Write the correct answer in each of the following:
Q.1. Which one of the following is a polynomial?
(a)
(b)
(c)
(d)
Correct Answer is option (c)
The power of x must be a nonnegative integer in a polynomial.
Hence, option C is correct.
As we can write option c as
Q.2. √2 is a polynomial of degree
(a) 2
(b) 0
(c) 1
(d) 1/2
Correct Answer is option (b)
We can write it as √2 = √2 × x^{0}
Thus, degree is 0.
Q.3. Degree of the polynomial 4x^{4 }+ 0x^{3} + 0x^{5} + 5x + 7 is
(a) 4
(b) 5
(c) 3
(d) 7
Correct Answer is option (a)
Degree of the polynomial 4x^{4 }+ 0x^{3} + 0x^{5} + 5x + 7 is 4
We know that, the degree of given polynomial 4x^{4 }+ 0 x x^{3} + 0 x x^{5} + 5x + 7 will be the highest power of variable that is 4.
Q.4. Degree of the zero polynomial is
(a) 0
(b) 1
(c) Any natural number
(d) Not defined
Correct Answer is option (d)
The degree of a polynomial is the highest degree of its monomials (individual terms) with nonzero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a nonnegative integer.Degree of zero polynomial is thus notdefined.
Q.5. If p (x ) = x^{2} –2√2 x+ 1 , then p(2√2) is equal to
(a) 0
(b) 1
(c) 4√2
(d)
Correct Answer is option (b)
p(x) = x^{2} −2√2 x + 1
p(2√2 ) = (2√2)^{2} − 2√2 × 2√2 +1
p(2√2) = 8 − 8 + 1
p(2√2) = 1
Hence, option B.
Q.6. The value of the polynomial 5x – 4x^{2} + 3, when x = –1 is
(a)  6
(b) 6
(c) 2
(d)  2
Correct Answer is option (a)
Given that: a polynomial as
P(x) = 5x − 4x^{2} + 3
To find value of polynomial at x = −1 or P(−1)
So we put x = −1 in polynomial,
P(−1) = 5 × (−1) − 4 × (−1)^{2} + 3
P(−1) = − 5 − 4 + 3
P(−1) = − 6
So value of polynomial at x = − 1 is − 6.
Q.7. If p(x) = x + 3, then p(x) + p(–x) is equal to
(a) 3
(b) 2x
(c) 0
(d) 6
Correct Answer is option (d)
Given, p(x) = x + 3
Therefore, p(−x) = −x + 3
Now, p(x) + p(−x) = (x+3) + (−x+3) = 6
Option D is correct.
Q.8. Zero of the zero polynomial is
(a) 0
(b) 1
(c) Any real number
(d) Not defined
Correct Answer is option (c)
Zero of the polynomial is any real number .e.g., Let us consider zero polynomial be 0(xk), where k is a real number for determining the zero
Put x − k = 0 ⇒ x = k
Hence, zero of the zero polynomial be any real number.
Q.9. Zero of the polynomial p(x) = 2x + 5 is
(a) (2/5)
(b) (5/2
(c) 2/5
(d) 5/2
Correct Answer is option (b)
Zero of a polynomial is the value of the variable for which the polynomial becomes 0.
Now, p(x) = 2x + 5.
For, p(x) = 0,
2x + 5 = 0.
or, x = −5/2.
Therefore, option D is correct.
Q.10. One of the zeroes of the polynomial 2x^{2} + 7x –4 is
(a) 2
(b) 1/2
(c) (1/2)
(d) –2
Correct Answer is option (b)
To find the zero of 2x 2 +7x−4 is same as solving the equation 2x^{2} + 7x − 4 = 0
⇒ 2x^{2} + 8x − x − 4 = 0
⇒ 2x(x+4) − 1(x+4) = 0
⇒ (2x−1)(x+4) = 0
⇒ (2x−1) = 0, (x + 4) = 0
⇒ x = 1/2 , −4
Therefore, option B is correct.
Q.11. If x^{51} + 51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50
Correct Answer is option (d)
By remainder theorem,Remainder =p(−1), since the divisor is x+1
p(x) = x ^{51} + 51
p(−1) = (−1)^{51} + 51 = −1 + 51
Hence, the remainder is 50
Q.12. If x + 1 is a factor of the polynomial 2x^{2} + kx, then the value of k is
(a) –3
(b) 4
(c) 2
(d) –2
Correct Answer is option (c)
(x + 1) is a factor of p (x)
x + 1 = 0
⇒ x = − 1
Therefore, p (− 1) = 0
∴ 2 (− 1)^{2} + k(− 1) = 0
⇒ 2 − k = 0 ⇒ k = 2
Hence, the value of k is 2.
Q.13. x + 1 is a factor of the polynomial
(a) x^{3} + x^{2 }– x + 1
(b) x^{3} + x^{2} + x + 1
(c) x^{4} + x^{3} + x^{2} + 1
(d) x^{4} + 3 x^{3} + 3x^{2} + x + 1
Correct Answer is option (b)
We know that if x + a is a factor of f(x) then, f(a) = 0.
Let f(x) = x^{4} + x^{3} + x^{2} + 1
Now, f(1) = (1)^{4} + (1)^{3} + (1)^{2} + 1
= 1  1 + 1 + 1
= 2 ≠ 0
So, f(x) is not a factor of x + 1.
Q.14. One of the factors of (25x^{2} – 1) + (1 + 5x)^{2} is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x
Correct Answer is option (d)
(25x^{2} – 1) + (1 + 5x)^{2}
= 25x^{2} − 1 + 1 + 25x^{2} + 10 x [Using identity, (a + b)^{2} = a^{2} + b^{2} + 2ab]
= 50x^{2} + 10 x = 10 x (5x + 1)
Hence, one of the factor of given polynomial is 10x.
