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**SEBA Class 9 Mathematics Chapter 2 Polynomials**

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**Solutions****SEBA Class 9 Mathematics Chapter 2 Polynomials****Polynomials**

**Polynomials****Chapter – 2**

Exercise 2.1 |

**Q.1. Which of the following expressions are polynomials in one vari-able and which are not? State reasons for your answer.**

**(i) 4x²-3x+7 **

Ans: Yes, 4 x²-3 x+7 is a polynomial of one variable. As x has degree 2 in it and only x is one variable.

**(ii) y²+√2**

Ans: Yes, y is only one variable.

**(iii) 3√t+t√2**

Ans: No as 3√t+r√2 can be written as 3,t¹/² +√2 Here, the exponent of t in 3 t ¹/² which is not a whole number.

**(iv) y+2/y **

Ans: No, as y+2/y can be written as y+2 y¹ where the exponent of y in 2/y is -1, which is not a whole number.

**(v) x¹⁰+y³+t⁵⁰**

Ans: Yes, It is a polynomial in three variables, x, y and t.

**Q.2. Write the coefficients of x² in each of the following:**

**(i) 2 + x² + x**

Ans: We have given that the equation 2 + x²+ x We can also write 1x²+ 1x + 2

Therefore, the coefficient of x² in this equation is 1.

**(ii) 2 – x² + x³**

Ans: We have given that the equation: 2 – x² + x³ We can also write x³- x² + 2

Therefore, the coefficient of x² in this equation is -1.

(i**ii) π/2 x² + x**

Ans: We have given that the equation π/2x² + x

Therefore, the coefficient of x² in this equation is 11/.7

**(iv) √2x- 1**

Ans: We have given that the equation √2x -1

We can also write ox²+ √2x) – 1

Therefore, the coefficient of x² in this equation is 0.

**(v) (2x – 3)(x² – 3x + 1)**

Ans: (2x – 3)(x² – 3x + 1)

= 2x(x²- 3x + 1) – 3(x² – 3x + 1)

= 2x³- 6x² + 2x – 3x² + 9x – 3

= 2x³ – 9x² + 11x – 3

∴ Coefficient x ^ 2 = – 9

**Q. 3. Give one example each of a binomial of degree 35, and a mono- mial of degree 100.**

Ans: We know that polynomials having only two terms are called binomials.

Therefore, the example of a binomial of degree 35 is ax³⁵ + b where a and b are any real number.

Again, the example of a monomial of degree 100 is ax¹⁰⁰ where a is any real number.

**Q.4. Write the degree of each of the following polynomials:**

**(i) 5x³ + 4x²+ 7x **

Ans: We know that the highest power of variable in a polyno- mial is called degree of the polynomial, In polynomial 5x³ + 4x² + 7x

The highest power of variable x is 3.

Therefore, the degree of the polynomial 5x³ + 4x² + 7x is 3.

**(ii) 4 – y²**

Ans: In polynomial 4 – y² the highest power of the variable y is 2.

Therefore, the degree of the polynomial 4 – y² is 2.

**(iii) 5t -√7**

Ans: In polynomial 5t – √5 the highest power of the variable t is 1.

Therefore, the degree of polynomial 5t -√5 is 1.

**(iv) 3hh.**

The only term here is 3 which can be written as 3x⁰ .So the highest power of the variable x is 0.

Therefore, the degree of the polynomial 3 is 0.

**Q.5. Classify the following as linear, quadratic and cubic polynomials.**

**(i) x²+ x**

Ans: In polynomial x² + x the highest power of the variable x is 2.

So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called a quadratic polynomial.

Therefore, the polynomial x² + x is a quadratic polynomial.

**(ii) x – x³**

Ans: In polynomial x – x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.

We know that the polynomial of degree 3 is called a cubic polynomial.

Therefore, the polynomial x – x³ is a cubic polynomial.

**(iii) y + y²+ 4**

Ans: In polynomial y + y^{2} + 4 the highest power of the variable y is 2. So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.

Therefore, the polynomial y + y^{2} + 4 is a quadratic polynomial.

**(iv) 1 + x**

Ans: In polynomial l + x the highest power of the variable x is 1. So the degree of the polynomial is 1.

We know that the polynomial of degree 1 is called linear polynomial.

Therefore, polynomial 1 + x is a linear polynomial.

**(v) 3t**

Ans: In polynomial 3t, the highest power of the variable t is 1. So, the degree of the polynomial is 1.

We know that the polynomial of degree 1 is called a linear polynomial.

Therefore, polynomial 3t is a linear polynomial

**(vi) r²**

Ans: In polynomial r² the highest power of the variable r is 2. So, the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.

Therefore, polynomial r ^ 2 is a quadratic polynomial.

**(vii) 7x²**

Ans: In polynomial 7x³ the highest power of the variable x is 3. So the degree of the polynomial is 3. We know that the polynomial of degree 3 is called a cubic polynomial.

Therefore, polynomial 7x³ is a cubic polynomial.

**Q. 6. Which one is a polynomial?**

**(a) x- + 3x – 1**

**(b) y + 1/y**

**(c) 3y/2 + x**

**(d) 2√x+ 1**

Ans: (c) 3y/2 + x

Exercise 2.2 |

**Q.1. Find the value of the polynomial 5x-4x²+3 at**

**(i) x = 0 **

Ans: Let P(x) = 5x-4 x² – 3

Therefore, P(0) = 5 (0) – 4 (0) + 3 = 0 – 0 + 3 = 3

So, the value of P(x) at x = 0 is 3.

