 â€“ 32

= 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}â€“10(âˆ’3)+24

= -27 â€“ 3(9) + 30 + 24

= -27 â€“ 27 + 30 + 24

= 54 â€“ 54

= 0

Let , x â€“ 4 = 0

â‡’ x = 4

Substitute the value of x in f(x)

f(4) = (4)^{3}â€“3(4)^{2}â€“10(4)+24

= 64 â€“ (3*16) â€“ 40 + 24

= 64 â€“ 48 â€“ 40 + 24

= 84 â€“ 84

= 0

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

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

**Sol :**

Here , f(x) = x^{3}â€“6x^{2}â€“19x+84

The factors given are (x + 4) , (x â€“ 3) and (x â€“ 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

â‡’ x = -4

Substitute the value of x in f(x)

f(-4) = (âˆ’4)^{3}â€“6(âˆ’4)^{2}â€“19(âˆ’4)+84

= -64 â€“ (6 * 16) â€“ ( 19 * (-4)) + 84

= -64 â€“ 96 + 76 + 84

= 160 â€“ 160

= 0

Let, x â€“ 3 = 0

â‡’ x = 3

Substitute the value of x in f(x)

f(3) = (3)^{3}â€“6(3)^{2}â€“19(3)+84

= 27 â€“ (6 * 9) â€“ ( 19 * 3) + 84

= 27 â€“ 54 â€“ 57 + 84

= 111 â€“ 111

= 0

Let, x â€“ 7 = 0

â‡’ x = 7

Substitute the value of x in f(x)

f(7) = (7)^{3}â€“6(7)^{2}â€“19(7)+84

= 343 â€“ (6 * 49) â€“ ( 19 * 7) + 84

= 343 â€“ 294 â€“ 133 + 84

= 427 â€“ 427

= 0

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

**Q10. For what value of a is (x â€“ 5) a factor of ^{x3}â€“3x^{2}+axâ€“10**

**Sol :**

Here, f(x) = x^{3}â€“3x^{2}+axâ€“10

By factor theorem

If (x â€“ 5) is the factor of f(x) then , f(5) = 0

â‡’ x â€“ 5 = 0

â‡’ x = 5

Substitute the value of x in f(x)

f(5) = 5^{3}â€“3(5)^{2}+a(5)â€“10

= 125 â€“ (3 * 25) + 5a â€“ 10

= 125 â€“ 75 + 5a â€“ 10

= 5a + 40

Equate f(5) to zero

f(5) = 0

â‡’ 5a + 40 = 0

â‡’ 5a = -40

â‡’

= -8

When a= -8 , (x â€“ 5) will be factor of f(x)

**Q11. Find the value of a such that (x â€“ 4) is a factor of 5x ^{3}â€“7x^{2}â€“axâ€“28**

**Sol :**

Here, f(x) = 5x^{3}â€“7x^{2}â€“axâ€“28

By factor theorem

If (x â€“ 4) is the factor of f(x) then , f(4) = 0

â‡’ x â€“ 4 = 0

â‡’ x = 4

Substitute the value of x in f(x)

f(4) = 5(4)^{3}â€“7(4)^{2}â€“a(4)â€“28

= 5(64) â€“ 7(16) â€“ 4a â€“ 28

= 320 â€“ 112 â€“ 4a â€“ 28

= 180 â€“ 4

Equate f(4) to zero, to find a

f(4) = 0

â‡’ 180 â€“ 4a = 0

â‡’ -4a = -180

â‡’ 4a = 180

â‡’ a = 180/4

â‡’ a = 45

When a = 45 , (x â€“ 4) will be factor of f(x)

**Q12. Find the value of a, if (x + 2) is a factor of 4x ^{4}+2x^{3}â€“3x^{2}+8x+5a**

**Sol :**

Here, f(x) = 4x^{4}+2x^{3}â€“3x^{2}+8x+5a

By factor theorem

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

â‡’ x + 2 = 0

â‡’ x = -2

Substitute the value of x in f(x)

f(-2) = 4(âˆ’2)^{4}+2(âˆ’2)^{3}â€“3(âˆ’2)^{2}+8(âˆ’2)+5a

= 4(16) + 2(-8) â€“ 3( 4) â€“ 16 + 5a

= 64 â€“ 16 â€“ 12 â€“ 16 + 5a

= 5a + 20

equate f(-2) to zero

f(-2) = 0

â‡’ 5a + 20 = 0

â‡’ 5a = -20

â‡’

â‡’ a = -4

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

**Q13. Find the value of k if x â€“ 3 is a factor of k ^{2}x^{3}â€“kx^{2}+3kxâ€“k**

**Sol :**

Let f(x) = k^{2}x^{3}â€“kx^{2}+3kxâ€“k

From factor theorem if x â€“ 3 is the factor of f(x) then f(3) = 0

â‡’ x â€“ 3 = 0

â‡’ x = 3

Substitute the value of x in f(x)

f(3) = k^{2}(3)^{3}â€“k(3)^{2}+3k(3)â€“k

= 27k^{2}â€“9k+9k â€“ k

= 27k^{2}â€“k

= k(27k â€“ 1)

Equate f(3) to zero, to find k

â‡’ f(3) = 0

â‡’ k(27k â€“ 1) = 0

â‡’ k = 0 and 27k â€“ 1 = 0

â‡’ k = 0 and 27k = 1

â‡’ k = 0 and k = 12/7

When k = 0 and 12/7 , (x â€“ 3) will be the factor of f(x)