Q.15. The value of 249^{2} – 248^{2} is
(a) 1^{2}
(b) 477
(c) 487
(d) 497
Correct Answer is option (d)
We know, a^{2} −b^{2} = (a + b) (a − b)
So, 249^{2} −248^{2} = (249 + 248) (249 − 248) = 497(1) = 497
Q.16. The factorisation of 4x^{2} + 8x + 3 is
(a) (x + 1) (x + 3)
(b) (2x + 1) (2x + 3)
(c) (2x + 2) (2x + 5)
(d) (2x –1) (2x –3)
Correct Answer is option (b)
4x^{2} + 8x + 3
⇒ 4x^{2} + 6x + 2x + 3
⇒ 2x(2x+3) + 1(2x+3)
⇒ (2x+3) (2x+1)
Q.17. Which of the following is a factor of (x + y)^{3} – (x^{3} + y^{3})?
(a) x^{2} + y^{2} + 2 xy
(b) x^{2} + y^{2} – xy
(c) xy^{2}
(d) 3xy
Correct Answer is option (d)
(x + y)^{3} −(x^{3 }+ y^{3})
= x^{3 }+ y^{3} + 3xy(x+y) − x^{3} − y^{3}
= 3xy(x+y)
Hence, 3xy and (x+y) are the factors of the equation
Q.18. The coefficient of x in the expansion of (x + 3)^{3} is
(a) 1
(b) 9
(c) 18
(d) 27
Correct Answer is option (d)
Now, (x + 3)^{3} = x^{3} + 3^{3} + 3x (3)(x + 3)
[using identity, (a + b)^{3} = a^{3 }+ b^{3}+ 3ab (a + b)]
= x^{3} + 27 + 9x (x + 3)
= x^{3} + 27 + 9x^{2} + 27x
Hence, the coefficient of x in (x + 3)^{3} is 27.
Q.19. If (x, y ≠ 0) , the value of x^{3} – y^{3} is
(a) 1
(b) –1
(c) 0
(d) 1/2
Correct Answer is option (c)
Consider the equation:
Simplify the above expression as follows:
x^{2} + y^{2} = xy
Now, x^{3}  y^{3} = (xy) (x^{2}+y^{2 }+ xy)
= (xy)(xy + xy) ...... [Substitute: x^{2} + y^{2} = –xy]
= (xy) × 0
= 0
Q.20. If 49x^{2} – b = then the value of b is
(a) 0
(b) 1/√2
(c) 1/4
(d) 1/2
Correct Answer is option (c)
Given (49x^{2} – b) =
......[Using identity, (a + b)(a – b) = a^{2} – b^{2}]⇒ (√b)^{2} = (1/2)^{2} ...... [Multiplying both sides by –1]
∴ b = 1/4
Q.21. If α + b + c = 0, then α^{3} + b^{3} + c^{3} is equal to
(a) 0
(b) αbc
(c) 3αbc
(d) 2αbc
Correct Answer is option (c)
Since, α^{3} + b^{3} + c^{3} −3αbc = (a + b + c)(a^{2} + b^{2} + c^{2} −bc − ca − ab)
Given, a + b + c = 0
∴ a^{3} + b^{3} + c^{3} − 3abc = 0
∴ a^{3} + b^{3} + c^{3} = 3abc
Option b is correct.
Q.1. Which of the following expressions are polynomials? Justify your answer:
(i) 8
Polynomial. Because, the exponent of the variable of 8 or 8x^{0} is 0 which is a whole number.
(ii)
Polynomial. Because, the exponent of the variable of is a whole number.
(iii) 1  √5x
Not a polynomial. Because, the exponent of the variable of 1 − √5x or 1 − √5x^{1/2} is 1/2 which is not a whole number.
(iv)
Polynomial. Because, the exponents of the variable of = 1/5x^{2} + 5x + 7 are whole numbers.
(v)
Not a polynomial. Because, the exponent of the variable of = x − 6 + 8x^{−1} is 1, which is not a whole number.
(vi) 1/x+1
Not a polynomial as the polynomial is expressed as a_{0} + a_{1}x + a_{2}x^{n} , where a_{0} , a_{1} , a_{2} ⋯ , a_{n} are constants. Now, the given function is in the form f(x) = p(x)/q(x) is a rational expression but not a polynomial.
(vii)
Polynomial. Because, the exponents of the variable of + 4 a − 7 are whole numbers.
(viii) 1/2x
Not a polynomial. Because, the exponent of the variable of 1/2x or 1/2^{x1} is 1, which is not a whole number.
Q.2. Write whether the following statements are True or False. Justify your answer.
(i) A binomial can have at most two terms
False
because a binomial has exactly two terms.
(ii) Every polynomial is a binomial
False
x^{3} + x + 1 is a polynomial but not a binomial.
(iii) A binomial may have degree 5
True
because a binomial is a polynomial whose degree is a whole number ≥ 1, so, degree can be 5 also.
(iv) Zero of a polynomial is always 0
False
because zero of a polynomial can be any real number.
(v) A polynomial cannot have more than one zero
False
A polynomial can have any number of zeroes. It depends upon the degree of the polynomial.
(vi) The degree of the sum of two polynomials each of degree 5 is always 5.
False
x^{5} + 1 and – x^{5} + 2 x + 3 are two polynomials of degree 5 but the degree of the sum of the two polynomials is 1.
Q.1. Classify the following polynomials as polynomials in one variable, two variables etc.
(i) x^{2} + x + 1
Here, the polynomial contains only one variable, i.e., x. Hence, the given polynomial is a polynomial in one variable.
(ii) y^{3} – 5 y
Here, the polynomial contains only one variable, i.e., y. Hence, the given polynomial is a polynomial in one variable.
(iii) xy + yz + zx
Here, the polynomial contains three variables, i.e., x, y and z. Hence, the given polynomial is a polynomial in three variable.
(iv) x^{2} – 2 xy + y^{2} + 1
Here, the polynomial contains two variables, i.e., x and y. Hence, the given polynomial is a polynomial in two variable.