**(ii) x = – 1 **

Ans: Let P(x)=5x-4x²+3

Therefore, P(-1)=5(-1)-4(-1)²+3=6

So the value of P(x) at x=-1 is -6

**(iii) x = 2**

Ans: Let P(x)=5x-4x²+3

Therefore, P(2)=5×2-4(2)²+3=10-16+3=-3

So, the value of P(x) at x=2 is-3

**Q.2. Find p(¹⁰) and p(²) for each of the following polynomials:**

**(i) p(y )= y² – y + 1**

Ans: We have given P(y) = y²-y+1

Therefore, the value of polynomial p(y) at y = 0 is

P(0)=0²-0+1=1

Again, the value of polynomial p(y) at y = 1 is P(1)=1²-1+1=1

p(2) = 2²-2+1=3

**(ii) p(t) = 2+1+2r²-t³**

Ans: We have given, p(t) = 2+t+2r²-t³

Therefore, the value of polynomial p(t) at 1 = 0 is

P(0)=2+0+2.(0)² – (0)³ = 2

Again, the value of polynomial p(t) at t = 1 is

p(1)=2+1+2(1)² – (1)³=4

Again, the value of polynomial p(t) at t = 2 is

p(2)=2+2+2 (2)² – (2)³ = 4

**(iii) p(x) = x³**

Ans: We have given, p(x) = x³

Therefore, the value of polynomial p(x) at x=0 is

P(0)=(0)³ = 0

Again the value of polynomial p(x) at x = 1 is

p(1) = (1)³ = 1

Again, the value of polynomial p(x) at x = 2 is is

P(2)=(2)³=8

**(iv) p(x)=(x-1)(x+1)**

Ans: We have given p(x)=(x-1)(x+1)

Therefore, the value of polynomial p(x) at x=0 is

P(0)=(0-1)(0+1)=(-1)x(+1)=-1

Again, the value of polynomial p(x) at x = 1 is

p(1) = (1-1)x(1+1)=0x2=0

Again, the value of polynomial p(x) at x = 2 is

p(2)=(2-1) (2+1)=1×3=3

**Q.3. Verify whether the following are zeros of the polynomial indi-cated against them:**

**(i) P(x)=3x+1,x=-1/3**

Ans: We have given that, p(x) = 3x + 1

Therefore, the value of polynomial p(x) at x = – 1/3 is

Yes x = – ⅓ is the zero of polynomial p(x)

**(ii) P(x) = 5x − π, x = -4/5**

Ans: x = – We have given that, p(x) = 5x – π

Therefore, the value of polynomial p(x) at x = 4/5 is

No, x = 4/5 is not the zero of p(x) = 5x – π.

**(iii) p(x) = x²-1, x = 1,-1**

Ans: We have given that, p(x) = x² – 1

Therefore, the value of polynomial p(x) at x = 1 is

p(1) = 1²- 1 = 0

Again, the value of polynomial p(x) at x = – 1 is

p(- 1) = (- 1)²- 1 = 0

Yes, x = 1, -1 are the zero of polynomial p(x) = x²- 1

**(iv) p(x) = (x+1)(x-2), x = – 1,2**

Ans: We have given that, p(x) = (x + 1)(x – 2)

Therefore, the value of polynomial p(x) at x = – 1 is

p(- 1) = (- 1 + 1)(- 1 – 2) = 0(- 3) = 0

Again, the value of polynomial p(x) at x = 2 is

p(2) = (2 + 1)(2 – 2) = 3 x 0 = 0

Yes x = – 1, 2 are the zero of polynomial

p(x) = x + 1(x – 2)

Yes x = – 1/3 is the zero of polynomial p(x)

**(v) p(x) = x², x=0**

Ans: We have given that, p(x) = x²

Therefore, the value of polynomial p(x) at x = 0 is

p(0) = 0² = 0

Yes x = 0 is the zero of polynomial p(x) = x²

**(vi) P(x)=lx+m, x=-m/1**

Ans: We have given that, p(x) = lx + m

Therefore, the value of polynomial p(x) at x = – m is

Yes, x = – m/l is the zero of polynomial p(x) = lx + m

**(vii) p(x)=3x²-3x²-1,x=1/√3,2-√3**

Ans: We have given that, p(x) = 3x² – 1** **

Therefore, the value of polynomial p(x) at x = – 1/√3 is

Again, the value of polynomial p(x) at x = 2/√3 is

Therefore, x = – 1/√3 is the zero of polynomial and p(x) and x = 2/√3

is not the zero of polynomial p(x).

**(viii) p(x) =2x+1,x=1/2**

Ans: We have given that, p(x) = 2x + 1

Therefore, the value of polynomial p(x) at x = 1/2 is

Therefore, x = 1/2 is not the zero of polynomial p(x).

**Q.4. Find the zero of the polynomial in each of the following cases:**

**(i) p(x) = x+5 **

Ans: We have given that, p(x) = x+5 ..(1)

To find the zero of polynomial p(x), we can take p(x) = 0…(2)

From equations (1) and (2)

x+5=0

∴ x=-5

Therefore, x = -5 is the zero of the polynomial p(x)=x+5.

**(ii) p(x) = x-5**

Ans: We have given that, p(x)=x-5 …(1)

To find the zero of polynomial p(x) we can take p(x) = 0 …(2)

From equations (1) and (2)

x-5=0

∴ x=5

Therefore x = 5 is the zero of polynomial p(x)=x-5.

