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**Exercise 2.4 1. Determine which of the following polynomials has (**

**(ii) x^{4} + x^{3} + x^{2} + x + 1**

(iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

**(iv) x^{3} - x^{2} - (2 + âˆš2)x + âˆš2**

**Answer**

(i) If (*x* + 1) is a factor of *p*(*x*) = *x*^{3} + *x*^{2} + *x* + 1, *p*(-1) must be zero.

Here, *p*(*x*) = *x*^{3} + *x*^{2} + *x* + 1 *p*(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

= -1 + 1 - 1 + 1 = 0

Therefore, *x* + 1 is a factor of this polynomial

(ii) If (*x* + 1) is a factor of *p*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1, *p*(-1) must be zero.

Here, *p*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1 *p*(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1

= 1 - 1 + 1 - 1 + 1 = 1

As, *p*(-1) â‰ 0

Therefore, *x* + 1 is not a factor of this polynomial

(iii)If (*x* + 1) is a factor of polynomial *p*(*x*) = *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1, *p*(- 1) must be 0. *p*(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1

= 1 - 3 + 3 - 1 + 1 = 1

As, *p*(-1) â‰ 0

Therefore, *x* + 1 is not a factor of this polynomial.

(iv) If (*x* + 1) is a factor of polynomial

*p*(*x*) = *x*^{3} - *x*^{2} - (2 + âˆš2)*x* + âˆš2, *p*(- 1) must be 0.

*p*(-1) = (-1)^{3} - (-1)^{2} - (2 + âˆš2) (-1) + âˆš2

= -1 - 1 + 2 + âˆš2 + âˆš2

=2âˆš2

As, *p*(-1) â‰ 0

Therefore,, *x* + 1 is not a factor of this polynomial.

**2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:**

(i) p(x) = 2x^{3} + x^{2} - 2x - 1, g(x) = x + 1

(ii) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

(iii) p(x) = x^{3} - 4 x^{2} + x + 6, g(x) = x - 3

**Answer**

(i) If *g*(*x*) = *x* + 1 is a factor of given polynomial *p*(*x*), *p*(- 1) must be zero.*p*(*x*) = 2*x*^{3} + *x*^{2} - 2*x* - 1*p*(- 1) = 2(-1)^{3} + (-1)^{2} - 2(-1) - 1

= 2(- 1) + 1 + 2 - 1 = 0

Hence, *g*(*x*) = *x* + 1 is a factor of given polynomial.

(ii) If *g*(*x*) = *x* + 2 is a factor of given polynomial *p*(*x*), *p*(- 2) must be 0.*p*(*x*) = *x*^{3} +3*x*^{2} + 3*x* + 1*p*(-2) = (-2)^{3} + 3(- 2)^{2} + 3(- 2) + 1

= -8 + 12 - 6 + 1

= -1

As, *p*(-2) â‰ 0

Hence *g*(*x*) = *x* + 2 is not a factor of given polynomial.

(iii) If *g*(*x*) = *x* - 3 is a factor of given polynomial *p*(*x*), *p*(3) must be 0.*p*(*x*) = *x*^{3} - 4*x*^{2} + *x* + 6*p*(3) = (3)^{3} - 4(3)^{2} + 3 + 6

= 27 - 36 + 9 = 0

Therefore,, *g*(*x*) = *x* - 3 is a factor of given polynomial.

**3. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:(i) p(x) = x^{2} + x + k**

(ii) p(x) = 2x^{2} + kx + âˆš2

(iii) p(x) = kx^{2} - âˆš2x + 1

(iv) p(x) = kx^{2} - 3x + k

(i) If

*p*(1) = 0

â‡’ (1)^{2} + 1 + *k* = 0

â‡’ 2 + *k* = 0

â‡’ *k* = - 2

Therefore, value of *k* is -2.

(ii) If *x* - 1 is a factor of polynomial *p*(*x*) = 2*x*^{2} + *kx* + âˆš2, then*p*(1) = 0

â‡’ 2(1)^{2} + *k*(1) + âˆš2 = 0

â‡’ 2 + *k + *âˆš2 = 0

â‡’ *k* = -2 - âˆš2 = -(2 + âˆš2)

Therefore, value of *k* is -(2 + âˆš2).

(iii) If *x* - 1 is a factor of polynomial *p*(*x*) = *kx*^{2} - âˆš2*x* + 1, then*p*(1) = 0

â‡’ *k*(1)^{2} - âˆš2(1) + 1 = 0

â‡’ *k* - âˆš2 + 1 = 0

â‡’ *k* = âˆš2 - 1

Therefore, value of *k* is âˆš2 - 1.