Q.2. Determine the degree of each of the following polynomials:
(i) 2x – 1
Degree of a polynomial in one variable = highest power of the variable in algebraic expression
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial 2x – 1 = 1
(ii) –10
There is no variable in the given term.
Let us assume that the variable in the given expression is x.
– 10 = –10x^{0}
Power of x = 0
Highest power of the variable x in the given expression = 0 Hence, degree of the polynomial – 10 = 0
(iii) x^{3} – 9 x + 3 x^{5}
Powers of x = 3, 1 and 5 respectively.
Highest power of the variable x in the given expression = 5 Hence, degree of the polynomial x^{3} – 9x + 3x^{5} = 5
(iv) y^{3} (1 – y^{4})
The equation can be written as, y^{3}(1 – y^{4}) = y^{3} – y^{7}
Powers of y = 3 and 7 respectively.
Highest power of the variable y in the given expression = 7 Hence, degree of the polynomial y^{3}(1 – y^{4}) = 7
Q.3. For the polynomial
write
(i) the degree of the polynomial
Powers of x = 3, 1, 2 and 6 respectively.
Highest power of the variable x in the given expression = 6 Hence, degree of the polynomial = 6
(ii) the coefficient of x^{3}
The given equation can be written as,
Hence, the coefficient of x3 in the given polynomial is 1/5.
(iii) the coefficient of x^{6 }
The coefficient of x^{6} in the given a polynomial is – 1
(iv) the constant term
Since the given equation can be written as,
The constant term in the given polynomial is 1/5 as it has no variable x associated with it.
Q.4. Write the coefficient of x^{2} in each of the following :
(i) (π/6)x + x^{2} – 1
(π/6) x + x^{2}−1 = (π/6) x + (1) x^{2}−1
The coefficient of x^{2} in the polynomial (π/6) x + x^{2}−1 = 1.
(ii) 3x – 5
3x – 5 = 0x^{2} + 3x – 5
The coefficient of x^{2} in the polynomial 3x – 5 = 0, zero.
(iii) (x –1) (3x – 4)
(x – 1)(3x – 4) = 3x^{2} – 4x – 3x + 4
= 3x^{2} – 7x + 4
The coefficient of x^{2} in the polynomial 3x^{2} – 7x + 4 = 3.
(iv) (2x – 5) (2x^{2} – 3x + 1)
(2x – 5) (2x^{2 }– 3x + 1)
= 4x^{3} – 6x^{2} + 2x – 10x^{2} + 15x– 5
= 4x^{3} – 16x^{2} + 17x – 5
The coefficient of x^{2} in the polynomial (2x – 5) (2x^{2} – 3x + 1) = – 16
Q.5. Classify the following as a constant, linear, quadratic and cubic polynomials:
(i) 2 – x^{2} + x^{3}
Powers of x = 2, and 3 respectively.
Highest power of the variable x in the given expression = 3 Hence, degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.
(ii) 3x^{3}
Power of x = 3.
Highest power of the variable x in the given expression = 3
Hence, the degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.
(iii) 5t – √7
Power of t = 1.
Highest power of the variable t in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.
(iv) 4 – 5y^{2}
Power of y = 2.
Highest power of the variable y in the given expression = 2 Hence, degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.
(v) 3
There is no variable in the given expression.
Let us assume that x is the variable in the given expression. 3 can be written as 3x0.
i.e., 3 = x^{0} Power of x = 0.
Highest power of the variable x in the given expression = 0 Hence, degree of the polynomial = 0
Since it is a polynomial of the degree 0, it is a constant polynomial.
(vi) 2 + x
Power of x = 1.
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.
(vii) y^{3} – y
Powers of y = 3 and 1, respectively.
Highest power of the variable x in the given expression = 3 Hence, degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.
(viii) 1 + x + x^{2}
Powers of x = 1 and 2, respectively.
Highest power of the variable x in the given expression = 2 Hence, degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.
(ix) t^{2}
Power of t = 2.
Highest power of the variable t in the given expression = 2
Hence, the degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.
(x) √2x – 1
Power of x = 1.
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.
Q.6. Give an example of a polynomial, which is:
(i) monomial of degree 1
Monomial = an algebraic expression that contains one term
An example of a polynomial, which is a monomial of degree 1 = 2t
(ii) binomial of degree 20
Binomial = an algebraic expression that contains two terms
An example of a polynomial, which is a binomial of degree 20 = x^{20} + 5
(iii) trinomial of degree 2
Trinomial = an algebraic expression that contains three terms
An example of a polynomial, which is a trinomial of degree 2 = y^{2} + 3y + 11
Q.7. Find the value of the polynomial 3x^{3} – 4x^{2} + 7 x – 5, when x = 3 and also when x = –3.