**(iii) p(x) = 2x+5 **

Ans: To find the zero of polynomial p(x), we can take p(x) = 0 …(2)

From equations (1) and (2)

Therefore, x = – 5/2 is the zero of polynomial p(x) = 2x+5

**(iv) p(x) = 3x – 2**

Ans: We have given that, p(x)=3x-2 …(1)

To find the zero of polynomial p(x) we can take p(x)=0 … (2)

Therefore, x=-5/3 is the zero of polynomial p(x)=3x-2

**(v) p(x) = 3x**

Ans: We have given that, p(x)= ax, a ≠ 0 …(1)

To find the zero of polynomial p(x) we can take p(x) = 0 …(2)

From equations (1) and (2)

ax = 0

∴ x = 0/a = 0

Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.

**(vi) p(x) = ax, a ≠ 0**

Ans: We have given that, p(x)=3x …(1)

To find the zero of polynomial p(x) we can take p(x)=0 …(2)

From equations (1) and (2), 3x=0 ∴ x=0

Therefore, x = 0 is the zero of polynomial p(x) = 3x

**(vii) p(x)=cx+d, c ≠ 0, c; d are real numbers.**

Ans: ) We have given that, p(x) = cx + d …(1)

c ≠ 0, c, d are real numbers.

To find the zero of polynomial p(x) we can take

p(x) = 0 …(2)

From equations (1) and (2)

cx + d = 0

x = – d/c x where c ≠ 0x and are real numbers

Therefore, x=-d/c, where c, d and c,d

are real numbers is the solution of polynomial

p(x) = cx + d.

Exercise 2.3 |

**Q. 1. Find the remainder when x³+ 3x²+ 3x+1 is divided by**

**(i) x + 1 **

Ans: Let p(x) = x³ + 3x²+ 3x + 1

(i) x + 1 = 0 Rightarrow x=-1

∴ Remainder = p(- 1) = (- 1)³ + 3 (- 1)² + 3(- 1) + 1 = – 1 + 3 – 3 + 1 = 0

**(ii) x – 1/2 **

Ans:

**(iii) x**

Ans: x.

Remainder = (0)³ +3(0)²+3(0)+1=1

(iv) x+π

x+π=0⇒x=-π

∴ Remainder = (-π)³+3(−π)²+3(−π)+1=−π³ +3π² −3π+1

**(iv) x + π**

Ans: x +** **π** **

X + = 0 ⇒ x = – ** **π

**(v) 5 + 2x**

Ans:** **5+2x

5 + 2 x = 0 ⇒ 2x = – 5 ⇒ x = – 5/2

∴ Remainder =

**Q.2. Find the remainder when x³-ax² +6x-a is divided by x-a**

Ans: Let p(x)=x³-ax²+6x-a

X – a = 0

∴ Remainder = (a)³ – a(a)² + 6(a)-a = a³-a³+6a-a=5a

**Q.3. Check whether 7+3x is a factor of 3x³ +7x**

Ans: 7+3x will be a factor of 3x³ +7x only if 7+3x divides 3x³ +7x leaving no remainder.

Let p(x)=3x³ +7x Now, 7+3x=0 3x=-7 ⇒x=-7/3

∴ Remainder =

∴ 7+ 3x is not a factor of 3x³ +7x

**Q.4. Find the quotient and reminder.**

**(i) x³-4x²+2x+5 is divided by x-2**

Ans:

∴ Quotient=x²-x-2

Reminder=1

**(ii) 4x³-2x²-3 is divided by 4x²-1**

Ans:

∴ Quotient=2x-1

Reminder=2x-4

**(iii) 3x³-5x²+10x-3 is divided by 3x+1**

Ans:

∴ Quotient=x²-2x+4

Reminder=-7

**(iv) x¹¹-5 is divided by x+1**

Ans:

∴ Quotient=x¹⁰-x⁹+x⁸-x⁷+x⁶-x⁵-x⁴-x³+x²-+1

Reminder=-6

**Q.5. (i) Divide 3x²- 2x – 40 by 3x + 10**

Ans:

∴ Quotient = 2x + 10

Reminder = -2x-140

**(ii) Divide 4 + 7x + 7x² + 2x³ by 2x + 1**

Ans:

∴ Quotient=x²+3x+2

Reminder= 2

**Q.6. (i) Is the polynomial – 14x² – 13x + 12 is completely divisible by 2x + 3**

Ans: (i) Let, P(x) = – 14x² – 13x + 12

∴ P(x) is completely divisible by 2x + 3

∴ 2x + 3 is a factor of P(x)

**(ii) Examine if x – 7 is a factor of x³+ 2x² – 3x + 4**

Ans: Let, P(x) = x³ + 2x² – 3x + 4

x – 7 =0 ⇒ x=7

∴ P(7) = 7³+ 2 (7)² – 3×7 + 4

= 343 + 98 – 21 + 4

= 445-21

= 424 ≠ 0

∴ x-7 is not a factor of P(x)

