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NCERT Class 9 Mathematics Chapter 2 Polynomials
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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.
(iii) 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 + y2 + 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 + y2 + 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
| 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
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 x 7 = 10000 + 1000 + 21 = 11021
(ii) 95 × 96
Ans: 95 x 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 + 8z x
(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 +2 ab + 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 -2 76³ – 144a²b +1 08ab²
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)