Given that,
p(x) = 3𝑥^{3} – 4𝑥^{2} + 7𝑥 – 5 According to the question, When x = 3,
p(x) = p(3)
p(x) = 3𝑥^{3} – 4𝑥^{2} + 7𝑥 – 5 Substituting x = 3,
p(3)= 3(3)^{3} – 4(3)^{2} + 7(3) – 5
p(3) = 3(3)^{3} – 4(3)^{2} + 7(3) – 5
= 3(27) – 4(9) + 21 – 5
= 81 – 36 + 21 – 5
= 102 – 41
= 61
When x = – 3, p(x) = p(– 3)
p(x) = 3𝑥^{3} – 4𝑥^{2} + 7𝑥 – 5 Substituting x = – 3,
p(– 3)= 3(– 3)^{3} – 4(– 3)^{2} + 7(– 3) – 5
p(– 3) = 3(–3)^{3} – 4(–3)^{2} + 7(–3) – 5
= 3(–27) – 4(9) – 21 – 5
= –81 – 36 – 21 – 5 = –143
Q.8. If p(x) = x^{2 }– 4 x + 3, evaluate : p(2) – p(–1) + p(1/2)
Given that,
p(𝑥) = 𝑥^{2} – 4𝑥 + 3
According to the question, When x = 2,
p(x) = p(2)
p(𝑥) = 𝑥^{2} – 4𝑥 + 3
Substituting x = 2,
p(2) = (2)^{2} – 4(2) + 3
= 4 – 8 + 3
= – 4 + 3
= – 1
When x = – 1, p(x) = p(– 1) p(𝑥) = 𝑥^{2} – 4𝑥 + 3
Substituting x = – 1,
p(– 1) = (– 1)^{2} – 4(– 1) + 3
= 1 + 4 + 3
= 8
When x = 1/2 , p(x) = p(1/2)
p(𝑥) = 𝑥^{2} – 4𝑥 + 3
Substituting x = 1/2,
p(1/2) = (1/2)^{2} – 4(1/2) + 3
= 1/4 – 2 + 3
= 1/4 + 1
= 5/4
Now,
p(2)− p(−1) + p(1/2) = – 1 – 8 + (5/4)
= – 9 + (5/4)
= ( – 36 + 5)/4
= – 31/4
Q.9. Find p(0), p(1), p(–2) for the following polynomials:
(i) p(x) = 10x – 4 x^{2} – 3
According to the question, p(x) = 10𝑥−4𝑥2 –3
When x = 0, p(x) = p(0)
Substituting x = 0,
p(0) = 10(0)−4(0)2 –3
= 0 – 0 – 3
= – 3
When x = 1, p(x) = p(1)
Substituting x = 1,
p(1) = 10(1) − 4(1)^{2} –3
= 10 – 4 – 3
= 6 – 3
= 3
When x = – 2, p(x) = p(– 2)
Substituting x = – 2,
p(– 2) = 10(– 2) − 4(– 2)^{2} –3
= – 20 – 16 – 3
= – 36 – 3
= – 39
(ii) p(y) = (y + 2) (y – 2)
According to the question, p(𝑦) = (y + 2) (y – 2)
When y = 0, p(y) = p(0)
Substituting y = 0, p(0) = (0 + 2) (0 – 2)
= (2)(– 2)
= – 4
When y = 1, p(y) = p(1)
Substituting y = 1, p(1)=(1 + 2) (1 – 2)
=(3) (– 1)
= – 3
When y = – 2, p(y) = p(– 2)
Substituting y = – 2,
p(– 2) =(– 2 + 2) (– 2 – 2)
= (0) (– 4)
= 0
Q.10. Verify whether the following are true or false:
(i) –3 is a zero of x – 3
False
Zero of x – 3 is given by, x – 3 = 0
⇒ x=3
(ii) – 1/3 is a zero of 3x + 1
True
Zero of 3x + 1 is given by, 3x + 1 = 0
⇒ 3x = – 1
⇒ x = – 1/3
(iii) – 4/5 is a zero of 4 –5y
False
Zero of 4 – 5y is given by, 4 – 5y =0
⇒ – 5y = – 4
⇒ y = 4/5
(iv) 0 and 2 are the zeroes of t^{2} – 2t
True
Zeros of t2 – 2t is given by, t2 – 2t = t(t – 2) = 0
⇒ t = 0 or 2
(v) –3 is a zero of y^{2} + y – 6
Zero of y^{2} + y – 6 is given by, y^{2} + y – 6 = 0
⇒ y^{2} + 3x – 2x – 6 = 0
⇒ y(y + 3) – 2(x + 3) = 0
⇒ (y – 2) (y + 3) = 0
⇒ y = 2 or – 3
Q.11. Find the zeroes of the polynomial in each of the following :
(i) p(x) = x – 4
Zero of the polynomial p(x) ⇒ p(x) = 0
P(x) = 0
⇒ x – 4 = 0
⇒ x = 4
Therefore, the zero of the polynomial is 4.
(ii) g(x) = 3 – 6x
Zero of the polynomial g(x) ⇒ g(x) = 0 g(x) = 0
⇒3 – 6x = 0
⇒ x = 3/6 =1/2
Therefore, the zero of the polynomial is 1/2
(iii) q(x) = 2x –7
Zero of the polynomial q(x) ⇒ q(x) = 0 q(x) = 0
⇒2x – 7 = 0
⇒ x = 7/2
Therefore, the zero of the polynomial is 7/2
(iv) h(y) = 2y
Zero of the polynomial h(y) ⇒ h(y) = 0 h(y) = 0
⇒ 2y = 0
⇒ y = 0
Therefore, the zero of the polynomial is 0
Q.12. Find the zeroes of the polynomial : p(x) = (x – 2)^{2} – ( x + 2)^{2}
We know that,
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)^{2 }− (x + 2^{)2} = 0
Expanding using the identity, a^{2} – b^{2} = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
⇒ – 8 x = 0
Therefore, the zero of the polynomial = 0
Q.13. By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x^{4} + 1; x –1
Performing the long division method, we get,
Hence, from the above long division method, we get, Quotient = x^{3} + x^{2} + x + 1
Remainder = 2.
Q.14. By Remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) p(x) = x^{3} – 2x^{2} – 4 x – 1, g(x) = x + 1
Let g(x) = x + 1
x + 1 = 0
x = –1
p(x) = x^{3} – 2x^{2} – 4x – 1
p(–1) = (–1)^{3} – 2(– 1)^{2} – 4(–1) – 1
= –1 – 2 × 1 + 4 – 1
= – 4 – 4 = 0
∴ Remainder = 0.
(ii) p(x) = x^{3} – 3x^{2} + 4 x + 50, g(x) = x – 3
p(x) = x^{3} – 3x^{2} + 4x + 50, g(x) = x – 3
Let g(x) = x – 3
x – 3 = 0
x = 3
p(3) = 3^{3} – 3(3^{2}) + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
∴ Remainder = 62.