**Q. 7. The polynomial ax³ +3x²+5x and x³-4x-a when divided by x-2, equal reminders are found. Find the value of a.**

Ans: Let, P(x)= ax³+3x² + 5x-4

F(x) = x³-4x-a

x-2=0⇒x=2

∴ P(2)=a (2)³+3(2)²+5(2) – 4

= 8a+ 12+ 10-4 = 8a+ 18

∴ F(2) = 2³-4(2)-a = 8-8-a=-a

According to question, 8a+ 18=-a⇒9a=-18⇒ a=-2

Exercise 2.4 |

**Q.1. Which of the following polynomials has (x+1) a factor:**

**(i) x³ + x²+x+1 **

Ans: (i) Here, p(x) = x³ + x² + x+1 and the zero of x+1 is -1

So, p(-1)=(-1)³ + (-1)² + (-1) + 1=-1 + 1-1 + 1=0

Here, remainder is 0, therefore x+1 is the factor of x³ + x²+x+1

**(ii) x⁴ + x³ + x² + x + 1**

(ii) Here, p(x)=x⁴+x³+x²+x+1 and the zero of x+1 is x-1

So, p(-1) = (-1)⁴ + (-1)² + (-1) (-1)+1=1-1 + 1 – 1 + 1 = 1

Here, remainder is 1, therefore x+1 is not the factor of

X⁴ + x³ + x² + x + 1

**(iii) x⁴ + 3x³ + 3x² + x + 1 **

Ans: Here, p(x) = x⁴ +3x³ +3x²+x+1 and the zero of x+1 is- 1

So,p(-1)=(-1)⁴+3(-1)³+3(-1)²+(-1)+1=1-3+3-1+1=1

Here, remainder is 1, therefore x+1 is not the factor of

x⁴+3x³ +3x²+x+1

**(iv) x³ – x² – (2+√2)x + √2**

Ans: Here, p(x)=x³-x²-(2+√2)(x)+ √2 and the zero of

x+1 is -1

So, p(-1)=(-1)³-(-1)²-(2+√2)(-1)+√2

=1-1+2+√2+√2=2√2

Here, remainder is 2√2, therefore x+1 is not the factor of

x³-x² – (2+ √2)x + √2.

**Q.2. Use the factor theorem to determine whether x³-x² – g(x) is a factor of p(x) in each of the following cases:**

**(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1**

Ans: (i) Here, p(x)=2x³ + x²-2x-1 and the zero of

g(x)=x+1 is -1

So, p(-1)=2(-1)³ +(-1)²-2(-1)-1=-2+1+2+1=0

Here, remainder is zero, therefore g(x)=x+1 is the factor of

p(x)=2x³+x²-2x-1

**(ii) p(x) = x³ – 3x² + 3x + 1 g(x) = x + 2**

Ans: Here, p(x) = x³+3x²+3x+1 and the zero of

g(x)=x+2 is-2.

So,p(-2)=(-2)³+3(-2)²+3(-2)+1=-8+12-6+1=-1

Here, remainder is -1, therefore g(x)=x+2 is not the factor of

p(x)=x³+3x²+3x+1

**(iii) p(x) = x³ – 4x² + x + 6 g(x) = x – 3**

Ans: Here, p(x) = x³ – 4x² + x + 6 and the zero of g(x) = x – 3 is 3

So, p(3) = (3)³ – 4(3)² + 3 + 6 = 27 – 36 + 3 + 6 = 0

Here, remainder is 0, therefore g(x)=x -3 is the factor of

p(x) = x³ + 4x² + x + 6

**Q.3. Find the value of k if x – 1 is a factor of p(x) in each of the following cases:**

**(i) p(x) = x² + x + k**

Ans: We have given that x-1 is the factor of p(x) = x² + x + k

So, (1)² + 1 + k = 0 or,

1+1+k = 0

2+k = 0

∴ k = -2

Therefore, the value of k in p(x) = x²+x+k is -2

**(ii) p(x) = 2x² + kx +√2 **

Ans: We have given that x-1 is the factor of

p(x)=2x² + kx + √2

∴ p(1)=0

Now, p(1)=2(1)² + k(1)+ √2

So, 0=2 + k + √2 or k=-2-√2

or, k=-(2+√2)

Therefore the value of k in p(x)=2x² + kx+√2 is

-(2+√2)

**(iii) p(x) = kx² – √2x + 1 **

Ans: We have given that x-1 is a factor of.

p(x)=kx² -√2x+1

∴ p(1)=0

Now, p(1)= k(1)² -√2(1)+1

So, k-√2+1=0

∴ k = √2-1

Therefore, the value of k in p(x)=kx²-√2x+1 is √2-1

**(iv) p(x) = kx² – 3x + k**

Ans: We have given that x-1 is the factor of

p(x) = kx² – 3x + k

So, k-3 + k = 0 or, 2k-3 = 0. k =0

∴ K=3/2

Therefore the value of k in p(x) = kx² – 3x + k -3/2

**Q.4. Factorise:**

**(i) 12x²7x + 1 **

Ans: 12x² -7x + 1

**(ii) 2x² + 7x + 3**

Ans: 2x² +7x + 3

= 12x² – 4x – 3x + 1 = 2x² + 6x + x + 3

= 4x(3x – 1) – 1(3x – 1) = 2x(x + 3)+1(x + 3)

=(3x – 1) (4x – 1) = (x + 3)(2x + 1)

**(iii) 6x² + 5x-6**

Ans: 6x² + 5x-6

= 6x² + 9x – 4x-6

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

= (2x + 3)(3x – 2)

**(iv) 3x²-x-4**

Ans: 3x² – x-4

= 3x² – 4x + 3x-4

= x(3x – 4) + 1(3x-4)

= (3x – 4)(x + 1)

**(v) 2x² +x-45**

Ans: 2x² + x-45

= 2x² + 10x – 9x-45

= 2x(x + 5)-9(x + 5)

= (x + 5) (2x – 9)

**(vi) y² + 18y + 65**

Ans: y² + 18y – 65

= y² + 13y + 5y + 65

= y(y + 13) + 5(y + 13)

= (y + 13) (y + 5)

**(vii) p² + 14p + 1**

Ans: p² + 14p+13

= p² + 13p + p + 13

= p(p + 13) +1(p + 13)

= (p + 13) (p + 1)