(iv) If *x* - 1 is a factor of polynomial *p*(*x*) = *kx*^{2} - 3*x* + *k*, then*p*(1) = 0

â‡’ *k*(1)^{2} + 3(1) + *k* = 0

â‡’ *k* - 3 + *k* = 0

â‡’ 2*k* - 3 = 0

â‡’ *k* = 3/2

Therefore, value of *k* is 3/2.

**4. Factorise: (i) 12 x^{2} + 7x + 1 (ii) 2x^{2} + 7x + 3 (iii) 6x^{2} + 5x - 6 (iv) 3x^{2} - x - 4 **

(i) 12

= 12

= 4

= (3

(ii) 2

= 2

= 2

= (

(iii) 6

= 6

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

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

(iv) 3*x*^{2} - *x* - 4

= 3*x*^{2} - 4*x *+ 3*x* - 4

= *x* (3*x* - 4) + 1 (3*x* - 4)

= (3*x* - 4) (*x* + 1)

**5. Factorise: (i) x^{3} - 2x^{2} - x + 2**

**(ii) x^{3} - 3x^{2} - 9x - 5**

(iii) x^{3} + 13x^{2} + 32x + 20

**(iv) 2 y^{3} + y^{2} - 2y - 1**

(i) Let

Factors of 2 are Â±1 and Â± 2

By trial method, we find that

So,

Now,

Therefore, (

*p*(*x*) = *x*^{3} - 2*x*^{2} - *x* 2*p*(*-1*) = (-1)^{3} - 2(-1)^{2} *-* (-1) 2 = -1 -2 1 2 = 0

Therefore, (*x* 1) is the factor of *p*(*x*)

Now, Dividend = Divisor Ã— Quotient + Remainder

*(x+1) (x ^{2} - 3x + 2)*

*= (x+1) (x ^{2} - x - 2x + 2)*

*= (x+1) {x(x-1) -2(x-1)}*

*= (x+1) (x-1) (x+2)*

(ii) Let *p*(*x*) = *x*^{3} - 3*x*^{2} - 9*x* - 5

Factors of 5 are Â±1 and Â±5

By trial method, we find that*p(5)* = 0

So, *(x-5)* is factor of *p*(*x*)

Now,*p*(*x*) = *x*^{3} - 2*x*^{2} - *x* + 2*p*(*5*) = (5)^{3} - 3(5)^{2} *-* 9(5) - 5 = 125 - 75 - 45 - 5 = 0

Therefore, (*x*-5) is the factor of *p*(*x*)

Now, Dividend = Divisor Ã— Quotient + Remainder

*(x-5) (x ^{2} + 2x + 1)*

*= (x-5) (**x ^{2} + x + x + 1*)

*= **(x-5)* {x(x+1) +1(x+1)}

*= **(x-5)* *(x+1) **(x+1)*

(iii) Let *p*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20

Factors of 20 are Â±1, Â±2, Â±4, Â±5, Â±10 and Â±20

By trial method, we find that*p(-1)* = 0

So, *(x+1)* is factor of *p*(*x*)

Now,*p*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20*p*(*-1*) = (-1)^{3} + 13(-1)^{2} *+ 32*(-1) + 20 = -1 + 13 - 32 + 20 = 0

Therefore, (*x*+1) is the factor of *p*(*x*)

Now, Dividend = Divisor Ã— Quotient + Remainder

*(x+1) (x ^{2} + 12x + 20)*

*= (x+1) (** x^{2} + 2x + 10x + 20*)

*= **(x-5)* {x(x+2) +10(x+2)}

*= **(x-5)* *(x+2) **(x+10)*

(iv) Let *p*(*y*) = 2*y*^{3} + *y*^{2} - 2*y* - 1

Factors of ab = 2Ã— (-1) = -2 are Â±1 and Â±2

By trial method, we find that*p(1)* = 0

So, *(y-1)* is factor of *p*(*y*)

Now,*p*(*y*) = 2*y*^{3} + *y*^{2} - 2*y* - 1*p*(*1*) = 2(1)^{3} + (1)^{2} *- 2*(1) - 1 = 2 +1 - 2 - 1 = 0

Therefore, (*y*-1) is the factor of *p*(*y*)

Now, Dividend = Divisor Ã— Quotient + Remainder

*(y-1) (2y ^{2} + 3y + 1)*

*= **(y-1)* (*2y^{2} + 2y + y + 1*)

*= ** (y-1)* {2y(y+1) +1(y+1)}

*= **(y-1)**(2y+1) **(y+1)*

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