(iii) p(x) = 4x^{3} – 12x^{2} + 14x – 3, g (x) = 2x – 1
Let us denote the given polynomials as
p(x) = 4x^{2 } 12x^{3}  12x^{2} + 14x  3,
g(x) = 2x  1
⇒ g(x) =
We have to find the remainder when f(x) is divided by g(x).
By the remainder theorem, when f(x) is divided by g(x) the remainder is
= 1/2  3 + 7  3
= 3/2
Now we will calculate the remainder by actual division
So the remainder by actual division is 3/2 .
(iv) p(x) = x^{3} – 6x^{2 }+ 2x – 4, g(x) = 1 – 3/2 x
Given, p(x) = x^{3 }– 6x^{2} + 2x – 4 and g(x) = 1  3/2x
Here, zero of g(x) is 2/3.
When we divide p(x) by g(x) using remainder theorem, we get the remainder p (2/3).
Hence, remainder is 136/27
Q.15. Check whether p(x) is a multiple of g(x) or not:
(i) p(x) = x^{3} – 5x^{2} + 4 x – 3, g(x) = x – 2
According to the question, g(x) = x – 2,
Then, zero of g(x), g(x) = 0
x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) =(2)^{3} – 5(2)^{2} + 4(2) – 3
= 8 – 20 + 8 – 3
= – 7 ≠ 0
Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0.
(ii) p(x) = 2x^{3} – 11x^{2} – 4 x + 5, g (x) = 2x + 1
According to the question, g(x)= 2𝑥 + 1
Then, zero of g(x), g(x) = 0
2x + 1 = 0
2x = – 1
x = –1/2
Therefore, zero of g(x) = – 1/2
So, substituting the value of x in p(x), we get,
p(–1/2) = 2 × ( – 1/2 )3 – 11 × ( – 1/2 )2 – 4 × ( – 1/2) + 5
= – 1/4  11/4 + 7
= 16/4
= 4 ≠ 0
Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0.
Q.16. Show that: (i) x + 3 is a factor of 69 + 11x – x^{2} + x^{3}.
According to the question,
Let p(x) = 69 + 11x − x^{2} + x^{3} and
g(x) = x + 3 g(x) = x + 3
zero of g(x) ⇒ g(x) = 0 x + 3 = 0
x = – 3
Therefore, zero of g(x) = – 3
So, substituting the value of x in p(x),
we get, p( – 3) = 69 + 11(– 3) –(–3)^{2} + (– 3)^{3}
= 69 – 69
= 0
Since, the remainder = zero, We can say that,
g(x) = x + 3 is factor of p(x) = 69 + 11x − x^{2} + x^{3}
(ii) 2x – 3 is a factor of x + 2x^{3 }– 9x^{2} + 12.
Let p(x) = 2x^{3}  9x^{2} + x + 12
We have to show that, 2x  3 is a factor of p(x).
= (81 – 81)/4
= 0
Hence, (2x – 3) is a factor of p(x).
Q.17. Determine which of the following polynomials has x – 2 a factor:
(i) 3x^{2} + 6x – 24
According to the question,
Let p(x) = 3𝑥^{2} + 6𝑥 − 24 and g(x) = x – 2 g(x) = x – 2
zero of g(x) ⇒ g(x) = 0 x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) = 3(2)^{2} + 6 (2) – 24
= 12 + 12 – 24
= 0
Since, the remainder = zero, We can say that,
g(x) = x – 2 is factor of p(x) = 3𝑥^{2} + 6𝑥 − 24
(ii) 4x^{2} + x – 2
According to the question,
Let p(x) = 4𝑥^{2} + 𝑥 − 2 and g(x) = x – 2 g(x) = x – 2
zero of g(x) ⇒ g(x) = 0 x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) = 4(2)^{2 }+ 2−2
= 16 ≠ 0
Since, the remainder = zero, We can say that,
g(x) = x – 2 is factor of p(x) = 4𝑥^{2} + 𝑥 − 2
Q.18. Show that p – 1 is a factor of p^{10} – 1 and also of p^{11} – 1.
Let f(p) = p^{10} – 1 and g(p) = p^{11} – 1
Putting p = 1 in f(p), we get
f(1) = 1^{10} − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p^{10} – 1)
Now, putting p = 1 in g(p), we get
g(1) = 1^{11} − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p^{11} – 1)
Q.19. For what value of m is x^{3} – 2mx^{2} + 16 divisible by x + 2 ?
According to the question,
Let p(x) = x^{3} – 2mx^{2} + 16, and g(x) = x + 2 g(x) = 0
⟹ x + 2 = 0
⟹ x = – 2
Therefore, zero of g(x) = – 2 We know that,
According to factor theorem,
if p(x) is divisible by g(x), then the remainder p(−2) should be zero. So, substituting the value of x in p(x), we get,
p( – 2) = 0
⟹ ( – 2)^{3} – 2m( – 2)^{2} + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
⟹ 8m = 8
⟹ m = 1
Q.20. If x + 2α is a factor of x^{5} – 4 α^{2}x^{3} + 2 x + 2α + 3, find α.
According to the question,
Let p(x) = x^{5} – 4a^{2}x^{3} + 2x + 2a + 3 and g(x) = x + 2a g(x) = 0
⟹ x + 2a = 0
⟹ x = – 2a
Therefore, zero of g(x) = – 2a We know that,
According to the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0 So, substituting the value of x in p(x), we get,
p ( – 2a) = (– 2a)^{5} – 4a^{2}(– 2a)3 + 2(– 2a) + 2a + 3 = 0
⟹ – 32a^{5} + 32a^{5} – 2a + 3 = 0
⟹ – 2a = – 3
⟹ a = 3/2
Q.21. Find the value of m so that 2x – 1 be a factor of 8x^{4} + 4x^{3} – 16x^{2} + 10x + m.