**(viii) -18 + 11x – x²**

Ans: -18+11x – x²

= -18 + 9x + 2x – x²

= -9(2 – x) + x(2 – x)

= (2 – x) (x – 9)

**(ix) 8a² – 22 ab + 15b²**

Ans: 8a²-22ab+15b²

= 8a²-12ab-10ab+15b²

= 4a(2a-3b)-5b(2a-3b)

= (2a-3b)(4a-5b)

**Q.5. Factorise:**

**(i) x³ – 2x² – x + 2**

Ans: Let p(x)=x³-2x²-x+2

The factors of 2 are ±1, ±2

By trial, we find that p(2)=0

So, (x-2) must be the factor of p(x)

Therefore,

x³-2x²-x+2=(x-2)(x²-1)=(x-2)(x+1)(x-1)

[∵ a²-b²=(a+b)(a-b)]

∵ x³-2x²-x+2=(x-2)(x-1)(x+1)

**(ii) x³ – 3x² – 9x – 5**

Ans: Let p(x)=x³-3x²-9x-5

The factors of -5 are ±1, ±5

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

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

Therefore, x³-3x²-9x-5

= (x+1)(x²-4x-5)=(x+1)[x²-5x+x-5]

= (x+1)[x(x-5)+1(x-5)]

∴ x³-3x²-9x-5= (x+1)(x+1)(x-5)

**(iii) x³ + 13x² + 32x + 20 **

Ans: Let p(x)=x³ +13x² +32x+20

The factor of 20 are ±1, ±2, ±4, ±5,±10,±20

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

So, (x+1) must be the factors of P(x).

Therefore,

x³ + 12x² + 32x + 20

= (x+1)(x²+12x+20)= (x+1)[x²+10x+2x+20]

= (x+1)[x(x+10)+2(x+10)]

∴ x³ +13x²+32x+20=(x+1)(x+10)(x+2)

**(iv) 2y³ + y²-2y-1**

Ans: Let p (y)=2y³ + y²-2y-1

The factor of -1, are ±1

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

So, (y-1) must be the factor of 2y³ +y²-2y-1

Therefore, 2y³ + y²-2y-1= (y-1) (2y²+3y+1)

= (y-1) [2y²+2y+y+1]=(y-1) [2x(x+1)+(y+1)]

= 2y³ + y²-2y-1=(y-1) (y+1) (2y+1)

**(v) x³ + x² – x – 1**

Ans: Let, P(x) = x³ + x²-x-1

P(1) = 1³+ 1²-1-1-2-2=0

∴ x-1 is a factor of P(x)

∴ P(x) = (x-1) (x² + 2x + 1) = (x-1) {(x)² +2. x. 1+1}²

= (x-1) (x+1)²

= (x − 1) (x + 1)(x+1)

**(vi) x³ + x² + x + 1**

Ans: Let, P(x) = x³+x²+x+1

P(-1) = (-1)³+(-1)² + (-1)+1

=−1 + 1 − 1 + 1=0

∴ (x+1) is a factor of P(x)

**(vii) x³+2x²-x-2**

Ans: Let, P(x) = x³+x²+x+1

P(-1) = (-1)³+(-1)² + (-1)+1

=−1 + 1 − 1 + 1=0

∴ (x+1) is a factor of P(x)

∴ P(x)=(x+1)(x² + 1)

**(viii) x³+3x²-7x-6**

Ans: Let, P(x) = x³+ 3x²- 7x – 6

P(2) = 2³+ 3(2)²- 7(2) – 6

= – 8 + 12 – 14 – 6 = 20 – 20 = 0

∴ x – 2 is a factor of P(x)

∴ P(x) = (x – 2)(x²+ 5x + 3)

**(ix) 3x³+5x²-16x-2**

Ans: Let, P(x) = 3x³ + 5x² – 16x – 12

P(- 3) = 3(- 3)³ + 5(- 3)²- 16(- 3) – 12

= – 81 + 45 + 48 – 12

= 93 – 93-0

∴ x + 3 is a factor of P(x)

∴ P(x) = (x + 3)(3x²- 4x – 4)

= (x + 3)(3x²- 6x + 2x – 4)

= (x + 3){3x(x – 2) + 2(x – 2)}

= (x + 3)(x – 2)(3x + 2)

**Q.6. If x + a is a common factor of the polynomials x²+ px + q and**

Ans: x + a is a common factor of the polynomials x²+ px + q and x² + mx + n

Let, F(x) = x²+ Px + q

G (x) = x²+ mn + n

x+a=0⇒x=-a

Now F a)=(-a)²+p(-a)+q=a²-pa+q

G(-a) =(-a)²+m(-a)+n=a²-ma+b

∴ a²- pa + q = a²- ma + n

⇒ – p +a + ma = n – q

⇒ a(m – p) = n – q a -q

Exercise 2.5 |

**Q.1. Use suitable identities to find the following products:**

**(i) (x + 4)(x + 10)**

Ans: (i) (x + 4)(x + 10)

[∵(x+a)(x+b)=x²+(a+b)x+ab]

Therefore, (x+4)(x+10)=x²+(4+10)x+4×10= x²+14x+40

**(ii) (x + 8)(x – 10)**

Ans: (x+8)(x-10) or (x+8)(x+(-10))

We know that, (x+a)(x+b)= x²+(a+b)x+ab

∴ (x+8)(x+(-10))= x² +(8+(-10))x+8x(-10)

= x²-2x-80

**(iii) (3x + 4)(3x – 5)**

Ans: (3x+4)(3x-5) or (3x+4) (3x+(-5))

We know that, (x+a)(x+b)= x²+(a+b)x+a.b

∴ (3x+4) (3x-5)=(3x)2 + (4+ ((5))3x+4x (-5)= 9x²-3x-20

Ans: We know that, (x+y)(x-y)=x²-y²

**(v) (3 – 2x)(3 + 2x)**

Ans: (3-2x) (3+2x)

We know that, (x-y)(x+y)= x²-y²

∴ (3-2x) (3+2x)=(3)² – (3)² – (2x)² = 9-4x²

**Q.2. Evaluate the following products without multiplying directly:**

**(i) 103 × 107**

Ans: 103×107

We can also write (100+3)x(100+7)

We know that, (x+1)(x+6)=x²+(a+b)x+a.b

∴ (100+3)(100+7)

= (100)²+(3+7)×100+3×7 = 10000+1000+21=11021

**(ii) 95 × 96 **

Ans: 95×96 = (90+5)x(90+6)

We know that, (x+a)(x+b) = x²+(a+b)x+a.b

∴ (90+5)x(90+6)

= (90)² +(5+6)×90+5×6=8,100+990+30= 9,120

**(iii) 104 × 96**

Ans:104×96

We can also write, (104+4)x(100-4)

We know that, (x + y)(x − y) = x²- y²

∴ (100+4)(100-4)= (100)²-(4)²=10000-16=9,984

**Q.3. Factorise the following using appropriate identities:**

**(i) 9x² + 6xy + y² **

Ans: 9x²+6xy+ y²

We can also write, (3x)² +2.3x. y+(y)²

We know that, a²+2ab+b² = (a+b)²

∴ (3x)²+2.3x.y + (y)²=(3x+y)²=(3x+y)

(3x+y)

**(ii) 4y² – 4y + 1 **

Ans: (4y)2-2.2y.1+ (1)²

We can also write, (2y)² – 2.2y.1+ (1)²

We know that, x²-2xy + y² = (x-y)²

∴ (2y)² – 2.2y.1+(1) = (2y – 1)=(2y – 1) (2y-1)

Ans:

We know that, a²-b² = (a+b)(a-b)

**Q.4. Expand each of the following using suitable identities:**

**(i) (x + 2y + 4z)² **

Ans: (i) We are given that, (x+2y+4z)²

We know that, (a+b+c)² = a²+b² +c²+2ab+2bc+2ca

∴ (x+2y+4z)²=(x)² +(2y)²+(4z)²+2.x.

2y+2.2y.4z+2.4z.x

= x²+4y²+16z²+4xy+16yz+8zx

**(ii) (2x – y + z)² **

Ans: We are given that, (2x-y+ z)²

We can also write, (2x + (y)+ z)²

Now, we know that, (a+b+c)²

= a²+b²+c²+2ab+2bc+2ca

∴ {2x+(-y)+z}² = (2x)²+(y)²+z²

+2.2x. (-y)+2. (y). z+2. z. 2x

= 4x² + y²+z²-4xy-2yz+4zx

**(iii) (-2x + 3y + 2z)² **

Ans: We are given that, (-2x + 3y + 2z)²

We can also write, {(-2x) + 3y + 2z}²

Now, we know that,

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

Therefore, (-2x + 3y + 2z)²=(-2x)² + (3y)² + (2z)²

+2.(-2x)(3y) + 2(3y)(2z) + 2.2z.(-2x)

= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx

**(iv) (3a-7b-c)²**

Ans: We are given that, (-3a-7b-c)²

We can also write, (3a +(-7b)+(-c)²

Now, we know that,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Therefore, (3a.+(-7b)+(-c))² = (3a)² +(-76)²

+(-c)² + 2.3a (-7b) +2. (-7b) (-c)+2.(c).3a

3a9a² + 49b² + c²-42ab +14bc – 6ca

**(v) (-2x+5y-3z)²**

Ans: We are given that, (-2x+5y-3z)²

We can also write, [(-2x+5y-3x)]²

Now, we know that, (a+b+c)²

= a² + b² + c² + 2ab + 2bc + 2ca

Therefore,

[(-2x)+5y + (-32)]² = (-2x)² +(5y)²

+ (-3z)² +2.(-2x). 5y+2.5y

(-3z) (-2x) z+2(-3z) (-2x)

= 4x²+25y²+9z2-20xy-30yz+12zx

Ans:

Now, we know that,

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

**Q.5. Factorise:**

**(i) 4x²+9y²+16z² +12xy-24yz-16xz**

Ans: We are given,

4x² +9y²+16z² +12xy-24yz-16xz

We can also write,

(2x)²+(3y)²+(-4z)² +2.2x. 3y+2.3y.

(-4z)+2.(-4z). 2x

Now we know that,

a²+b²+c²+2ab+2bc+2ca = (a+b+c)²

Therefore,

(2x)² + (3y)² + (-4z)² + 2.2x3y + 23y(-4z) + 2.(-4z). 2x

=(2x + 3y + (-4z))² = (2x + 3y-4z)² = (2x + 3y-4z)(2x + 3y – 4z)

**(ii) 2x² + y² + 8z² – 2√2 xy + 4√2xy + 4√2yz – 8xz**

Ans: We are given that,

2x² + y²+8z²-2√2xy +4√2yz-8xz

We can also write,

(-√2x)²+(y)² +(2√2z)²+2(-√2x). y+2. y. (2√2z)

+2. (2√2z). (-√2z)

Now, we know that,

a²+b²+c²+2ab+2bc+2ca = (a+b+c)²

Therefore,

(-√2x)²+ (y)²+ (2√2z)²+2.(-√2x)y

+2.y.(2√2 z)+2. (2√2z)(- √2x)