Let p(x) = 8x^{4} + 4x^{3} – 16x^{2} + 10x + m
Since, 2x  1 is a factor of p(x), then put p(1/2) = 0
⇒ 1 + 1 + m = 0
∴ m = – 2
Hence, the value of m is – 2.
Q.22. If x + 1 is a factor of ax^{3} + x^{2} – 2 x + 4 a – 9, find the value of a.
Let p(x) = ax^{3} + x^{2} – 2x + 4a – 9
Since, x + 1 is a factor of p(x), then put p(– 1) = 0
∴ a(1)^{3} + (1)^{2}  2(1) + 4a  9 = 0
⇒ a + 1 + 2 + 4a  9 = 0
⇒ 3a  6 = 0
⇒ 3a = 6
⇒ a = 6/3 = 2
Hence, the value of a is 2.
Q.23. Factorise:
(i) x2 + 9 x + 18
x^{2} + 9x + 18 = x^{2} + 3x + 6x + 18
= x(x + 3) + 6(x + 3)
= (x + 3)(x + 6)
(ii) 6x^{2} + 7 x – 3
6x^{2} + 7x − 3 = 6x^{2} + 9x − 2x − 3
= 3x(2x + 3) − 1(2x+3)
= (3x − 1)(2x + 3)
(iii) 2x^{2} – 7 x – 15
2x^{2 }− 7x − 15 = 2x^{2 }− 10x + 3x − 15
= 2x(x − 5) + 3(x − 5)
= (x − 5)(2x + 3)
(iv) 84 – 2r – 2 r^{2}
84 − 2r − 2r^{2 }= −(2r^{2} + 2r − 84)
= −(2r^{2} + 14r + 12r − 84)
= −(2r(r + 7) − 12(r + 7))
= −(r + 7)(2r − 12)
Q.24. Factorise:
(i) 2x^{3} – 3x^{2} – 17x + 30
2x^{3} − 3x^{2} − 17x + 30
= 2x^{3} + 2x^{2} − 12x − 5x^{2} − 5x + 30
= 2x(x^{2} + x − 6) − 5(x^{2} + x  6)
= (x^{2} + x−6)(2x − 5)
= (x^{2} + 3x − 2x − 6)(2x − 5)
= [x(x + 3) − 2(x + 3)](2x − 5)
= (x + 3)(x − 2)(2x − 5)
(ii) x^{3} – 6x^{2} + 11x – 6
Let p(x) = x^{3}  6x^{2} + 11x  6
By trial, we find that
p(1) = (1)^{3}  6(1)^{2 }+ 11(1)  6 = 0
∴ By converse of factor theorem, (x  1) is a factor of p(x).
Now, x^{3}  6x^{2} + 11x  6
= x^{2} (x  1) 5x (x  1) + 6 (x  1)
= (x  1) (x^{2}  5x + 6)
= (x  1) {x^{2}  2x  3x + 6}
= (x  1) {x(x  2)3 (x  2)}
= (x  1)(x  2)(x  3)
(iii) x^{3} + x^{2} – 4x – 4
Let x + 1 = 0
∴ x = 1
On substituting value of x in the expression
∴ f(1) = (1)3 + (1)2  4(1) 4 = 0
Clearly x + 1 is a factor of
f(x) = x^{3} + x^{2}  4x  4
∴ f(x) = (x + 1) (x^{2}  4) ...(By actual division)
= (x + 1) (x  2) (x + 2)
(iv) 3x^{3} – x^{2} – 3x + 1
Let f(x) = 3x^{3}  x^{2 } 3x + 1 be the given polynomial.
Now, putting x = 1, we get
f(1) = 3(1)^{3}  (1)^{2}  3(1) + 1
= 3  1  3 + 1 = 0
Therefore, (x1) is a factor of polynomial f(x).
Now,
f(x) = 3x^{2}(x1) + 2x(x1)  1(x  1)
= (x  1){3x^{2} + 2x  1}
= (x  1){3x^{2 }+ 2x  1}
= (x  1)(x + 1)(3x  1)
Hence (x  1), (x + 1) and (3x  1 )are the factors of polynomial f(x).
Q.25. Using suitable identity, evaluate the following:
(i) 103^{3}
= (100 + 3)^{3}
= (100)3 + (3)^{3} + 3 x 100 x 3 x (100 + 3)
= 1000000 + 27 + 900 (103)
= 1000027 + 92700
= 1092727
(ii) 101 × 102
= (100 + 1) (100 + 2)
= (100) + 100(1 + 2) + 1 x 2
= 10000 + 300 + 2
= 10302
(iii) 999^{2}
= (1000  1)^{2}
= (1000)2 + (1)^{2}  2 x 1000 x 1
= 1000000 + 1  2000
= 998001
Q.26. Factorise the following:
(i) 4x^{2} + 20x + 25
4x^{2} + 20x + 25 = (2x)^{2} + 2 × 2x × 5 + (5)^{2}
= (2x+5)^{2} ........[Using identity, a^{2} + 2ab + b^{2} = (a + b)^{2}]
(ii) 9y^{2} – 66yz + 121z^{2}
Using (a−b)^{2} = a^{2} + b^{2} −2ab
= (3y)^{2} + (11z)^{2} −2 × 3y × 11
= (3y−11z)^{2}
(iii)
Using a^{2} − b^{2} = (a+b) (a−b)
Q.27. Factorise the following:
(i) 9x^{2} – 12x + 3
⇒ 9x^{2 }− 3x − 9x + 3
⇒ 3x(3x − 1) − 3(3x −1)
⇒ (3x − 3)(3x − 1)
⇒ 3(x − 1)(3x − 1)
(ii) 9x^{2} – 12x + 4
9x^{2} – 12x + 4 = (3x)^{2}  2 × 3x × 2 + (2)^{2}
= (3x2)^{2} .....[Using identity, (a – b)^{2} = a^{2} – 2ab + b^{2}]
= (3x2) (3x2)
Q.28. Expand the following :
(i) (4a – b + 2 c)^{2}
We have,
(4a − b + 2c)^{2}
=(4a)^{2} +(−b)^{2} +(2c)^{2} + 2(4a)(−b) + 2(−b)(2c) + 2(2c)(4a)
[∵ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}]
=16a^{2} + b^{2} + 4c^{2} −8ab − 4ac + 16ca
(ii) (3a – 5b – c)^{2}
(3a − 5b − c)^{2}
= (3a)^{2 }+ (−5b)^{2 }+ (−c)^{2} + 2(3a)^{2} − 5b+2(−5b)(−c) + 2(−c)(3a)
[∵ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}]
= 9a^{2} + 25b^{2} + c^{2 }− 30ab + 10bc − 6ca.