= ((-√2x)+ y + 2√2z)² = (-√2x+ y + 2√2z)

(-√2x + y + 2√2z)

**Q.6. Write the following cubes in expanded form:**

**(i) (2x + 1)³**

Ans: We are given that, (2x + 1)³

We know that, (a + b)³ = a³ + b³ + c³ + 3ab + (a + b)

Therefore, (2x+1)³ =(2x)³ +(1)³ +3. (2x)(1)(2x + 1)

= 8x³ + 1 + 6x(2x + 1) = 8x³ + 12x²+ 6x + 1

**(ii) (2a – 3b)³**

Ans: We are (2a – 3b)³

We know that, (a – b)³= a³ – b³ – 3ab(a – b)

∴ (2a – 3b)²= 8a³ – 27b³ – 18ab(2a – 3b)

= 8a³- 27b³- 336a²* b + 54ab²

Ans:

Ans: We know that,(a+b)³=a³+b²+3ab(a+b)

Therefore,

Ans: We know that, (a – b)³= a³- b³ – 3ab(a – b)

Therefore,

**Q.7. Evaluate the following using suitable identities:**

**(i) (99)³ **

Ans: We are given, (99)³

We can also write, (100-1)³

We know that, (a-b)³ = a³-b³-3ab (a-b)

∴ (100-1)³ =(100)³ – (1)³-3.100.1(100-1)

= 10,00,000-1-30000+300=9,70,299

**(ii) (102)³**

Ans: We are given that, (102)³

Now, we know that, (a+b)³= a³+b³+3ab (a+b)

∴ (100+2)³=(100)³+(2)³ +3.100.2 (100+2)

= 10,00,000+8+600 (100+2)

= 10,00,000+8+60000+1200 = 10,61,208

**(iii) (998)³**

Ans: We are given that, (998)³

We can also write, (1000-2)³

Now, we know that, (a-b)³ = a³-b³-3ab (a-b)

∴ (1000-2)³=(1000)³-(2)³-3.1000 2(1000-2)

= 1,00,00,000-8-6000 (100-2) = 99,40,11,992

**Q.8. Factorise each of the following:**

**(i) 8a³ +b³ +12a²b+6ab² **

Ans: (i) We are given that, 8a³ +b³ +12a²b+6ab²

We can also write, (2a)³ + (b)³+6ab (2a+b)

or (2a)³ +b²+3.2a.b (2a+b)

We know that, a³ +b³ +3ab (a+b)=(a+b)³

∴ (2a)³+(b)³ +3.2a.b (2a+b)=(2a+b)³

**(ii) 8a³-b³-12a²b+6ab²**

Ans: We are given that, 8a³-b³-12a²b+bab²

We can also write, (2a)³-(b)³-6ab (2a-b)

or (2a)³-b³-3.2a.b (2a-b)

We know that, a³-b³-3ab (a-b)=(a-b)³

∴ (2a)³-(b)³-3.2a.b (2a-b)=(2a-b)³

**(iii) 27 – 125a³ – 135a + 225a²**

Ans: We are given that, 27-125a³-135a+225a²

We can also write, (3)³-(5a)³-45a (3-5a)

or, (3)³-(5a)³-3.3.5a (3-5a)

We know that, a³-b³-3ab (a-b) = (a+b)³

∴ (3)³ – (5a)³ – 3.3.5a (3 – 5a)

=(3-5a)³=(3-5a) (3-5a) (3-5a)

**(iv) 64a³-276³-144a²b+108ab²**

Ans: We are given that, 64a³-27b³-144a²b+108ab²

We can also write, (4a)³ -(36)³-36ab (4a-3b)

or, (4a)³-(3b)³-3.4a. 3b (4a-3b)

Now, we know that, a³-b³-3ab (a – b)=(a-b)³

∴ (4a)³-(36)³-3.4a.3b (4a-3b)=(4a-3b)³

Ans: we are given that,

we can also write,

we know that a³-b³-3ab(a-b) =(a-b)³

**Q.9. Verify:**

**(i) x³ + y³ = (x + y) (x²-xy + y²)**

Ans: R.H.S.= (x+y)(x²-xy + y²)

By actual multiplication we have

X³ -x²y + xy² – xy² + x²y + y³ = x³ + y³ = L.HS.

**(ii) x³-y³ = (x-y)(x² + xy + y²)**

Ans: R.H.S.= (x+y)(x²-xy + y²)

by actual multiplication, we have,

X³- x²y + xy² – xy² + x²y + y³ = x³ + y³ = L.H.S

**Q. 10. Factorise each of the following:**

**(i) 27y³ +125z³ **

Ans: We are given, 27 y³ +125z³

We can also write, (3y)³ +(5z)³

We know that, a³+b³=(a+b) (a²-ab+b²)

Therefore, (3y)³ + (5z)³ = (3y+5z) ((3y)² – (3y). (5z)+(5z)²)

= (3y + 5z) (9y² -15yz + 25z²)

**(ii) 64m³ – 34.3n³**

Ans: We are given, 64m³-343n³

We can also write, (4m)³ -(7n)³

We know that, a³-b³ = (a-b)(a²+ab+b²)

Therefore,

(4m)³-(7n)³=(4m-7n) ((4m)²+(4m) (7n)+(7n)²)

= (4m-7n) (16m² +28 mn+49n²)

**Q. 11. Factorise: 27x³+ y³+z³-9xyz**

Ans: We have given, 27x³ + y³ + z³-9xyz

We can also write, (3x)³ +(y)³ + (x)³ – 3. (3x). (y). (z)

We know that, a³ + b³ + c³ – 3abc = (a + b + c)

(a² + b² + c² – ab – bc – ca)

Therefore,

(3x)³+(v)³+ (z)³-3. (3x) (y)(z)=(3x+y+z) [(3x)²+(y)² +(z)²-(3x). (y)-(y)(z)-(z) (3x)]

= (3x+y+z) (9x² + y²+z²-3xy-yz-3zx)

**Q.12. Verify that**

x² + y²+z² −3xyz=1/2(x+y+z) [(x−y)² +(v−z)² +(z−x)²]

Ans: We are given L.H.S. is x³ + y³ +z³-3xyz

We know that, x³ + y³+z³-3xyz

=(x+y+z) (x² + y²+z²-xy-yz-zx)

=(x+y+z)2×1/2(x²+y²+z²-xy-yz-zx)

= 1/2 (x+y+z) (2x²+2y²+2z² -2xy-2yz-2zx)

=1/2(x + y + z) (( x − y)² + (y + z)² + (z−x)²)

Hence verified.

**Q. 13. (i) If x+y+z = 0, show that x³+ y³+z³ = 3xyz**

Ans: We know that, x³ + y³ + z³-3xyz

= (x + y + z) (x² + y²+z²-xy-yz-zx)

Here, we have given x+y+z=0 ∴ x³+y³ +z³-3xyz

= 0x (x² + y²+z²-xy-yz-zx)

or, x³ + y³ + z³-3xyz = 0

∴ x³+y³+z³ = 3xyz

**(ii) If a+b+c=0, then show that**

Ans: a²(b+c)+b²(c+a)+c²(a + b) + 3abc = 0

Ans: a+b+c=0

∴ a+b=-c

b+c=-a

c+a=-b

L.H.S. =a²(b+c)+ b²(c+a)+ c²(a+b)+3abc

=a²(-a)+ b²(-b) + c²(-c) + 3abc

=-a³-b³-c³+ 3abc

=-(a³+b³+c³-3abc) [ ∵ a+b+c=0]

=0=R. H.S.

**(iii) If 2a-b+c=0, then show that 4a²b² + c² + 4ac = 0**

Ans: 2a-b+c=0 ⇒2a+c=b

⇒ (2a+c)²=(b)² [squaring both sides]

⇒ 4a²+2.2a.c+c²= b²

⇒ 4a²-b²+c²+4ac=0 Proved.

**(iv) If a+b+c=0, then show that (b+c)² (c+a)² (a+b)²**

Ans: a+b+c=0

∴ a + b=-c

b+c=-a

c+a=-b

**(v) If a² + b² + c² -ab-bc-ca = 0 then show that, a=b=c**

Ans: a²+b²+c²-ab-bc-ca=0

**Q.14. Without actually calculating the cubes, find the value of each of the following:**

**(i) (-12)³+(7)³+(5)³ **

Ans: (i) We are given that, (-12)³+(7)³ +(5)³

First we check the value of x+y+z we have -12+7+5=0

We know that if x+y+z=0 then x³+y³ + z³ = 3xyz

∴ (-12)³+(7)³ +(5)³ =3x(-12)x7x5 = -1260

**(ii) (28)³+(-15)³+(-13)³**

Ans: (28)³+(-15)³+(-13)³

First we check the value of x+y+z we have,

28+(-15)+(-13)=0

We know that if x + y + z = 0 then, x³ + y³+z³ = 3xyz,

x³+y³+z³ = 3xyz ∴ (28)³+(-15)³+(-13)³

= 3x28x(-15)x(-13)=16,380

**Q.15. Give possible expressions of the length and breadth of each of the following rectangles in which their areas are given.**

**(i) Area: 25a²-35a+12 **

Ans: We know that, Area of rectangle = length X breadth …(1)

But, we have given that

Area of rectangle = 35a² +35a+12

From equations (1) and (2)

lxb=25a²-35a+12=25a²-35a+12 …(2)

= (5a-4) (5a-3)

∴ Length (I)=(5a-3)

**(ii) Area: 35y²+13y-12**

Ans: We know that, Breadth (b) = (5a-4)

Area of rectangle = length x breadth …(1)

But, we have given that

Area of rectangle = 35y²+13y-12 …(2)

From equations (1) and (2)

Length x breadth= 35y²+ 13y – 12 = (5y + 4)(7y -3)

or, Length X breadth = (5y + 4) (7y -3)

Therefore, length = (5y + 4), breadth =(7y -3)

**Q.16. What are the possible expression for the dimension of the cuboids whose volumes are given below:**

**(i) volume= 3x²- 12x **

Ans: (i) We know that, volume of cuboid= length X breadth X height

But we have given that,

volume of cuboid = 3x²-12x =3x² -12x

or, length X breadth X height =3×x× (×–4)

∴ length of cuboid =3

breadth of cuboid = X height of cuboid = (x-4)

**(ii) volume = 12ky² + 8ky -20k?**

Ans: We know that, volume of cuboid = length x breadth X height. (1)

But, we have given that

volume of cuboid= 12ky²+ 8ky- 20k … (2)

From equation (1) and (2)

Length x breadth x height = 12ky² + 8ky – 20k

= 4k(3y² +2y-5) = 4k [y(3y +5) – 1(3y+ 5)]

or, length X breadth X height = 4k(3y + 5) (y-1)

∴ length of cuboid = 4k

breadth of cuboid = (3y + 5)

and height of cuboid = (y-1)