(iii) (– x + 2y – 3z)^{2}
We know that,
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3z)^{2} = x^{2} + (2y)^{2} + (3z)^{2} + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x^{2} + 4y^{2} + 9z^{2} + 4xy + 12y^{2} + 6zx
Q.29. Factorise the following :
(i) 9x^{2} + 4 y^{2} + 16z^{2} + 12xy – 16yz – 24xz
= (3x)^{2} + (2y)^{2} + (−4z)^{2} + 2(3x)(2y) + 2(2y)(−4z) + 2(−4z)(3x)
= (3x + 2y−4z)^{2}= (3x + 2y − 4z)(3x + 2y − 4z)
(ii) 25x^{2} + 16y^{2} + 4z^{2} – 40xy + 16yz – 20xz
25x^{2} +16y^{2} +4z^{2} −40xy + 16yz − 20xz
= (−5x)^{2} +(4y)^{2} +(2z)^{2} + 2(−5x)(4y) + 2(4y)(2z) + 2(−5x)(2z)
Suitable identities is (x + y + z)^{3} = x^{3} + y^{3} + z^{3} + 2xy + 2yz + 2xz
Therefore, (−5x)^{2} +(4y)^{2} +(2z)^{2} + 2(−5x)(4y) + 2(4y)(2z) + 2(−5x)(2z)
= (−5x + 4y + 2z)^{2}
(iii) 16x^{2} + 4y^{2} + 9z^{2} – 16xy – 12yz + 24 xz
16x^{2} + 4y^{2} + 9z^{2} −16xy −12yz + 24xz
Using identity,(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
16x^{2} + 4y^{2} + 9z^{2 }− 16xy − 12yz + 24xz
=(4x)^{2} + (−2y)^{2} + (3z)^{2} + 2(4x)(−2y) + 2(−2y)(3z) + 2(3z)(4x)
=(4x − 2y + 3z)^{2}
Q.30. If a + b + c = 9 and ab + bc + ca = 26, find a^{2} + b^{2} + c^{2}.
Given, a + b + c = 9 and ab + bc + ca = 26.
We know, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
⇒(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca).
Putting the values here, we get,
(9)^{2} = a^{2} + b^{2} + c^{2} + 2(26)
⇒ 81 = a^{2} + b^{2} + c^{2} + 52
⇒ a^{2} + b^{2} + c^{2} = 81 − 52 = 29.
Hence, option D is correct.
Q.31. Expand the following :
(i) (3a – 2b)^{3}
= (3a)^{3} – (2b)^{3} – 3(3a)(2b) [3a – 2b]
= 27a^{3} – 8b^{3 –} 18ab (3a – 2b)
= 27a^{3} – 8b^{3} – 54a^{2}b + 36ab^{2}
= 27a^{3 }– 54a^{2}b + 36ab^{2} – 8b^{3}
(ii) (1/x + y/3)^{3}
......[Using identity, (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)]
Q.32. Factorise the following
(i) 1 – 64a^{3 }– 12a + 48a^{2}
= (1)^{3} −(4a)^{3} − 3(1)^{2} (4a) + 3(1)(4a)^{2}
Suitable identities is x^{3} − y^{3 }^{ }3x^{2} y + 3xy^{2} = (x−y) 3
∴(1)^{3} −(4a)^{3} −3(1)^{2} (4a) + 3(1)(4a)^{2}
= (1−4a)^{3}
(ii)
Q.33. Find the following products :
(i)
= (x/2)^{3} + (2y)^{3} [(a + b)(a^{2} − ab + b^{2}) = a^{3} + b^{3}]
= x^{3}/8 + 8y^{3}
(ii) (x^{2} – 1) (x^{4} + x^{2} + 1)
Given (x^{2}  1) (x^{4 }+ x^{2} + 1)
We shall use the identity (a  b) (a^{2} + ab + b^{2}) = a^{3}  b^{3}
We can rearrange the (x^{2}  1) (x^{4} + x^{2} + 1) as
(x^{2} − 1) [(x^{2})^{2 }+ (x^{2}) (1) + (1)^{2}]
=(x^{2})^{3}  (1)^{3}
=(x^{2}) × (x^{2}) × (x^{2})  (1) × (1) × (1)
= x^{6}  1^{3}
= x^{6}  1
Hence the Product value of (x^{2}  1)(x^{4} + x^{2} + 1) is x^{6}  1.
Q.34. Factorise :
(i) 1 + 64x^{3}
We know, (a^{3} + b^{3}) = a^{3} + a^{2}b − a^{2}b − ab^{2} + ab^{2} + b^{3}
= (a^{3} + a^{2}b) − (a^{2}b + ab^{2}) + (ab^{2} + b^{3})
= a^{2}(a + b) − ab(a + b) + b^{2}(a + b)
= (a + b)(a^{2} − ab + b^{2})
Now, given, 1 + 64x^{3}
=1^{3} + (4x)^{3}
= (1 + 4x)(1 − 4x(1) + (4x^{2}))
= (1 + 4x)(1 − 4x + 16x^{2})
(ii) a^{3} –2√2b^{3}
(a^{3}  b^{3}) = a^{3}  a^{2}b + a^{2}b − ab^{2} + ab^{2}  b^{3}
= (a^{3}  a^{2}b) + (a^{2}b  ab^{2}) + (ab^{2}  b^{3})
= a^{2}(a  b) + ab(a  b) + b^{2}(a  b)
= (a  b)(a^{2} + ab + b^{2})
Now, given a^{3} − 2√2b^{3}
=(a − √2b)(a^{2} + a(√2b) + (√2b)^{2})
= (a − √2b)(a^{2} + √2ab + 2b^{2})
Q.35. Find the following product: (2x – y + 3z) (4x^{2} + y^{2} + 9z^{2} + 2xy + 3yz – 6xz)
= (2x − y + 3z)(4x^{2} + y^{2} + 9z^{2} + 2xy + 3yz − 6xz)
= (a + b + c)(a^{2} + b^{2} + c^{2}  ab − bc − ac)
= a^{3} + b^{3} + c^{3}−3abc
= (2x)^{3} + (−y)^{3} + (3z)^{3} − 3.(2x)(−y)(3z)
= 8x^{3} − y^{3} + 27z^{3} + 18xyz
Q.36. Factorise :
(i) a^{3} – 8b^{3} – 64c^{3} – 24abc
Here the given expression can be written as,
a^{3} − 8b^{3} − 64c^{3} −24abc = (a)^{3} + (−2b)^{3} + (−4c)^{3} −3(a)(−2b)(−4c)
Comparing with the given identity,
x^{3} + y^{3} + z^{3} −3xyz ≡ (x + y + z)(x^{2} + y^{2} + z^{2} − xy − yz − xz)
We get factor as
= (a − 2b − 4c)[(a)^{2} − (−2b)^{2} + (−4c)^{2} − (a)(−2b) − (−2b)(−4c) − (−4c)(a)]
= (a − 2b − 4c)(a^{2} + 4b^{2} + 16c^{2} + 2ab − 8bc + 4ca)
(ii) 2√2a^{3} + 8b^{3} – 27c^{3} + 18√2 abc.
We know that,
a^{3} + b^{3} + c^{3} −3abc = (a + b + c)(a^{2} + b^{2} + c^{2} − ab − bc − ca)
Thus, 2√2a^{3} + 8b^{3} − 27c^{3} + 18√2 abc
Here, a = √2 a, b = 2b and c = −3c
= (√2a + 2b − 3c)[(√2a)^{2} + (2b)^{2} + (−3c)^{2} − √2a(2b) − (2b)(−3c)−(−3c)(√2a)]
= (√2a + 2b − 3c)(2a^{2} + 4b^{2} + 9c^{2} − 2√2ab + 6bc + 3√2ac)
Q.37. Without actually calculating the cubes, find the value of :
(i)
Let
We know that a^{3} + b^{3} + c3 = 3abc when a + b + c = 0
=  5/12
(ii) (0.2)^{3} – (0.3)^{3} + (0.1)^{3}
Given,
0.2^{3} − 0.3^{3} + 0.1^{3}= (0.1 × 2)^{3} − (0.1 × 3)^{3} + (0.1 × 1)^{3}
= 0.1^{3}(2^{3 }− 3^{3} + 1^{3})
= 0.1^{3}(−18) = −0.018
Q.38. Without finding the cubes, factorise
(x – 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3}
We know that, a^{3} + b^{3} + c^{3} − 3abc = (a + b + c) (a^{2 }+ b^{2 }+ c^{2} − ab − bc − ca)
Also, if a + b + c = 0,
then a^{3} + b^{3} + c^{3} = 3abc .....condition(1)
Here, we see that a + b + c = (x − 2y) + (2y − 3z) + (3z − x) = 0∴ Using condition (1), we get
⇒ (x − 2y)^{3} + (2y − 3z)^{3} + (3 z − x)^{3}
= 3(x − 2y) (2y − 3z) (3z − x)
Q.39. Find the value of
(i) x^{3} + y^{3 }– 12xy + 64, when x + y = – 4
We know that, a^{3} + b^{3} + c^{3 }= (a + b + c) (a^{2 }+ b^{2 }+ c^{2} − ab − bc − ca) + 3abc
If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc
Now, given x^{3} + y^{3} −12xy + 64 and
x + y = −4
=> x + y + 4 = 0
Here, a = x, b = y, c = 4 and a + b + c = x + y + 4 = 0
Therefore x^{3} + y^{3} + 64 = 3xy(4)
= 12xyz
Now, x^{3} + y^{3} + 64 − 12xyz = 12xyz − 12xyz = 0
(ii) x^{3} – 8y^{3} – 36xy – 216, when x = 2y + 6
x^{3 }− 8y^{3 }− 36xy − 216
= x^{3} + (−2y)^{3} + (−6)^{3} − 3(x)(−2y)(−6)
= (x−2y−6)(x^{2} + 4y^{2 }+ 36 + 2xy − 12y + 6x)
= 0 × (x^{2 }+ 4y^{2} + 36 + 2xy − 12y + 6x) = 0
Q.40. Give possible expressions for the length and breadth of the rectangle whose area is given by 4a^{2} + 4a –3.
Given
Area = 4a^{2 }+ 4a − 3.
We know that
Area of rectangle = length × breadth
So, to find the possible expressions for the length and breadth we have to factorise the given expression.
Using the method of splitting the middle term,4a + 4a−3
= 4a^{2} + 6a − 2a − 3
= 2a(2a+3)−1(2a+3)
= (2a−1)(2a+3)
∴ length × breadth =(2a−1)(2a+3)
Hence, the possible expressions for the length and breadth of the rectangle are :
length = (2a−1) and breadth =(2a+3) or, length = (2a + 3) and breadth = (2a − 1).
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5. How do you multiply polynomials? 
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NCERT Exemplar Solutions: Polynomials  2 Doc  6 pages 
Test: Polynomials  1 Test  25 ques 
Algebraic Identities Doc  6 pages 

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